Answer:
V = 523 in^3
Step-by-step explanation:
The volume of a sphere is given by
V = 4/3 pi r^3
V = 4/3 ( 3.14) * 5^3
V = 523.33333repeating
Rounding to the nearest inch^3
V = 523 in^3
Answer:
[tex] 523.6 {in}^{3} [/tex]
Step-by-step explanation:
[tex]v = \frac{4}{3} \pi {r}^{3} \\ = \frac{4}{3} \pi \times 5 \times 5 \times 5 \\ = 523.6 {in}^{3} [/tex]
16.50 and pays 20.00 in cash the change due is
Answer:
Change due is 3.50
Step-by-step explanation:
20.00-16.50 is 3.50
Answer: $3.50
Step-by-step explanation:
You subtract 20 from 16.50.
If cot Theta = Two-thirds, what is the value of csc Theta? StartFraction StartRoot 13 EndRoot Over 3 EndFraction Three-halves StartFraction StartRoot 13 EndRoot Over 2 EndFraction Eleven-thirds
Answer:
csctheta= [tex]\frac{\sqrt{13} }{3}[/tex]
Step-by-step explanation:
answer is provided on top
The value of the [tex]\rm cosec \theta = \frac{\sqrt{13} }{3}[/tex]. Cosec is found as the ratio of the hypotenuse and the perpendicular.
What is trigonometry?The field of mathematics is concerned with the relationships between triangles' sides and angles, as well as the related functions of any angle
The given data in the problem is;
[tex]\rm cot \theta = \frac{2}{3}[/tex]
The [tex]cot \theta[/tex] is found as;
[tex]\rm cot \theta = \frac{B}{P} \\\\ \rm cot \theta = \frac{2}{3} \\\\ B=2 \\\\ P=3 \\\\[/tex]
From the phythogorous theorem;
[tex]\rm H=\sqrt{P^2+B^2} \\\\ \rm H=\sqrt{2^2+3^2} \\\\ H=\sqrt{13} \\\\[/tex]
The value of the cosec is found as;
[tex]\rm cosec \theta = \frac{H}{P} \\\ \rm cosec \theta = \frac{\sqrt{13} }{3}[/tex]
Hence the value of the [tex]\rm cosec \theta = \frac{\sqrt{13} }{3}[/tex].
To learn more about the trigonometry refer to the link;
https://brainly.com/question/26719838
Consider the differential equation:
2y'' + ty' − 2y = 14, y(0) = y'(0) = 0.
In some instances, the Laplace transform can be used to solve linear differential equations with variable monomial coefficients.
If F(s) = ℒ{f(t)} and n = 1, 2, 3, . . . ,then
ℒ{tnf(t)} = (-1)^n d^n/ds^n F(s)
to reduce the given differential equation to a linear first-order DE in the transformed function Y(s) = ℒ{y(t)}.
Requried:
a. Sovle the first order DE for Y(s).
b. Find find y(t)= ℒ^-1 {Y(s)}
(a) Take the Laplace transform of both sides:
[tex]2y''(t)+ty'(t)-2y(t)=14[/tex]
[tex]\implies 2(s^2Y(s)-sy(0)-y'(0))-(Y(s)+sY'(s))-2Y(s)=\dfrac{14}s[/tex]
where the transform of [tex]ty'(t)[/tex] comes from
[tex]L[ty'(t)]=-(L[y'(t)])'=-(sY(s)-y(0))'=-Y(s)-sY'(s)[/tex]
This yields the linear ODE,
[tex]-sY'(s)+(2s^2-3)Y(s)=\dfrac{14}s[/tex]
Divides both sides by [tex]-s[/tex]:
[tex]Y'(s)+\dfrac{3-2s^2}sY(s)=-\dfrac{14}{s^2}[/tex]
Find the integrating factor:
[tex]\displaystyle\int\frac{3-2s^2}s\,\mathrm ds=3\ln|s|-s^2+C[/tex]
Multiply both sides of the ODE by [tex]e^{3\ln|s|-s^2}=s^3e^{-s^2}[/tex]:
[tex]s^3e^{-s^2}Y'(s)+(3s^2-2s^4)e^{-s^2}Y(s)=-14se^{-s^2}[/tex]
The left side condenses into the derivative of a product:
[tex]\left(s^3e^{-s^2}Y(s)\right)'=-14se^{-s^2}[/tex]
Integrate both sides and solve for [tex]Y(s)[/tex]:
[tex]s^3e^{-s^2}Y(s)=7e^{-s^2}+C[/tex]
[tex]Y(s)=\dfrac{7+Ce^{s^2}}{s^3}[/tex]
(b) Taking the inverse transform of both sides gives
[tex]y(t)=\dfrac{7t^2}2+C\,L^{-1}\left[\dfrac{e^{s^2}}{s^3}\right][/tex]
I don't know whether the remaining inverse transform can be resolved, but using the principle of superposition, we know that [tex]\frac{7t^2}2[/tex] is one solution to the original ODE.
[tex]y(t)=\dfrac{7t^2}2\implies y'(t)=7t\implies y''(t)=7[/tex]
Substitute these into the ODE to see everything checks out:
[tex]2\cdot7+t\cdot7t-2\cdot\dfrac{7t^2}2=14[/tex]
A major traffic problem in the Greater Cincinnati area involves traffic attempting to cross the Ohio River from Cincinnati to Kentucky using Interstate 75. Let us assume that the probability of no traffic delay in one period, given no traffic delay in the preceding period, is 0.9 and that the probability of finding a traffic delay in one period, given a delay in the preceding period, is 0.6. Traffic is classified as having either a delay or a no-delay state, and the period considered is 30 minutes.Required:a. Assume that you are a motorist entering the traffic system and receive a radio report of a traffic delay. What is the probability that for the next 60 minutes (two time periods) the system will be in the delay state?b. What is the probability that in the long run the traffic will not be in the delay state?c. An important assumption of the Markov process model presented here has been the constant or stationary transition probabilities as the system operates in the future. Do you believe this assumption should be questioned for this traffic problem? Explain.
Answer:
a) 0.36
b) 0.3
c) Yes
Step-by-step explanation:
Given:
Probability of no traffic delay in one period, given no traffic delay in the preceding period = P(No_Delay) = 0.9
Probability of finding a traffic delay in one period, given a delay in the preceding period = P(Delay) = 0.6
Period considered = 30 minutes
a)
Let A be the probability that for the next 60 minutes (two time periods) the system will be in the delay state:
As the Probability of finding a traffic delay in one period, given a delay in the preceding period is 0.6 and one period is considered as 30 minutes.
So probability that for the next two time periods i.e. 30*2 = 60 minutes, the system in Delay is
P(A) = P(Delay) * P(Delay) = 0.6 * 0.6 = 0.36
b)
Let B be the probability that in the long run the traffic will not be in the delay state.
This statement means that the traffic will not be in Delay state but be in No_Delay state in long run.
Let C be the probability of one period in Delay state given that preceding period in No-delay state :
P(C) = 1 - P(No_Delay)
= 1 - 0.9
P(C) = 0.1
Now using P(C) and P(Delay) we can compute P(B) as:
P(B) = 1 - (P(Delay) + P(C))
= 1 - ( 0.6 + 0.10 )
= 1 - 0.7
P(B) = 0.3
c)
Yes this assumption should be questioned for this traffic problem because it implies that traffic will be in Delay state for the 30 minutes and just after 30 minutes, it will be in No_Delay state. However, traffic does not work like this in general and it makes this scenario unrealistic. Markov process model can be improved if probabilities are modeled as a function of time instead of being presented as constant (for 30 mins).
10. A sample of 60 mutual funds was taken and the mean return in the sample was 13% with a standard deviation of 6.9%. The return on a particular index of stocks (against which the mutual funds are compared) was 11.5%. Therefore, the test statistic is 1.68. When testing the hypothesis that the average return on actively-managed mutual funds is higher than the return on an index of stocks, if the critical value is 1.96, what is your conclusion concerning the null hypothesis
Answer:
In this question, we shall be accepting the null hypothesis H0 since the critical value is greater than the test statistic value
Step-by-step explanation:
Here in this question, we want to make a conclusion about the null hypothesis H0.
To make or give the correct conclusion about the null hypothesis in this case, we shall need to compare the absolute value of the test statistic used against the value of the critical value.
Hence, we draw a conclusion if the test statistic is larger or smaller than the critical value.
From the value given in the question, we can see that the test statistic given as 1.68 is lesser in value compared to the critical value given as 1.96.
In this kind of case, the conclusion that we shall be drawing is that we will accept the null hypothesis H0 and reject the alternative hypothesis
help pls:Find all the missing elements
Step-by-step explanation:
Using Sine Rule
[tex] \frac{ \sin(a) }{ |a| } = \frac{ \sin(b) }{ |b| } = \frac{ \sin(c) }{ |c| } [/tex]
[tex] \frac{ \sin(42) }{5} = \frac{ \sin(38) }{a} [/tex]
[tex]a = \frac{5( \sin(38))}{ \sin(42) } [/tex]
[tex]a = 4.6[/tex]
[tex] \frac{ \sin(42) }{5} = \frac{ \sin(100) }{b} [/tex]
[tex]b= \frac{5( \sin(100))}{ \sin(42) } [/tex]
[tex]b = 7.4[/tex]
What is the 25th term in the following arithmetic sequence? -7, -2, 3, 8, ...
Answer:
108.
Step-by-step explanation:
-7, -2, 3, 8 is an arithmetic sequence with a1 (first term) = -7 and common difference (d) = 5.
The 24th term = a1 + (24 - 1)d
= -7 + 23 * 5
= -7 + 115
= 108.
Triangle+ Triangle + Triangle = 30 Triangle + circle + circle = 20 Circle + Square + Square = 13 Triangle + circle x half square = ?
Answer:
Below
Step-by-step explanation:
Let T be triangle, C the circle and S the square.
● T + T + T = 30
● 3T = 30
Divide both sides by 3
● 3T/3 = 30/3
● T = 10
So the triangle has a value of 10.
●30 T + C + C = 20C + S + S = 13T +C ×S/2
Add like terms together
●30 T + 2C = 20C +2S= 13T + C×S/2
Replace T by its value (T=10)
● 300 + 2C = 20C + 2S = 130 + C×S/2
Take only this part 20C + 2S = 130 + C × S/2
● 20C + 2S = 130 + C×S/2 (1)
Take this part (300+2C = 20C+2S) and express S in function of C
● 20C + 2S = 300 + 2C
Divide everything by 2 to make easier
● 10 C + S = 150+ C
● S = 150+C-10C
● S = 150-9C
Replace S by (5-9C) in (1)
● 20C + 2S = 130 + C×S/2
● 20C + 2(150-9C) = 130 +C× (150-9C)/2
● 20C + 300-18C= 130 + C×(75-4.5C)
● 2C + 300 = 130 + 75 -4.5C^2
● 2C +300-130 = 75C - 4.5C^2
● 2C -75C + 170 = -4.5C^2
● -73C + 170 = -4.5C^2
Multiply all the expression by -1
● -4.5C^2 +73C+ 170= 0
This is a quadratic equation, so we will use the discriminant method.
Let Y be the discriminant
● Y = b^2-4ac
● b = 73
● a = -4.5
● c = 170
● Y = 73^2 - 4×(-4.5)×170= 8389
So the equation has two solutions:
● C = (-b +/- √Y) /2a
√Y is approximatively 92
● C = (-73 + / - 92 )/ -9
● C = 18.34 or C = -2.11
Approximatively
● C = 18 or C = -2
■■■■■■■■■■■■■■■■■■■■■■■■■
● if C = 18
30T + 2C = 300 + 36 = 336
● if C = -2
30T + 2C = 300-4 = 296
Please answer this correctly without making mistakes
Answer:
1/8
Step-by-step explanation:
3/8-1/8-1/8=1/8
you pick a card at random from an ordinary deck of 52 cards. If the card is an ace, you get 9 points; if not, you lose a point
Answer: a = 9, b = 48, c = -1
Step-by-step explanation:
"a" represents the points you receive if an Ace is picked. It is given that you get 9 points ----> a = 9
"b" represents the number of cards that are Not an Ace. 4 cards in the deck are Aces so 52 - 4 = 48 cards are Not an Ace -----> b = 48
"c" represents the points you receive if Not an Ace is picked. It is given that you lose 1 point ----> c = -1
Answer:
Here is the rest of the page
Step-by-step explanation:
Suppose you have a bag with the following in it: 5 one dollar bills, 4 fives, 3 tens, 5 twenties, and 3 fifties. Assuming the experiment requires drawing one bill from the bag at random, complete the probability distribution for this experiment.
Required:
What is the probability of drawing 9 dollars or less in a single draw?
Answer:
(a) Probility Distribution
Outcome probability
$1 5/15 = 1/3
$5 4/15
$10 3/15 = 1/5
$20 5/15 = 1/3
(b) P($9 or less) = 3/5
Step-by-step explanation:
(a) Probility Distribution
Outcome probability
$1 5/15 = 1/3
$5 4/15
$10 3/15 = 1/5
$20 5/15 = 1/3
Any other denomination
0
(b)
ways to draw $9 or less in a single draw
P($1) = 1/3
P($5) = 4/15
P($9 or less) = P($1) + P($5) = 1/3 + 4/15 = 9/15 = 3/5
A machine that produces ball bearings has initially been set so that the true average diameter of the bearings it produces is 0.500 in. A bearing is acceptable if its diameter is within 0.004 in. of this target value. Suppose, however, that the setting has changed during the course of production, so that the bearings have normally distributed diameters with a mean 0.499 in. and standard deviation 0.002 in. What percentage of bearings will now not be acceptable
Answer:
the percentage of bearings that will not be acceptable = 7.3%
Step-by-step explanation:
Given that:
Mean = 0.499
standard deviation = 0.002
if the true average diameter of the bearings it produces is 0.500 in and bearing is acceptable if its diameter is within 0.004 in.
Then the ball bearing acceptable range = (0.500 - 0.004, 0.500 + 0.004 )
= ( 0.496 , 0.504)
If x represents the diameter of the bearing , then the probability for the z value for the random variable x with a mean and standard deviation can be computed as follows:
[tex]P(0.496\leq X \leq 0.504) = (\dfrac{0.496 - \mu}{\sigma} \leq \dfrac{X -\mu}{\sigma} \leq \dfrac{0.504 - \mu}{\sigma})[/tex]
[tex]P(0.496\leq X \leq 0.504) = (\dfrac{0.496 - 0.499}{0.002} \leq \dfrac{X -0.499}{0.002} \leq \dfrac{0.504 - 0.499}{0.002})[/tex]
[tex]P(0.496\leq X \leq 0.504) = (\dfrac{-0.003}{0.002} \leq Z \leq \dfrac{0.005}{0.002})[/tex]
[tex]P(0.496\leq X \leq 0.504) = (-1.5 \leq Z \leq 2.5)[/tex]
[tex]P(0.496\leq X \leq 0.504) = P (-1.5 \leq Z \leq 2.5)[/tex]
[tex]P(0.496\leq X \leq 0.504) = P(Z \leq 2.5) - P(Z \leq -1.5)[/tex]
From the standard normal tables
[tex]P(0.496\leq X \leq 0.504) = 0.9938-0.0668[/tex]
[tex]P(0.496\leq X \leq 0.504) = 0.927[/tex]
By applying the concept of probability of a complement , the percentage of bearings will now not be acceptable
P(not be acceptable) = 1 - P(acceptable)
P(not be acceptable) = 1 - 0.927
P(not be acceptable) = 0.073
Thus, the percentage of bearings that will not be acceptable = 7.3%
One number is twice another. The sum of their reciprocals is 3/2 . Find the numbers.
Answer:
The two numbers are 1 and 2.
Step-by-step explanation:
Let the two numbers be a and b.
One number is twice another, so let's let b=2a.
Their reciprocals are 3/2. Thus:
[tex]\frac{1}{a}+\frac{1}{b} =\frac{3}{2}[/tex]
Substitute and solve for a:
[tex]\frac{1}{a}+\frac{1}{2a} =\frac{3}{2}\\[/tex]
Combine the fractions by forming a common denominator by multiplying the left term by 2:
[tex]\frac{2}{2a} +\frac{1}{2a}=\frac{3}{2}[/tex]
Combine and cross-multiply:
[tex]3/2a=3/2\\6a=6\\a=1\\b=2(1)=2[/tex]
Thus, the two numbers are 1 and 2.
20 points!
Please help.
BRAINLIST AND A THANK YOU AND 5 stars WILL BE REWARDED PLS ANSER
Answer:
The first picture's answer would be (6, 21)
Step-by-step explanation:
You have to find the points on the 8th and the 9th day, and then you would add them together, and then divide by two finding the average, which would be 24 and 18, so when added, you get 42, divided by 2 you get 21. You look on the graph for the point with 21, and you find it is on 6.
The entire graph of the function h is shown below write the domain and range of h using interval notation.
you can only see values of [tex] x[/tex] Ranging from $-3$ to $3$ and they're included, so domain is $[-3,3]$
and $y$ values ranging from $-2$ to $4$ but $-2$ is not included so range is $(-2,4]$
if f(x)=3-2x and g(x)= 1/x+5 what is the value of (f/g) (8)
Answer:
Step-by-step explanation:
(f/g) = (3 - 2x ) / (1/x + 5) You could go to the trouble to simplify all of this, but the easiest way is to just put in the 8 where you see an x
(f/g)8 = (3 - 2*8) / (1/8 + 5)
(f/g)/8 = (3 - 16 / (5 1/8) 1/8 = 0.125
(f/g) 8 = - 13 / ( 5.125)
(f/g)8 = - 2.54
i need help quick!!!
Answer: A,C, and D
Step-by-step explanation:
Answer:
the answer to this question may be option B, C and D
BRAINLIEST IF CORRECT!!! and 15 points solve for z -cz + 6z = tz + 83
Answer:
z = 83/( -c+6-t)
Step-by-step explanation:
-cz + 6z = tz + 83
Subtract tz from each side
-cz + 6z -tz= tz-tz + 83
-cz + 6z - tz = 83
Factor out z
z( -c+6-t) = 83
Divide each side by ( -c+6-t)
z( -c+6-t)/( -c+6-t) = 83/( -c+6-t)
z = 83/( -c+6-t)
In a Gallup poll of randomly selected adults, 66% said that they worry about identity theft. For a group of 1013 adults, the mean of those who do not worry about identify theft is closest to ________.
Answer: 669
Step-by-step explanation:
Given, In a Gallup poll of randomly selected adults, 66% said that they worry about identity theft.
i.e. The proportion of adults said that they worry about identity theft. (p) = 0.66
Sample size : n= 1013
Then , Mean for the sampling distribution of sample proportion = np
= (1013) × (0.66)
= 668.58 ≈ 669 [Round to the nearest whole number]
Hence, the mean of those who do not worry about identify theft is closest to 669 .
Find the odds in favor and the odds against a randomly selected person from Country X, age 25 and over, with the stated amount of education. four years (or more) of college
Answer:
25 : 63 and 63 : 25
Step-by-step explanation:
This is a complete question
The table shows the educational attainment of the population of Country X, ages 25 and over. Use the data in the table, expressed in millions, to solve the problem. of 10 questions ge 1: Ages 25 and Over, in Miltions 4 Years igh College 4 Years High School (Less than College School Only 4years) Cor Moce) Total Male 29 19 25 89 Female 11 28 23 Total 2 57 42 50 [176 Find the odds in favor and the odds against a randomty selected person from Country X.age 25 and over, with the stated amount of education. four years (or more) of college 21:67, 67:21 63:88, 88:63 25:63, 63:25 25:88, 88:25
According to the question, the relevant data provided in the question for the solution is as follows
Four years or more of college
Number of students = 50
Total = 176 students
Number of students does not belong = 126
So odds in favor is
= 50 : 126
= 25 : 63
And automatically out against the favor is 63 : 25
The lengths of pregnancies in a small rural village are normally distributed with a mean of 265 days and a standard deviation of 14 days. In what range would we expect to find the middle 50% of most lengths of pregnancies
Answer:
the middle 50% of most lengths of pregnancies ranges between 255.62 days and 274.38 days
Step-by-step explanation:
Given that :
Mean = 265
standard deviation = 14
The formula for calculating the z score is [tex]z = \dfrac{x -\mu}{\sigma}[/tex]
x = μ + σz
At middle of 50% i.e 0.50
The critical value for [tex]z_{\alpha/2} = z_{0.50/2}[/tex]
From standard normal table
[tex]z_{0.25}=[/tex] + 0.67 or -0.67
So; when z = -0.67
x = μ + σz
x = 265 + 14(-0.67)
x = 265 -9.38
x = 255.62
when z = +0.67
x = μ + σz
x = 265 + 14 (0.67)
x = 265 + 9.38
x = 274.38
the middle 50% of most lengths of pregnancies ranges between 255.62 days and 274.38 days
Suppose log subscript a x equals 3, log subscript a y equals 7, and log subscript a z equals short dash 2. Find the value of the following expression. log subscript a open parentheses fraction numerator x cubed y over denominator z to the power of 4 end fraction close parentheses
Answer:
24Step-by-step explanation:
Given the following logarithmic expressions [tex]log_ax = 3, log_ay = 7, log_az = -2[/tex], we are to find the value of [tex]log_a(\frac{x^3y}{z^4} )[/tex]
[tex]from\ log_ax = 3, x = a^3\\\\from\ log_ay = 7,y = a^7\\\\from\ log_az = -2, z = a^{-2}[/tex]Substituting x = a³, y = a⁷ and z = a⁻² into the log function [tex]log_a(\frac{x^3y}{z^4} )[/tex] we will have;
[tex]= log_a(\frac{x^3y}{z^4} )\\\\= log_a(\dfrac{(a^3)^3*a^7}{(a^{-2})^4} )\\\\= log_a(\dfrac{a^9*a^7}{a^{-8}} )\\\\= log_a\dfrac{a^{16}}{a^{-8}} \\\\= log_aa^{16+8}\\\\= log_aa^{24}\\\\= 24log_aa\\\\= 24* 1\\\\= 24[/tex]
Hence, the value of the logarithm expression is 24
A chemical company makes two brands
of antifreeze. The first brand is 30% pure
antifreeze, and the second brand i$ 80% pure
antifreeze. In order to obtain 80 gallons of a
mixture that contains 70o£ pure antifreeze, hov
mabry gallons of each band ot antifneze must
bo used?
Answer:
16 bags for the first(30% pure) and 64 bags of the second(80% pure)
Step-by-step explanation:
If they are mixed in a ratio of x bags to y bags
(0.3x+0.8y)/(x+y) = 0.7
0.3x + 0.8y = 0.7(x+y)
Multiply both sides with 10
3x + 8y = 7(x+y)
4x = y ——(1)
x + y = 80 ——(2)
Solve simultaneously
x + 4x = 80
5x = 80
x = 16 bags
y = 4x = 64 bags
602/100 into a decimal describe plz
Answer:
6.02
six point zero two
Step-by-step explanation:
Answer:
602 / 100= 6,02
Step-by-step explanation:
602 to divide 100 = 6,02
A test is being conducted to test the difference between two population means using data that are gathered from a matched pairs experiment. If the paired differences are normal, then the distribution used for testing is the:
Answer:
Student t-distribution.
Step-by-step explanation:
In this scenario, a test is being conducted to test the difference between two population "means" using data that are gathered from a matched pairs experiment. If the paired differences are normal, then the distribution used for testing is the student t-distribution.
In Statistics and probability, a student t-distribution can be defined as the probability distribution which can be used to estimate population parameters when the population variance is not known (unknown) and the sample population is relatively small. The student t-distribution is a statistical distribution which was published in 1908 by William Sealy Gosset.
A student t-distribution has a similar curve with the normal distribution curve, except that it is fatter and a little bit shorter.
What information do you need in order to determine the total distance Sam drives versus the actual displacement between his starting and ending points?
Answer:
his path
Step-by-step explanation:
In order to determine the total distance driven from one place to another, you need to know the path taken.
99 litres of gasoline oil is poured into a cylindrical drum of 60cm in diameter. How deep is the oil in the drum?
Answer:
35 cm
Step-by-step explanation:
The volume of a cylinder is given by ...
V = πr²h
We want to find h for the given volume and diameter. First, we must convert the given values to compatible units.
1 L = 1000 cm³, so 99 L = 99,000 cm³
60 cm diameter = 2 × 30 cm radius
So, we have ...
99,000 cm³ = π(30 cm)²h
99,000/(900π) cm = h ≈ 35.01 cm
The oil is 35 cm deep in the drum.
Can I have help with 43 and 44 I need to see how to do them thanks.
Answer:
see explanation
Step-by-step explanation:
(43)
3[tex]x^{5}[/tex] - 75x³ ← factor out 3x³ from each term
= 3x³(x² - 25) ← this is a difference of squares and factors in general as
a² - b² = (a - b)(a + b) , thus
x² - 25 = x² - 5² = (x - 5)(x + 5)
Thus
3[tex]x^{5}[/tex] - 75x³ = 3x³(x - 5)(x + 5)
(44)
81c² + 72c + 16 ← is a perfect square of the form
(ac + b)² = a²c² + 2abc + b²
Compare coefficients of like terms
a² = 81 ⇒ a = [tex]\sqrt{81}[/tex] = 9
b² = 16 ⇒ b = [tex]\sqrt{16}[/tex] = 4
and 2ab = 2 × 9 × 4 = 72
Thus
81c² + 72c + 16 = (9c + 4)²
1. 3x^5 -75x³
=3x³(x²-25)
=3x³(x²-5²)
=3x³(x-5)(x+5)
2. 81c²+72c+16
=81c²+36c+36c+16
=9c(9c+4)+4(9c+4)
=(9c+4)(9c+4)
=(9c+4)²
Claire has to go to the movie theater the movie starts at 4:15 pm it is a 25min walk to the theater from her home what time dose the have to leave the house to get there on time
Answer:
claire has to leave at 3:50 from her house.
Answer:
She needs to leave by 3:50 to get there on time.
Step-by-step explanation:
4:15 - 0:25 = 3:50.
