[tex]\rightarrow\sf {5a}^{2} + {b(a}^{2} + 5) + {b}^{2} [/tex]
Solution:[tex]\rightarrow\sf {5a}^{2} + {b(a }^{2} + 5) + {b}^{2} \\ = \sf {5a}^{2} + {ba}^{2} + b \times 5 + {b}^{2} \\ = \large\boxed{\sf{\red{ {5a}^{2} + {ba}^{2} + 5b + {b}^{2} }}}[/tex]
Answer:[tex]\rightarrow\large\boxed{\sf{\red{ {5a}^{2} + {ba}^{2} + 5b + {b}^{2} }}}[/tex]
[tex]\color{red}{==========================}[/tex]
✍︎ꕥᴍᴀᴛʜᴅᴇᴍᴏɴǫᴜᴇᴇɴꕥ
✍︎ꕥᴄᴀʀʀʏᴏɴʟᴇᴀʀɴɪɴɢꕥ
3x+4 number of terms
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Answer:
2
Step-by-step explanation:
In this expression, the terms are the parts of the sum. They are 3x and 4. There are 2 terms.
4b^2+300=0 this is a quadratic equation that I am trying to solve including any solutions with imaginary numbers I will include a picture
Answer:
b= 5i square root of 3
b = -5i square root of 3
Step-by-step explanation:
4b^2+300=0
4b^2 = -300
b^2 = -75
b = square root of -75
b = -75^1/2
^1/2 means square root
b = 25^1/2 * 3^1/2 * i
b= 5i square root of 3
b = -5i square root of 3
Please help me quick I’ll give brainliest
A closed, rectangular-faced box with a square base is to be constructed using only 36 m2 of material. What should the height h and base length b of the box be so as to maximize its volume
Answer:
[tex]b=h=\sqrt{6}[/tex] m
Step-by-step explanation:
Let
Bas length of box=b
Height of box=h
Material used in constructing of box=36 square m
We have to find the height h and base length b of the box to maximize the volume of box.
Surface area of box=[tex]2b^2+4bh[/tex]
[tex]2b^2+4bh=36[/tex]
[tex]b^2+2bh=18[/tex]
[tex]2bh=18-b^2[/tex]
[tex]h=\frac{18-b^2}{2b}[/tex]
Volume of box, V=[tex]b^2h[/tex]
Substitute the values
[tex]V=b^2\times \frac{18-b^2}{2b}[/tex]
[tex]V=\frac{1}{2}(18b-b^3)[/tex]
Differentiate w. r.t b
[tex]\frac{dV}{db}=\frac{1}{2}(18-3b^2)[/tex]
[tex]\frac{dV}{db}=0[/tex]
[tex]\frac{1}{2}(18-3b^2)=0[/tex]
[tex]\implies 18-3b^2=0[/tex]
[tex]\implies 3b^2=18[/tex]
[tex]b^2=6[/tex]
[tex]b=\pm \sqrt{6}[/tex]
[tex]b=\sqrt{6}[/tex]
The negative value of b is not possible because length cannot be negative.
Again differentiate w.r.t b
[tex]\frac{d^2V}{db^2}=-3b[/tex]
At [tex]b=\sqrt{6}[/tex]
[tex]\frac{d^2V}{db^2}=-3\sqrt{6}<0[/tex]
Hence, the volume of box is maximum at [tex]b=\sqrt{6}[/tex].
[tex]h=\frac{18-(\sqrt{6})^2}{2\sqrt{6}}[/tex]
[tex]h=\frac{18-6}{2\sqrt{6}}[/tex]
[tex]h=\frac{12}{2\sqrt{6}}[/tex]
[tex]h=\sqrt{6}[/tex]
[tex]b=h=\sqrt{6}[/tex] m
ASAP What is the rule for this relation? i will give brainliest
Answer:
your selected answer is right
Order the following decimals. State your method of choice and your reasons for choosing it. Explain how you know this order is accurate.
Answer:
.40 is the greatest .350 is the second greatest and last but not least .3456 is the lowest
Step-by-step explanation:
The awnser for this question
What is graph for the equation y=-4x+1
Answer: The line starts at 1 positive, then from there go -4 (so go to the left) then 1 down from that point.
Step-by-step explanation: the problem is supposed to have been Y= -4/1 +1
Order these numbers from least to greatest.
5.772 , 11/2, 5 6/11, 5.77
Answer:
6/11, 11/2, 5.77, 5.772
Step-by-step explanation:
If y- 1 equals 10 then y
Answer:
11
Step-by-step explanation:
y-1=10
Any figure that crosses equal sign, the operational sign changes.
y=10+1
y= 11
15×115-(-3)}(4-4)÷3{5+(-3)×(-6
Answer:
15×115+3{0÷3}5-3×(-6)
15×115+3of 0 of 5-3×(-6)
15×115+0 of 5-3×(-6)
15×115+0+18
1725+0+18
1743
if triangle TAN has vertices T(0, 2), A(-1,3), and N(-2,-4), which of the following coordinates is N' of the dilation from the origin using the scale factor 3?
Answer:
(-6,-12)
Step-by-step explanation:
A dilation makes a figure gets bigger so just multiply 3 to point N to find N prime.
[tex] - 2 \times 3 = - 6[/tex]
[tex] - 4 \times 3 = - 12[/tex]
So our new coordinates is
(-6,-12)
Answer:
(-6,-12)
Step-by-step explanation:
A dilation makes a figure gets bigger so just multiply 3 to point N to find N prime.
So our new coordinates is
(-6,-12)
Step-by-step explanation:
what is the value of the expression 5²5
Answer:
5^2×5
=25×5
=125
hope this will help you
Answer:
125
Step-by-step explanation:
hope this helps you
=25×5
=125
Andrew wants to build a square garden and needs to determine how much area he has for planting the perimeter of the garden is between 12 and 14 feet what is the range if the possible areas
Answer:
9 ft^2 and 12.25 ft^2
Step-by-step explanation:
We need to figure out the area for a square with a perimeter of 12 feet and 14 feet.
A square has four sides that are all equal in length, therefore:
12/4 = 3
14/4 = 3.5
3 and 3.5 are the individual side lengths of the garden, so to find the area, we just multiply those numbers by themselves (since it is a square garden).
3*3 = 9
3.5*3.5 = 12.25
Therefore, the answer is 9 ft^2 and 12.25 ft^2
a student takes two subjects A and B. Know that the probability of passing subjects A and B is 0.8 and 0.7 respectively. If you have passed subject A, the probability of passing subject B is 0.8. Find the probability that the student passes both subjects? Find the probability that the student passes at least one of the two subjects
Answer:
0.64 = 64% probability that the student passes both subjects.
0.86 = 86% probability that the student passes at least one of the two subjects
Step-by-step explanation:
Conditional Probability
We use the conditional probability formula to solve this question. It is
[tex]P(B|A) = \frac{P(A \cap B)}{P(A)}[/tex]
In which
P(B|A) is the probability of event B happening, given that A happened.
[tex]P(A \cap B)[/tex] is the probability of both A and B happening.
P(A) is the probability of A happening.
In this question:
Event A: Passing subject A
Event B: Passing subject B
The probability of passing subject A is 0.8.
This means that [tex]P(A) = 0.8[/tex]
If you have passed subject A, the probability of passing subject B is 0.8.
This means that [tex]P(B|A) = 0.8[/tex]
Find the probability that the student passes both subjects?
This is [tex]P(A \cap B)[/tex]. So
[tex]P(B|A) = \frac{P(A \cap B)}{P(A)}[/tex]
[tex]P(A \cap B) = P(B|A)P(A) = 0.8*0.8 = 0.64[/tex]
0.64 = 64% probability that the student passes both subjects.
Find the probability that the student passes at least one of the two subjects
This is:
[tex]p = P(A) + P(B) - P(A \cap B)[/tex]
Considering [tex]P(B) = 0.7[/tex], we have that:
[tex]p = P(A) + P(B) - P(A \cap B) = 0.8 + 0.7 - 0.64 = 0.86[/tex]
0.86 = 86% probability that the student passes at least one of the two subjects
Draw a line representing the "rise" and a line representing the "run" of the line. State the slope of the line in simplest form.
Answer:
See attachment showing the rise and run
Slope = 1
Step-by-step explanation:
In the diagram attached below, the rise is represented by the blue line, while the run is represented by the red line.
Rise = 4 units
Run = 4 units
It's a positive slope because the line slopes upwards from left to right
Slope = rise/run = 4/4
Slope = 1
A display case of toy rings are marked 5 for $1. If Zach wants to buy 50 toy rings, how much will Zach spend (not including tax)
Answer:
$10
Step-by-step explanation:
5 toys = $1
Zach wants 50 of these
50 ÷ 5 = 10
10 x 1 = 10
= $10
Answer:
10 dollars
Step-by-step explanation:
We can use a ratio to solve
5 rings 50 rings
---------- = --------------
1 dollar x dollars
Using cross products
5*x = 1 * 50
5x = 50
Divide by 5
5x/5 = 50/5
x = 10
What’s the equation of the line
Answer:
[tex]y = - \frac{1}{3}x + 5[/tex]
Step-by-step explanation:
Consider two points through which the line passes.
Let it be ( 0 , 5 ) and ( 6 , 3 )
Step 1 : Find slope
[tex]Slope, m = \frac{y_2 - y_ 1 }{x_2 - x_1}[/tex]
[tex]= \frac{3-5}{6-0} \\\\=\frac{-2}{6}\\\\= -\frac{1}{3}[/tex]
Step 2 : Find the equation of the line passing through the points.
[tex]( y - y_1) = m (x - x_1)\\\\(y - 5) = -\frac{1}{3} ( x - 0) \\\\y = -\frac{1}{3}x + 5[/tex]
what graph shows the solution to the equation below log3(x+2)=1
Answer:
The solution to the equation log3(x+2)=1 is given by x=1
Step-by-step explanation:
We are given that
[tex]log_3(x+2)=1[/tex]
We have to find the graph which shows the solution to the equation log3(x+2)=1.
[tex]log_3(x+2)=1[/tex]
[tex]x+2=3^1[/tex]
Using the formula
[tex]lnx=y\implies x=e^y[/tex]
[tex]x+2=3[/tex]
[tex]x=3-2[/tex]
[tex]x=1[/tex]
The length of the box is 15 centimeters, the breadth of the box is 20 centimeter, the height of a box, 20 centimeter fine its volume. Step by step
Answer:
volume=length×width×height
v=15×20×20
v=6000
The number of measles cases increased 26.3% to 321 cases this year. What was the number of cases prior to the increase? Express your answer rounded correctly to the nearest whole number.
Answer:
The right answer is "[tex]x\simeq 254[/tex]".
Step-by-step explanation:
Let the number of earlier case will be "x".
Now,
⇒ [tex]x+x\times \frac{26.3}{100}=321[/tex]
or,
⇒ [tex]x+x\times 0.263=321[/tex]
By taking "x" common, we get
⇒ [tex]x(1+0.263)=321[/tex]
⇒ [tex]x=\frac{321}{1.263}[/tex]
⇒ [tex]=254.15[/tex]
or,
⇒ [tex]x\simeq 254[/tex]
Use the given graph of f to state the value of each quantity, if it exists. (If an answer does not exist, enter DNE.)
The x y-coordinate plane is given.
The function enters the window in the second quadrant, goes up and right becoming less steep, crosses the y-axis at approximately y = 3.2, changes direction at the approximate point (0.7, 3.3), goes down and right becoming more steep, and stops at the closed point (2, 3).
The function starts again at the open point (2, 1), goes up and right becoming more steep, goes up and right becoming less steep, passes through the open point (4, 4), changes direction at the approximate point (4.2, 4.1), goes down and right becoming more steep, and exits the window in the first quadrant.
(a) lim x → 2− f(x)
(b) lim x → 2+ f(x)
(c) lim x → 2 f(x)
(d) f(2)
(e) lim x → 4 f(x)(f) f(4)
Answer:
Hence the answer is given as follows,
Step-by-step explanation:
Graph of y = f(x) given,
(a) [tex]\lim_{x\rightarrow 2^{-}}f(x)=3[/tex]
(b) [tex]\lim_{x\rightarrow 2^{+}}f(x)=1[/tex]
(c) [tex]\lim_{x\rightarrow 2}f(x)= DNE \left \{ \therefore \lim_{x\rightarrow 2^{-}} f(x)\neq \lim_{x\rightarrow 2^{+}}f(x) \right.[/tex]
(d) [tex]f(2)=3[/tex]
(e) [tex]\lim_{x\rightarrow 4}f(x) = 4[/tex]
(f) [tex]f(4)= DNE.[/tex]{ Hole in graph}
Hence solved.
Find the area of the shaded region in terms of .
Please help :)
9514 1404 393
Answer:
50π cm²
Step-by-step explanation:
The radius of the larger circle is 10 cm, so its area is ...
A = πr² = π(10 cm)² = 100π cm²
The area of each smaller circle is ...
A = π(5 cm)² = 25π cm²
Then the shaded area is ...
shaded = large circle - 2 × small circle
shaded = 100π cm² - 2(25π cm²) = 50π cm²
The population of a bacteria colony is growing exponentially, doubling every 6 hours. If there are 150 bacteria currently present, how many (to the nearest ten bacteria) will be present in 10 hours
Answer:
If rounded to the nearest 10 bacteria, then it would be 500 bacteria.
Step-by-step explanation:
First multiply 150 by two in order to get 300, that leaves 4 hours to figure out. From there you can figure out the rest by seeing that 4 is 2/3 of 6. I converted it into the decimal number .66. Multiply 300 by .66 to get 198 and then add it to 300 to get 498. Then just round it up to the nearest 10 bacteria which leaves you with the final answer of 500 bacteria.
Given sets X, Y, Z, and U, find the set Xn(X - Y) using the listing method.
X = {d, c, f, a}
Y = {d, e, c}
Z ={e, c, b, f, g}
U = {a, b, c, d, e, f, g}
Answer:
{f, a}
Step-by-step explanation:
Given the sets:
X = {d, c, f, a}
Y = {d, e, c}
Z ={e, c, b, f, g}
U = {a, b, c, d, e, f, g}
To obtain the set X n (X - Y)
We first obtain :
(X - Y) :
The elements in X that are not in Y
(X - Y) = {f, a}
X n (X - Y) :
X = {d, c, f, a} intersection
(X - Y) = {f, a}
X n (X - Y) = elements in X and (X - Y)
X n (X - Y) = {f, a}
Type the correct answer in each box. Use numerals instead of words.
Multiply the expressions.
If a = 1, find the values of b, c, and d that make the given expression equivalent to the expression below.
Answer:
a=1, b=9, c=-2, d=4
Step-by-step explanation:
Can you help me answer this question? Screenshot is added.
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Answer:
(c)
Step-by-step explanation:
[tex]\displaystyle\sqrt[3]{xy^5}\sqrt[3]{x^7y^{17}}=\sqrt[3]{x^{1+7}y^{5+17}}=\sqrt[3]{x^6x^2y^{21}y}=\sqrt[3]{x^6y^{21}}\sqrt[3]{x^2y}\\\\=\boxed{x^2y^7\sqrt[3]{x^2y}}[/tex]
what is the sum of a 7 term geometric series if the first term is 6 the last term is -24576 and the common ratio is -4
Answer:
Sum = 19,662
Step-by-step explanation:
Given that this is a finite geometric series (meaning it stops at a specific term or in this case -24,576), we can use this formula:
[tex]\frac{a(1-r^n)}{1-r}[/tex], where a is the first term, r is the common ratio, and n is the number of terms.
Substituting for everything and simplifying gives us:
[tex]\frac{6(1-(-4)^7)}{1-(-4)} \\\\\frac{6(16385}{5}\\ \\\frac{98310}{5}\\ \\19662[/tex]
A study was conducted to investigate the effectiveness of hypnotism in reducing pain. Results for randomly selected subjects are given below. At the 1% level of significance, test the claim that the sensory measurements are lower after hypnotism (scores are in cm. on a pain scale). Assume sensory measurements are normally distributed. Note: You do not need to type these values into Minitab Express; the data file has been created for you.Before 6.6 6.5 9.0 10.3 11.3 8.1 6.3 11.6 After 6.8 2.4 7.4 8.5 8.1 6.1 3.4 2.0
Answer:
sensory measurement are lower after hypnotism
Step-by-step explanation:
Given the data :
Before 6.6 6.5 9.0 10.3 11.3 8.1 6.3 11.6
After 6.8 2.4 7.4 8.5 8.1 6.1 3.4 2.0
The difference ;
After - Before, d = 0.2, - 4.1, - 1.6, - 1.8, - 3.2, - 2, - 2.9, - 9.6
Hypothesis :
H0 : μd = 0
H0 : μ < 0
The test statistic ;
T = μd / sd/√n
Where, xd = mean of difference
sd = standard deviation of difference
n = sample size
Mean of difference, μd = Σx/n = - 3.13
Standard deviation of difference, sd = 2.91
T = - 3.13 / 2.91/√8
T = - 3.13 / 1.0288403
T = - 3.042
α = 0.01
The Pvalue using a Pvalue calculator ;
Degree of freedom, df = n - 1 ; 8-1 = 7
Pvalue(-3.042, 7) = 0.00939
Pvalue < α ; we reject the null and conclude that sensory measurement are lower after hypnotism
The accompanying data represent the homework scores for material for a random sample of students in a college algebra course.
36
47
54
58
60
66
66
68
69
70
72
75
77
77
78
78
78
79
79
79
79
79
80
82
84
85
86
86
86
87
89
89
91
92
92
93
93
94
96
99
(a) Construct a relative frequency distribution with a lower class limit of the first class equal to 30 and a class width of 10.
(b) What is the probability a randomly selected student fails the homework (scores less than 70)? (The standard deviation is 13.64)
Simplify your answer to two decimal places.
Answer:
[tex]\begin{array}{ccc}{Class} & {Frequency} & {Relative\ Frequency} &{30-39} & {1} & {0.025} & {40-49} & {1} & {0.025} & {50 - 59} & {2} & {0.050} & {60 - 69} & {5} & {0.125} & {70 - 79} & {13} & {0.325} & {80 - 89} & {10} & {0.250} & {90 - 99} & {8} & {0.200} &{Total} & {40} & {1}\ \end{array}[/tex]
[tex]P(x < 70) = 0.225[/tex]
Step-by-step explanation:
Given
[tex]Lower = 30[/tex]
[tex]Width = 10[/tex]
Solving (a): The relative frequency table
First, we construct the frequency table using the given parameters.
[tex]\begin{array}{cc}{Class} & {Frequency} &{30-39} & {1} & {40-49} & {1} & {50 - 59} & {2} & {60 - 69} & {5} & {70 - 79} & {13} & {80 - 89} & {10} & {90 - 99} & {8} & {Total} & {40}\ \end{array}[/tex]
The relative frequency (RF) is calculated as:
[tex]RF = \frac{Frequency}{Total}[/tex]
Using the above formula to calculate the relative frequency, the relative frequency table is:
[tex]\begin{array}{ccc}{Class} & {Frequency} & {Relative\ Frequency} &{30-39} & {1} & {0.025} & {40-49} & {1} & {0.025} & {50 - 59} & {2} & {0.050} & {60 - 69} & {5} & {0.125} & {70 - 79} & {13} & {0.325} & {80 - 89} & {10} & {0.250} & {90 - 99} & {8} & {0.200} &{Total} & {40} & {1}\ \end{array}[/tex]
Solving (b): [tex]P(x < 70)[/tex]
To do this, we add up the relative frequencies of classes less than 70.
i.e.
[tex]P(x < 70) = [30 - 39] + [40 - 49] + [50 - 59] + [60 - 69][/tex]
So, we have:
[tex]P(x < 70) = 0.025 + 0.025 + 0.050 + 0.125[/tex]
[tex]P(x < 70) = 0.225[/tex]
