Answer: t is the dependent variable and x the independent variable
Step-by-step explanation: t is the dependent variable it depends upon the values of x that we put in.
x is the independent variable as we can use whatever values we want for x it is not affected by t
Given: ABCD is a parallelogram with AE = 9x−5, AC = 14x + 34. Find AC
The value of AC according to given equation of Parallelogram is 188 units.
What is parallelogram?
In elementary geometry, a parallelogram may be a quadrilateral with 2 pairs of parallel sides. the alternative or facing sides of a quadrangle square {measure} of equal length
Main body:
according to question:
AE = 9X-5
AC = 14X+34
as E is midpoint of AC so , we can say
2AE = AC
2(9x-5)= 14x+34
18x-10= 14x+34
4x = 44
x = 11
Now we need to find AC = 14x+34
= 14*11+34
= 188 units
Hence the value of AC according to given equation of Parallelogram is 188 units.
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A line is graphed on the coordinate grid to the right.
Which equation describes the line?
A.y=x/2+1/3
B.y = x/3+1/2
C.y=x/2+15 1/2
D.y=x/3+15 1/2
The equation of the line would be y = x/3 + 1/2.
Option (B) is correct.
What is the slope-intercept form of the line?
The slope-intercept equation is used to find the general equation of a straight line using its slope and the point where it intersects the y-axis. The slope intercept form equation is given as, y = mx + b.
As we can see in the graph the line cuts the y-axis at y = 1/2
so the y-intercept of the line would be c = 1/2.
And when x=4, y = 2
So we can prepare the equation as y = x/3 + 1/2.
Hence, the equation of the line would be y = x/3 + 1/2.
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The equation of the line would be y = x/3 + 1/2.
Option (B) is correct.
What is the slope-intercept form of the line?
The slope-intercept equation is used to find the general equation of a straight line using its slope and the point where it intersects the y-axis. The slope intercept form equation is given as, y = mx + b.
As we can see in the graph the line cuts the y-axis at y = 1/2
so the y-intercept of the line would be c = 1/2.
And when x=4, y = 2
So we can prepare the equation as y = x/3 + 1/2.
Hence, the equation of the line would be y = x/3 + 1/2.
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two different two-digit whole numbers are selected at random. what is the probability that their product is less than 200. express your answer as a common fraction. (hints: (l) there are 90 different two-digit numbers, (2) the pair {10, 11} produces the smallest product and the pair {11, 18} produces the largest product less than 200).
The probability that the product of the two two-digit numbers is less than 200 is given as follows:
43/8010.
How to calculate the probability?A probability is calculated as the division of the number of desired outcomes in the context of the experiment by the number of total outcomes.
There are 90 different two-digit numbers, hence the number of total outcomes for the product is of:
90 x 89 = 8010.
(the numbers have to be different)
The desired outcomes which result in a product of less than 200 are of given as follows:
10 multiplied by 9 numbers, from 11 to 19.11 multiplied by 10, 12, 13, 14, 15, 16, 17, 18. (8 numbers).12 multiplied by 10, 11, 13, 14, 15, 16. (6 numbers).13 multiplied by 10, 11, 12, 14, 15 (5 numbers).14 multiplied by 10, 11, 12, 13. (4 numbers).15 multiplied by 10, 11, 12, 13. (4 numbers).16 multiplied by 10, 11, 12. (3 numbers).17 multiplied by 10, 11. (2 numbers).18 multiplied by 10, 11. (2 numbers).Hence the number of desired outcomes is given as follows:
9 + 8 + 6 + 5 + 4 + 4 + 3 + 2 + 2 = 43.
Meaning that the probability is of:
43/8010.
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I need. Help with this question
Derivative of function is 5(2x + 4) * (x² + 4x + 6)⁴.
What is differentiation?
In arithmetic, the derivative of a function of a true variable measures the sensitivity modification of the perform price with relation to a change in its argument. Derivatives are a elementary tool of calculus.
Main body:
by using product rule;
Let = u = x² + 4x + 6
⇒ du/dx = 2x + 4 .
Now y = u⁵
⇒ dy/dx = 5u⁴ .
dy/dx = dy/dx * du/dx = 5u⁴ * (2x + 4)
= 5 * (x² + 4x + 6)⁴ * (2x + 4)
= 5(2x + 4) * (x² + 4x + 6)⁴
hence , derivative of function is 5(2x + 4) * (x² + 4x + 6)⁴.
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What is the freezing point, in C, of a 2.75 m solution of C8H18 in benzene?
The freezing point of the 2.75 m solution of octane in benzene is -8.56 °C.
The freezing point of a solution is the temperature at which the solution becomes a solid. The freezing point of a solution is lower than the freezing point of the pure solvent because the solute particles interfere with the movement of the solvent molecules, which slows down the freezing process.
To determine the freezing point of a solution, we can use the freezing point depression equation:
ΔTf = Kf x molality
where ΔTf is the change in freezing point, Kf is the freezing point depression constant for the solvent, and molality is the concentration of the solute in the solution expressed in moles of solute per kilogram of solvent.
To find the freezing point of a 2.75 m solution of C8H18 (octane) in benzene, we need to know the freezing point depression constant for benzene, which is 5.12 °C/m. We can then use the equation above to calculate the change in freezing point:
ΔTf = 5.12 °C/m x 2.75 m = 14.06 °C
To find the freezing point of the solution, we need to subtract the change in freezing point from the freezing point of the pure solvent. The freezing point of pure benzene is 5.5 °C, so the freezing point of the 2.75 m solution of octane in benzene is:
5.5 °C - 14.06 °C = -8.56 °C
This means that the freezing point of the 2.75 m solution of octane in benzene is -8.56 °C. At this temperature, the solution will become a solid.
Is there anyone that can help me with a finance question?
Answer:
Yes, there are many people who can help you with a finance question. Some of the people who can help you include: financial advisors, accountants, financial planners, and financial analysts. Additionally, there are many online resources available such as personal finance forums, websites, and blogs.
Step-by-step explanation:
Check the binomial distribution to see whether it can be approximated by the normal distribution. Round p and q to 1 decimal place, as needed. n = 95 P = 0.96 9 -0.04 np - and ng Is a normal approximation appropriate ? Yes No
As per the binomial distribution, the value of the normal approximation is 0.6573
The term binomial distribution refers the discrete probability distribution that gives only two possible results in an experiment, either success or failure.
Here we have given that n = 95 P = 0.96 and q = 0.04
Now, here we have to check the binomial distribution to see whether it can be approximated by the normal distribution.
While we looking into the given question we know that the value of n = 95 P = 0.96.
Then as per the binomial distribution formula, the normal distribution is calculated as,
=> P(X=1) = 95C4 * (0.96)⁴ * (1-0.96)⁹⁵⁻⁴
When we simplify this one then we get the value as 0.6573
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Which statement correctly demonstrates using limits to determine a vertical asymptote of g (x) = StartFraction 2 (x + 4) squared Over x squared minus 16 EndFraction
There is a vertical asymptote at x = 4 because Limit of g (x) as x approaches 4 minus = infinity and limit of g (x) as x approaches 4 plus = negative infinity
There is a vertical asymptote at x = 4 because Limit of g (x) as x approaches 4 minus = infinity and limit of g (x) as x approaches 4 plus = infinity
There is a vertical asymptote at x = –4 because Limit of g (x) as x approaches 4 minus = infinity and limit of g (x) as x approaches 4 plus = negative infinity
There is a vertical asymptote at x = –4 because Limit of g (x) as x approaches 4 minus = negative infinity and limit of g (x) as x approaches 4 plus = infinity
The correct option that describes the vertical asymptote is; B: There is a vertical asymptote at x = 4 because Limit of g (x) as x approaches 4 minus = infinity and limit of g (x) as x approaches 4 plus = infinity
How to find the vertical asymptote of a function?A vertical asymptote of a graph is defined as a vertical line x = a where the graph tends toward positive or negative infinity as the inputs approach a.
A vertical asymptote is a value of x for which the function is not defined, that is, it is a point which is outside the domain of a function;
In a graph, these vertical asymptotes are given by dashed vertical lines.
An example is a value of x for which the denominator of the function is 0, and the function approaches infinity for these values of x.
We are given the function;
g(x) = 2(x + 4)²/(x² - 16)
Simplifying the denominator gives;
(x² - 16) = (x + 4)(x - 4)
Thus, our function is;
g(x) = 2(x + 4)²/[(x + 4)(x - 4)]
(x + 4 ) will cancel out to give;
g(x) = 2(x + 4)/(x - 4)
Vertical asymptote:
Point in which the denominator is 0, so:
(x - 4) = 0
x = 4
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Let f:R→S be a surjective homomorphism of rings with identity.
(a) If R is a PID, prove that every ideal in S is principal.
(b) Show by example that S need not be an integral domain.
Every ideal of S is principal when f:R⇒S be a surjective homomorphism of rings with identity.
In a homomorphism, corresponding elements of two systems behave very similarly in combination with other corresponding elements. For example, let G and H be groups. The elements of G are denoted g, g′,…, and they are subject to some operation ⊕.
In algebra, a homomorphism is a structure-preserving map between two algebraic structures of the same type (such as two groups, two rings, or two vector spaces). The word homomorphism comes from the Ancient
Let f:R⇒S be a surjective homomorphism of rings with identity.
We have to find if R is a PID, prove that every ideal in S is principal.
We know that,
Let I be the ideal of S
Since f is sufficient homomorphism.
So, f⁻¹(I) is an ideal of R.
Since R is PID so ∈ r ∈ R such that
f⁻¹(I) = <r>
I = <f(r)>
Therefore,
Every ideal of S is principal when f:R⇒S be a surjective homomorphism of rings with identity.
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A train travels at 80 miles per hour. An equation can be written that compares the time (t) with the distance (d). What is the domain and range?
1. The domain is distance (d) and the range is time (t).
2. The domain is time (t) and the range is distance (d).
3. The domain is time (t) and the range is 80.
4. The domain is 80 and the range is time (t).
The required answer is the domain is time (t) and the range is a distance (d) i.e. Option 2.
What are domain and range?
The value range that can be plugged into a function is known as its domain. In a function like f, this set represents the x values f(x). The collection of values that a function can take on is known as its range. The values that the function outputs when we enter an x value are in this set.
From the given question, and the above definition of domain and range,
the time (t) acts as an x-values or input value and the distance (d) acts as a y-value or output value
Hence, the domain is time (t) and the range is a distance (d) i.e. Option 2.
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in the morning there were t people at the beach. later at noon, 2/7 of the people left the beach and there were 30 people left on the beach. find t
Answer:
42
Step-by-step explanation:
If 2/7 left that means that 5/7 are still there.
5/7x = 30 Multiple both sides by 7/5
x = [tex]\frac{30}{1}[/tex] x [tex]\frac{7}{5}[/tex]
x = 42
