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2 years ago

13

Margaret invested $2000 into an account earning 4% interest compounded continuously. How much will her investment be worth after

5 years?

1 answer:

mr_godi [17]2 years ago

8 0

Answer:

$400

Step-by-step explanation:

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Which of the following equations is used to calculate the probability of event A happening, given that event B has occurred?

Naddika [18.5K]

Answer:

B

Step-by-step explanation:

Not much to explain

6 0

3 years ago

Measure the diagonal of a square whose area is exactly one square foot. Is the diagonal length closer to 1.5 or 1.4 feet? Explai

kondor19780726 [428]

It is closer to 1.4

Step-by-step explanation:

This is a 45 45 90 triangle

To find the hypotenuse of this triangle it is any side length times the square root of two.

Formula: Square Root = sqrt

Hypotenuse = SideLength*sqrt(2) ; 1*sqrt(2) = 1.412...

Rounded to the nearest tenths it is closer to 1.4 than 1.5

8 0

3 years ago

Tim's company offers a reimbursement package of $0.65 per mile plus $145

zalisa [80]

Answer:

C = 0.65x + 145

Step-by-step explanation:

Assume:

Number of miles = x

Annual maintenance = $145

Total reimbursement = C = $0.65*x + $145 = $0.65x + $145

4 0

3 years ago

C=a/5t what does t equal?

-Dominant- [34]

Photo has the answer! hope it helped

8 0

3 years ago

The graph of an inverse trigonometric function passes through the point (1, pi/2). Which of the following could be the equation

rodikova [14]

Answer: C) $y=sin^-1 x$

Step-by-step explanation:

Since, the graph of an inverse trigonometric function will pass through the point $(1,\frac{\pi}{2})$ ,

If this point satisfies the function,

For the function $y=cos^{-1} x$

If x = 1

$y=cos^{-1}1=0$

Thus, $(1,\frac{\pi}{2})$ is not satisfying function $y=cos^{-1} x$ ,

⇒ The graph of $y=cos^{-1} x$ is not passing through the point $(1,\frac{\pi}{2})$

For the function $y=cot^{-1}x$

If x = 1

$y=cot^{-1}1=\frac{\pi}{4}$

Thus, $(1,\frac{\pi}{2})$ is not satisfying function $y=cot^{-1}x$ ,

⇒ The graph of $y=cot^{-1}x$ is not passing through the point $(1,\frac{\pi}{2})$

For the function $y=sin^{-1} x$

If x = 1

$y=sin^{-1}1=\frac{\pi}{2}$

Thus, $(1,\frac{\pi}{2})$ is satisfying function $y=sin^{-1} x$ ,

⇒ The graph of $y=sin^{-1} x$ is passing through the point $(1,\frac{\pi}{2})$ .

For the function $y=tan^{-1}x$

If x = 1

$y=tan^{-1}1=\frac{\pi}{4}$

Thus, $(1,\frac{\pi}{2})$ is not satisfying function $y=cos^{-1} x$ ,

⇒ The graph of $y=tan^{-1} x$ is not passing through the point $(1,\frac{\pi}{2})$ .

Hence, Option C is correct.

3 0

3 years ago