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3 years ago

You deposit $5000 in a bank account that pays 3.5% annual

Mathematics

1 answer:

miskamm [114]3 years ago

6 0

Answer:

Its A i just took the test

Step-by-step explanation:

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Determine the intercepts of the line.<br> y + 5 = 2(x + 1)<br> y-intercept:<br> x-intercept:

Airida [17]

Answer:

(

)

Find the x-intercepts.

x-intercept(s):

(

)

Find the y-intercepts.

y-intercept(s):

(

−

)

List the intersections.

x-intercept(s):

(

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y-intercept(s):

(

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7 0

3 years ago

Tarinda's corn plant was 7- 1/3 inches tall at the end of Week 1. The corn plant was 18 1/2 inches tall at the end of Week 2. Ho

liberstina [14]

Answer:

The corn plant grew by 11.17 inches during week 2.

Step-by-step explanation:

7 and 1/3 inches tall at the end of Week 1.

The height at the end of week 1, as a simple fraction, is:

$7 + \frac{1}{3} = \frac{3*7 + 1}{3} = \frac{22}{3}$

The corn plant was 18 1/2 inches tall at the end of Week 2.

At the end of week 2, the height will be of:

$18 + \frac{1}{2} = \frac{18*2+1}{2} = \frac{37}{2}$

How many inches did the corn plant grow during Week 2?

Height at the end of week 2 subtracted by the height at the end of week 1. So

$\frac{37}{2} - \frac{22}{3} = \frac{3*37-2*22}{6} = \frac{67}{6} = 11.17$

The corn plant grew by 11.17 inches during week 2.

5 0

3 years ago

What does a star mean after a number

Triss [41]

A star is written to show multiplication. It is a replacement for a dot or a x which is what you probably learned to be the "times" sign. In upper level maths, the "x" is replaced with a " * " to make it more legible and to avoid confusion with the variable "x".

8 0

3 years ago

Read 2 more answers

Determine whether each integral is convergent or divergent. If it is convergent evaluate it. (a) integral from 1^(infinity) e^(-

AVprozaik [17]

Answer:

a) So, this integral is convergent.

b) So, this integral is divergent.

c) So, this integral is divergent.

Step-by-step explanation:

We calculate the next integrals:

$\int_1^{\infty} e^{-2x} dx=\left[-\frac{e^{-2x}}{2}\right]_1^{\infty}\\\\\int_1^{\infty} e^{-2x} dx=-\frac{e^{-\infty}}{2}+\frac{e^{-2}}{2}\\\\\int_1^{\infty} e^{-2x} dx=\frac{e^{-2}}{2}\\$

So, this integral is convergent.

$\int_1^{2}\frac{dz}{(z-1)^2}=\left[-\frac{1}{z-1}\right]_1^2\\\\\int_1^{2}\frac{dz}{(z-1)^2}=-\frac{1}{1-1}+\frac{1}{2-1}\\\\\int_1^{2}\frac{dz}{(z-1)^2}=-\infty\\$

So, this integral is divergent.

$\int_1^{\infty} \frac{dx}{\sqrt{x}}=\left[2\sqrt{x}\right]_1^{\infty}\\\\\int_1^{\infty} \frac{dx}{\sqrt{x}}=2\sqrt{\infty}-2\sqrt{1}\\\\\int_1^{\infty} \frac{dx}{\sqrt{x}}=\infty\\$

So, this integral is divergent.

4 0

3 years ago

Consider this expression:

Tema [17]

8(4a+2b)
The equivalent is 32a+16b (B)

4 0

3 years ago