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3 years ago

Someone please help me with this I need the answers so I will be prepared for the test

Mathematics

1 answer:

3241004551 [841]3 years ago

6 0

I can't see the pages to be able to help you sorry

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Chris tried to rewrite the expression \left( 4^{-2} \cdot 4^{-3} \right)^{3}(4

crimeas [40]

We have been given an expression $\left( 4^{-2} \cdot 4^{-3} \right)^{3}$ . We have been given steps how Chris tried to solve the given expression. We are asked to choose the correct option about Chris's work.

Let us simplify our given expression.

Using exponent property, $a^m\cdot a^n=a^{m+n}$ , we cab rewrite our given expression as:

$\left( 4^{-2+(-3)} \right)^{3}$

$\left( 4^{-5} \right)^{3}$

Now we will use exponent property $(a^m)^n=a^{m\cdot n}$ to further simplify our expression.

$\left( 4^{-5} \right)^{3}= 4^{-5\cdot 3}$

$\left( 4^{-5} \right)^{3}= 4^{-15}$

Therefore, Chris made mistake in step 2.

8 0

4 years ago

Question 27 of 28 1 Point What is the slope of the line shown below? (3,-2) (2,-4)

Gwar [14]

Answer:

The slope is 2

Step-by-step explanation:

To find the slope given two points, we use the formula

m = (y2-y1)/(x2-x1)

= (-4--2)/(2-3)

= (-4+2)/(2-3)

= -2/-1

= 2

6 0

3 years ago

What is the common denominator of x+5/x+8 =1+6/x+1

Kobotan [32]

It’s going to be x+8

3 0

2 years ago

Read 2 more answers

A tour company has a boat that sits at most 40 passengers. For a cruise on the Hudson River they charge $12 per adult and $5 per

EleoNora [17]

Answer:

Step-by-step explanation:

For the adults:

$12 times the 14 adults which is: $168

Then for the kids:

$5 times the 9 kids which is: $45

Add next:

$168 plus the $45 equals $213

A(14)+b(5)=c

5 0

3 years ago

PLZ HELP!!! Use limits to evaluate the integral.

Marrrta [24]

Split up the interval [0, 2] into n equally spaced subintervals:

$\left[0,\dfrac2n\right],\left[\dfrac2n,\dfrac4n\right],\left[\dfrac4n,\dfrac6n\right],\ldots,\left[\dfrac{2(n-1)}n,2\right]$

Let's use the right endpoints as our sampling points; they are given by the arithmetic sequence,

$r_i=\dfrac{2i}n$

where $1\le i\le n$ . Each interval has length $\Delta x_i=\frac{2-0}n=\frac2n$ .

At these sampling points, the function takes on values of

$f(r_i)=7{r_i}^3=7\left(\dfrac{2i}n\right)^3=\dfrac{56i^3}{n^3}$

We approximate the integral with the Riemann sum:

$\displaystyle\sum_{i=1}^nf(r_i)\Delta x_i=\frac{112}n\sum_{i=1}^ni^3$

Recall that

$\displaystyle\sum_{i=1}^ni^3=\frac{n^2(n+1)^2}4$

so that the sum reduces to

$\displaystyle\sum_{i=1}^nf(r_i)\Delta x_i=\frac{28n^2(n+1)^2}{n^4}$

Take the limit as n approaches infinity, and the Riemann sum converges to the value of the integral:

$\displaystyle\int_0^27x^3\,\mathrm dx=\lim_{n\to\infty}\frac{28n^2(n+1)^2}{n^4}=\boxed{28}$

Just to check:

$\displaystyle\int_0^27x^3\,\mathrm dx=\frac{7x^4}4\bigg|_0^2=\frac{7\cdot2^4}4=28$

4 0

3 years ago