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3 years ago

8 times as many as 8 is 64.A)8x8=64B)64x8=8C)8x64=8

Mathematics

2 answers:

kifflom [539]3 years ago

7 0

Answer:

Option A is correct

$8 \times 8 = 64$

Step-by-step explanation:

Given the statement:

$\text{8 times as many as 8 is 64}$

"8 times as many as 8" translated to $8 \times 8$

then;

we can write this as:

$8 \times 8 = 64$

Therefore, $\text{8 times as many as 8 is 64}$ is, $8 \times 8 = 64$

Ivanshal [37]3 years ago

5 0

Answer:
8*8 = 64 which is the first option

Explanation:
8 times as many as 8 means that we will repeat the number 8 for 8 times. This means that:
8 times as many as 8 = 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 = 8 * 8

This quantity is equivalent to 64.
Therefore:
8*8 = 64

Hope this helps :)

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PLEASE HELP ASAP In this task, you will practice finding the area under a nonlinear function by using rectangles. You will use g

mrs_skeptik [129]

Answer:

a) $1280 u^{2}$

b) $1320 u^{2}$

c) $\frac{4000}{3} u^{2}$

Step-by-step explanation:

In order to solve this problem we must start by sketching the graph of the function. This will help us visualize the problem better. (See attached picture)

You can sketch the graph of the function by plotting as many points as you can from x=0 to x=20 or by finding the vertex form of the quadratic equation by completing the square. You can also do so by using a graphing device, you decide which method suits better for you.

So we are interested in finding the area under the curve, so we divide it into 5 rectangles taking a right hand approximation. This is, the right upper corner of each rectangle will touch the graph. (see attached picture).

In order to figure the width of each rectangle we can use the following formula:

$\Delta x=\frac{b-a}{n}$

in this case a=0, b=20 and n=5 so we get:

$\Delta x=\frac{20-0}{5}=\frac{20}{5}=4$

so each rectangle must have a width of 4 units.

We can now calculate the hight of each rectangle. So we figure the y-value of each corner of the rectangles. We get the following heights:

h1=64

h2=96

h3=96

h4= 64

h5=0

so now we can use the following formula to find the area under the graph. Basically what the formula does is add the areas of the rectangles:

$A=\sum^{n}_{i=1} f(x_{i}) \Delta x$

which can be rewritten as:

$A=\Delta x \sum^{n}_{i=1} f(x_{i})$

So we go ahead and solve it:

$A=(4)(64+96+96+64+0)$

so:

$A= 1280 u^{2}$

B) The same procedure is used to solve part B, just that this time we divide the area in 10 rectangles.

In order to figure the width of each rectangle we can use the following formula:

$\Delta x=\frac{b-a}{n}$

in this case a=0, b=20 and n=10 so we get:

$\Delta x=\frac{20-0}{10}=\frac{20}{10}=2$

so each rectangle must have a width of 2 units.

We can now calculate the hight of each rectangle. So we figure the y-value of each corner of the rectangles. We get the following heights:

h1=36

h2=64

h3=84

h4= 96

h5=100

h6=96

h7=84

h8=64

h9=36

h10=0

so now we can use the following formula to find the area under the graph. Basically what the formula does is add the areas of the rectangles:

$A=\sum^{n}_{i=1} f(x_{i}) \Delta x$

which can be rewritten as:

$A=\Delta x \sum^{n}_{i=1} f(x_{i})$

So we go ahead and solve it:

$A=(2)(36+64+84+96+100+96+84+64+36+0)$

so:

$A= 1320 u^{2}$

In order to find part c, we calculate the area by using limits, the limit will look like this:

$\lim_{n \to \infty} \sum^{n}_{i=1} f(x^{*}_{i}) \Delta x$

so we start by finding the change of x so we get:

$\Delta x =\frac{b-a}{n}$

$\Delta x =\frac{20-0}{n}$

$\Delta x =\frac{20}{n}$

next we find $x^{*}_{i}$

$x^{*}_{i}=a+\Delta x i$

so:

$x^{*}_{i}=0+\frac{20}{n} i=\frac{20}{n} i$

and we find $f(x^{*}_{i})$

$f(x^{*}_{i})=f(\frac{20}{n} i)=-(\frac{20}{n} i)^{2}+20(\frac{20}{n} i)$

cand we do some algebra to simplify it.

$f(x^{*}_{i})=-\frac{400}{n^{2}}i^{2}+\frac{400}{n}i$

we do some factorization:

$f(x^{*}_{i})=-\frac{400}{n}(\frac{i^{2}}{n}-i)$

and plug it into our formula:

$\lim_{n \to \infty} \sum^{n}_{i=1}-\frac{400}{n}(\frac{i^{2}}{n}-i) (\frac{20}{n})$

And simplify:

$\lim_{n \to \infty} \sum^{n}_{i=1}-\frac{8000}{n^{2}}(\frac{i^{2}}{n}-i)$

$\lim_{n \to \infty} -\frac{8000}{n^{2}} \sum^{n}_{i=1}(\frac{i^{2}}{n}-i)$

And now we use summation formulas:

$\lim_{n \to \infty} -\frac{8000}{n^{2}} (\frac{n(n+1)(2n+1)}{6n}-\frac{n(n+1)}{2})$

$\lim_{n \to \infty} -\frac{8000}{n^{2}} (\frac{2n^{2}+3n+1}{6}-\frac{n^{2}}{2}-\frac{n}{2})$

and simplify:

$\lim_{n \to \infty} -\frac{8000}{n^{2}} (-\frac{n^{2}}{6}+\frac{1}{6})$

$\lim_{n \to \infty} \frac{4000}{3}+\frac{4000}{3n^{2}}$

and solve the limit

$\frac{4000}{3}u^{2}$

4 0

3 years ago

If I average a speed of 60 miles per hour while driving, how many hours of driving time will it take me to reach Arizona?

Mumz [18]

Answer:

Well where do you start

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

What is the recursive formula of the geometric sequence? 1, 5, 25, 125, 625, ...

Ilia_Sergeevich [38]

The recursive formula of the geometric sequence is given by option D; an = (1) × (5)^(n - 1) for n ≥ 1

<h3>How to determine recursive formula of a geometric sequence?</h3>

Given: 1, 5, 25, 125, 625, ...

first term, a = 1

Common ratio, r = 25/5

= 5

n = number of terms

an = a × r^(n - 1)

= 1 × 5^(n - 1)

an = (1) × (5)^(n - 1) for n ≥ 1

Learn more about recursive formula of geometric sequence:

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2 years ago

An equation using three variables is graphed in coordinate space

balandron [24]

This is not possible. False.

(BLASPHEMY I SAY!)

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3 years ago

Triangle ABC , m∠A=45° and c = 16, find b.

zmey [24]

Using the cosine function:

$\begin{gathered} \cos (\theta)=\frac{_{\text{ }}adjacent}{_{\text{ }}hypotenuse} \\ so\colon \\ \cos (45)=\frac{b}{16} \end{gathered}$

Solve for b:

$\begin{gathered} b=16\cdot\cos (45) \\ b=8\sqrt[]{2} \end{gathered}$

5 0

1 year ago