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2 years ago

Aliza needs to run at a rate faster than 8.8 feet

Mathematics

1 answer:

Mnenie [13.5K]2 years ago

3 0

Answer:

Why does Aliza need ro run at a rate faster than 8.8 feet?

Step-by-step explanation:

And what was the question

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Two solutions of different concentrations of acid are mixed creating 40 mL of a solution that is 32% acid. One-quarter of the so

Georgia [21]

In order to construct this equation, we will use the variables:
V to represent mixture volume (40 ml)
C to represent mixture concentration (0.32)
v₁ to represent volume of first solution (40 / 4 = 10 ml)
c₁ to represent concentration of first solution (0.2)
v₂ to represent the volume of the second solution (40 * 3/4 = 30 ml)
c₂ to represent the concentration of the second solution

We know that the total amount of substance, product of the volume and concentration, in the final solution is equal to the individual amounts in the two given solutions. Thus:
VC = v₁c₁ + v₂c₂
40(0.32) = 10(0.2) + 30c

6 0

2 years ago

Read 2 more answers

The following formula for the sum of the cubes of the first n integers is proved in Appendix E. Use it to evaluate the limit in

Marina86 [1]

Answer:

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

And when we apply the limit we got that:

$\lim_{n\to\infty} (1+ \frac{2}{n} +\frac{1}{n^2}) =1$

Step-by-step explanation:

Assuming this complete problem: "The following formula for the sum of the cubes of the first n integers is proved in Appendix E. Use it to evaluate the limit . 1^3+2^3+3^3+...+n^3=[n(n+1)/2]^2"

We have the following formula in order to find the sum of cubes:

$\lim_{n\to\infty} \sum_{n=1}^{\infty} i^3$

We can express this formula like this:

$\lim_{n\to\infty} \sum_{n=1}^{\infty}i^3 =\lim_{n\to\infty} [\frac{n(n+1)}{2}]^2$

And using this property we need to proof that: 1^3+2^3+3^3+...+n^3=[n(n+1)/2]^2

$\lim_{n\to\infty} [\frac{n(n+1)}{2}]^2$

If we operate and we take out the 1/4 as a factor we got this:

$\lim_{n\to\infty} \frac{n^2(n+1)^2}{n^4}$

We can cancel $n^2$ and we got

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

We can reorder the terms like this:

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

We can do some algebra and we got:

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

We can solve the square and we got:

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

And when we apply the limit we got that:

$\lim_{n\to\infty} (1+ \frac{2}{n} +\frac{1}{n^2}) =1$

3 0

2 years ago

After the bank cashed a check Maureen wrote for $60, her balance was -$14.The equation b+(-60)=-14 can be used to represent this

dem82 [27]

we are given

$b+(-60)=-14$

Since, we have to solve for b

$b-60=-14$

so, we will isolate b ony one side

so, we will add 60 on both sides

$b-60+60=-14+60$

$b=-14+60$

$b=46$

so,

Answer:

Maureen should have added 60 to both sides.

5 0

2 years ago

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Given the function f(x) = −5x2 − x + 20, find f(3). −28 −13 62 64

Katyanochek1 [597]

Answer:

Option (a) is correct.

The value of given function $f(x)=-5x^2-x+20$ at x = 3 is -28

Step-by-step explanation:

Given : Function $f(x)=-5x^2-x+20$

We have to find the value of given function at x = 3

Consider the given function $f(x)=-5x^2-x+20$

Since, we have tof ind the value of function at x = 3

Put x = 3 in given function f(x)

$f(3)=-5(3)^2-(3)+20$

Simplify, we have,

$f(x)=-45-3+20=-28$

Thus, the value of given function $f(x)=-5x^2-x+20$ at x = 3 is -28

5 0

2 years ago

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If p and q are nonzero integers, which pair of points must lie in the same quadrant?

Tems11 [23]

Hello there.
<span>
If p and q are nonzero integers, which pair of points must lie in the same quadrant?
</span>
<span>B. (p, q) and (2p, 2q)
</span>

8 0

2 years ago

Read 2 more answers