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3 years ago

11

Anastasia uses the equation p = 0.7(rh + b) to estimate the amount of take-home pay, p, for h hours worked at a rate of r dollar

s per hour and any bonus received, b.What is an equivalent equation solved for h?

2 answers:

poizon [28]3 years ago

8 0

Answer:

$h=\frac{p-0.7b}{0.7r}$

Step-by-step explanation:

The equation given is $p=0.7(rh+b)$

Using distributive property on this, we can expand. Distributive property is given as:

$a(b+c)=ab+ac$

<em>Using this, we have:</em>

$p=0.7rh+0.7b$

<em>Rearranging and solving for $h$ gives us:</em>

$p-0.7b=0.7rh\\\frac{p-0.7b}{0.7r}=h$

This is the formula with $h$ as the subject.

Hitman42 [59]3 years ago

4 0

P = 0.7 ( r h + b )
p = 0.7 r h + 0.7 b
0.7 r h = p - 0.7 b
h = ( p - 0.7 b ) / 0.7 r
h = 10 ( p - 0.7 b ) / 7 r
h = ( 10 p - 7 b ) / 7 r

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Whats the answer ? evaluate 7^3 - 25

gladu [14]

Answer: 318

Step-by-step explanation:

7 to the 3rd power is 343, 343 minus 25 is 318

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3 years ago

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A(n - 3) +8= bn for n please help me

Papessa [141]

Given equation: A(n - 3) +8= bn.

Solution: On the left side of the equation we have A(n-3).

We don't have any sign in between A and parenthesis (n-3).

So, we need to multiply A and (n-3).

We need to apply distributive property to multiply A and (n-3).

Distributing A over (n-3), we get

A(n-3) = A*n - 3*A = An -3A.

Substituting this value in original equation,

An -3A +8= bn.

We need to solve it for n, so we get n terms on a side.

We have An on left side, we need to get rid n from left side.

Subtracting An from both sides, we get

An -3A +8-An= bn-An.

-3A +8 = bn - An.

We can see n is a common factor on rigth side in bn-an.

Factoring out n on right side from bn-an.

-3A +8 = (b - A)n.

Dividing both sides by (b-A),

$\frac{(-3A+8)}{(b-A)} = \frac{(b-A)n}{(b-A)}$

On right (b-A) paranthiss cancelled and we get n on right side.

$\frac{(-3A+8)}{(b-A)} =n$ Final answer.

So, that would be our final answer

n = (-3A+8)/(b-A)

7 0

4 years ago

I'll give you more points but you have to answer all the parts to the question. I need this plsssss ty lol. Show all work too pl

alexira [117]

Answer:

a) 34mi

b) 10mi

c) 44mi

d) by Calculating the amount of miles it would take John to go to school plus the amount of miles it would take Sara to go to school. Then add the sum of the amount of miles it took for both to go to school.

Step-by-step explanation:

a) explanation:

$a^{2} +b^2=c^2$

$16^2+30^2=c^2$

$256 + 900 = c^2$

$c^2=1156$

$\sqrt{1156}=34mi$

b) explanation:

$a^{2} +b^2=c^2$

$6^2+8^2=c^2$

$36+64=c^2$

$c^2=100$

$\sqrt{100} =10mi$

c) explanation: we already have all the data from John's house to Sara's House.

John's House to Sara's House = John's House to School + Sara's House to School.

Sara's House to School (Question #1) = 34mi

John's House to School (Question #2) = 10mi

34(mi) + 10(mi) = 44(mi)

d) explanation: explanation is on the answer

7 0

3 years ago

Prove or disprove (from i=0 to n) sum([2i]^4) <= (4n)^4. If true use induction, else give the smallest value of n that it doe

ddd [48]

Answer:

The statement is true for every n between 0 and 77 and it is false for $n\geq 78$

Step-by-step explanation:

First, observe that, for n=0 and n=1 the statement is true:

For n=0: $\sum^{n}_{i=0} (2i)^4=0 \leq 0=(4n)^4$

For n=1: $\sum^{n}_{i=0} (2i)^4=16 \leq 256=(4n)^4$

From this point we will assume that $n\geq 2$

As we can see, $\sum^{n}_{i=0} (2i)^4=\sum^{n}_{i=0} 16i^4=16\sum^{n}_{i=0} i^4$ and $(4n)^4=256n^4$ . Then,

$\sum^{n}_{i=0} (2i)^4 \leq(4n)^4 \iff \sum^{n}_{i=0} i^4 \leq 16n^4$

Now, we will use the formula for the sum of the first 4th powers:

$\sum^{n}_{i=0} i^4=\frac{n^5}{5} +\frac{n^4}{2} +\frac{n^3}{3}-\frac{n}{30}=\frac{6n^5+15n^4+10n^3-n}{30}$

Therefore:

$\sum^{n}_{i=0} i^4 \leq 16n^4 \iff \frac{6n^5+15n^4+10n^3-n}{30} \leq 16n^4 \\\\ \iff 6n^5+10n^3-n \leq 465n^4 \iff 465n^4-6n^5-10n^3+n\geq 0$

and, because $n \geq 0$ ,

$465n^4-6n^5-10n^3+n\geq 0 \iff n(465n^3-6n^4-10n^2+1)\geq 0 \\\iff 465n^3-6n^4-10n^2+1\geq 0 \iff 465n^3-6n^4-10n^2\geq -1\\\iff n^2(465n-6n^2-10)\geq -1$

Observe that, because $n \geq 2$ and is an integer,

$n^2(465n-6n^2-10)\geq -1 \iff 465n-6n^2-10 \geq 0 \iff n(465-6n) \geq 10\\\iff 465-6n \geq 0 \iff n \leq \frac{465}{6}=\frac{155}{2}=77.5$

In concusion, the statement is true if and only if n is a non negative integer such that $n\leq 77$

So, 78 is the smallest value of n that does not satisfy the inequality.

Note: If you compute $(4n)^4- \sum^{n}_{i=0} (2i)^4$ for 77 and 78 you will obtain:

$(4n)^4- \sum^{n}_{i=0} (2i)^4=53810064$

$(4n)^4- \sum^{n}_{i=0} (2i)^4=-61754992$

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4 years ago

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N76 [4]

x is equal to - 2/3 after collecting like terms

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3 years ago