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Viefleur [7K]
3 years ago
12

In a race, Jason's position was -11 1/5 feet relative to the leader after 1 3/4 minutes. On average, how much did Jason's positi

on relative to the leader change per minute?
Respond with your answer as a mixed number in simplified form.
Mathematics
1 answer:
Nutka1998 [239]3 years ago
3 0

Answer:

B. -39

Step-by-step explanation:

your answer is-39

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Howard needs to buy pens and pencils for work. Pens cost $0.40 each and pencils cost 50.25 each. He needs to buy at least 340
Korolek [52]

Answer: what is the answer

Step-by-step explanation:

4 0
3 years ago
A rectangular playground is to be fenced off and divided in two by another fence parallel to one side of the playground. Seven h
Arisa [49]

Answer:

Step-by-step explanation:

Suppose the dimensions of the playground are x and y.

The total amount of the fence used is given and it is 780 ft. In terms of x and y this would be 3x+2y=780 (we add 3x because we want it to be cut in the middle). Therefore,  y= 780/2-3/2x. Now, the total area (A )to be fenced is

A=x*y= x*(390-3/2x)=-3/2 x^2+390x

Calculating the derivative of A and setting it equals to 0 to find the maximum

A'= -3x+390=0

This yields x=130.

Therefore y=780/2-3/2*130=195

Thus, the maximum area is 130*195=25,350ft^2

8 0
3 years ago
True or false the reciprocal of 5 is 1/5<br>シ​
rodikova [14]

Answer:

True

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
Why is the answer undefined????? Please explain to me!! Thanks!!!!!
Alja [10]
If you input: 1/0.00000001 in a calculator, the result will be a very large number
so as you get closer to zero, the result of 1/x larger until it becomes infinite (or undefined as some people like to call it) at 1/0
4 0
3 years ago
Could someone help me, please?
k0ka [10]

We can rewrite the expression under the radical as

\dfrac{81}{16}a^8b^{12}c^{16}=\left(\dfrac32a^2b^3c^4\right)^4

then taking the fourth root, we get

\sqrt[4]{\left(\dfrac32a^2b^3c^4\right)^4}=\left|\dfrac32a^2b^3c^4\right|

Why the absolute value? It's for the same reason that

\sqrt{x^2}=|x|

since both (-x)^2 and x^2 return the same number x^2, and |x| captures both possibilities. From here, we have

\left|\dfrac32a^2b^3c^4\right|=\left|\dfrac32\right|\left|a^2\right|\left|b^3\right|\left|c^4\right|

The absolute values disappear on all but the b term because all of \dfrac32, a^2 and c^4 are positive, while b^3 could potentially be negative. So we end up with

\dfrac32a^2\left|b^3\right|c^4=\dfrac32a^2|b|^3c^4

3 0
3 years ago
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