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2 years ago

12

To train for a race, Scott's want to run 20 miles this week . he runs 1/4 of the miles on Monday. If he runs 2 miles on Tuesday

how many miles will he still need to run to meet his goal

1 answer:

quester [9]2 years ago

6 0

Answer:

Scott will need to run 13 miles

Step-by-step explanation:

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For number 6, evaluate the definite integral.

maks197457 [2]

$\bf \displaystyle \int\limits_{0}^{28}\ \cfrac{1}{\sqrt[3]{(8+2x)^2}}\cdot dx\impliedby \textit{now, let's do some substitution}\\\\
-------------------------------\\\\
u=8+2x\implies \cfrac{du}{dx}=2\implies \cfrac{du}{2}=dx\\\\
-------------------------------\\\\$

$\bf \displaystyle \int\limits_{0}^{28}\ \cfrac{1}{\sqrt[3]{u^2}}\cdot \cfrac{du}{2}\implies \cfrac{1}{2}\int\limits_{0}^{28}\ u^{-\frac{2}{3}}\cdot du\impliedby 
\begin{array}{llll}
\textit{now let's change the bounds}\\
\textit{by using } u(x)
\end{array}\\\\
-------------------------------\\\\
u(0)=8+2(0)\implies u(0)=8
\\\\\\
u(28)=8+2(28)\implies u(28)=64$

$\bf \\\\
-------------------------------\\\\
\displaystyle \cfrac{1}{2}\int\limits_{8}^{64}\ u^{-\frac{2}{3}}\cdot du\implies \cfrac{1}{2}\cdot \cfrac{u^{\frac{1}{3}}}{\frac{1}{3}}\implies \left. \cfrac{3\sqrt[3]{u}}{2} \right]_8^{64}
\\\\\\
\left[ \cfrac{3\sqrt[3]{(2^2)^3}}{2} \right]-\left[ \cfrac{3\sqrt[3]{2^3}}{2} \right]\implies \cfrac{12}{2}-\cfrac{6}{2}\implies 6-3\implies 3$

3 0

3 years ago

You weigh six packages and find the weights to be 20, 12, 52, 16, 48, and 44 ounces. If you include a package that weighs 60 oun

kvv77 [185]

Answer:

median

Step-by-step explanation:

The first step is to sort the numbers from largest to smallest (you could go the other way as well).

60 52 48 44 20 16 12

The median is the middle number 44 in this case. The median has 3 numbers on its left and 3 on its right (in this case). 44 has 60 52 48 or its left and 20 16 and 12 on its right.

The mean is the average. We'll call it the new average

The mean is 60 + 52 + 48 + 44 + 20 + 16 + 12 = 252

The new mean is 252 / 7 = 36

=====================

The old median is the average between the middle 2 numbers (before you added 60 you had 6 numbers)

52 + 48 + 44 + 20 + 16 + 12 = the average of 44 + 20 which 64/2 = 32

The old mean was (52 + 48 + 44 + 20 + 16 + 12)/6 = 32.

=======================

The median increased from 44 to 32 = 12

The mean went from 32 to 36 which is 4

Answer: the median increased more

3 0

3 years ago

What is the square root of <br><img src="https://tex.z-dn.net/?f=%20%5Csqrt%7B163%7D%20" id="TexFormula1" title=" \sqrt{163} " a

crimeas [40]

Hi,

$\sqrt{163} = 12.76$

There are no other simple steps to reduce the square root of one hundred and sixty-three.

The actual square root value is ≈12.767145334803704

But we have taken only significant figures and it is as 12.76.

Hope it helps! :)

5 0

3 years ago

I need help plz help

STatiana [176]

The answer would be 78. C

8 0

3 years ago

Read 2 more answers

Given that f(x) = 5x^2-3x+7 and f(g(x))=(5x^4)/9 + (17x^2)/3 + 21. Find all possible values for the sum of the coefficients in t

shtirl [24]

steps

(x) = 5x^2-3x+7

refine

x=5x^2-3x+7

Switch sides

5x^2-3x+7-x=x-x

Subtract x from both sides

5x^2-4x+7=0

Solve with the quadratic formula

8 0

3 years ago