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3 years ago

If your pace on a treadmill is 40.0 meters per minute, how many seconds will it take for you to walk a distance of 7220 feet?

Mathematics

2 answers:

Amiraneli [1.4K]3 years ago

5 0

Answer:

55 minutes

Step-by-step explanation:

Lelechka [254]3 years ago

4 0

In this question , it is given that

your pace on a treadmill is 40.0 meters per minute.

In means 40 meters are travelled in one minute .

And 1 meter equals to 3.28 feet .

So 3.28*40 feet are travelled in one minute, that is

131.2 feets are travelled in 1 minute.

Therefore, 7220 feet are travelled in

$= \frac{7220}{131.2} minutes = 55 minutes$

You might be interested in

Find the 70th term of the arithmetic sequence -6,5,16, ...

aliina [53]

Answer:

70th term is 753

Step-by-step explanation:

a70= ?

n= 70 (nth term, term to find)

a1= -6 (first term)

d= 11 (common difference)

a70= a1 + (n - 1)d

a70= -6 + 69(11)

a70= -6 + 759

a70= 753

8 0

3 years ago

Read 2 more answers

27 divided by 53 minus 6 plus 53 times -2 squared

mamaluj [8]

Answer:

$\left(\left(\frac{27}{53}\right)-6\right)+\left(53\left(-\left(2^2\right)\right)\right)=-\frac{11527}{53}\quad \left(\mathrm{Decimal:\quad }\:-217.49056\dots \right)$

Step-by-step explanation:

Considering the expression

27 divided by 53 minus 6 plus 53 times -2 squared

Which can be written as

$\left(\left(\frac{27}{53}\right)-6\right)+\left(53\cdot \left(-\left(2^2\right)\right)\right)$

So, solving the expression

$\left(\left(\frac{27}{53}\right)-6\right)+\left(53\cdot \left(-\left(2^2\right)\right)\right)$

$\mathrm{Remove\:parentheses}:\quad \left(a\right)=a$

$=\frac{27}{53}-6-53\cdot \:2^2$

$\mathrm{Convert\:element\:to\:fraction}:\quad \:6=\frac{6\cdot \:53}{53}$

$=-\frac{6\cdot \:53}{53}+\frac{27}{53}$

$\mathrm{Since\:the\:denominators\:are\:equal,\:combine\:the\:fractions}:\quad \frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c}$

$=\frac{-6\cdot \:53+27}{53}$

$=\frac{-291}{53}$ ∵ $-6\cdot \:53+27=-291$

$\mathrm{Apply\:the\:fraction\:rule}:\quad \frac{-a}{b}=-\frac{a}{b}$

$=-2^2\cdot \:53-\frac{291}{53}$

$=-212-\frac{291}{53}$ ∵ $53\cdot \:2^2=212$

$\mathrm{Convert\:element\:to\:fraction}:\quad \:212=\frac{212\cdot \:53}{53}$

$=-\frac{212\cdot \:53}{53}-\frac{291}{53}$

$\mathrm{Since\:the\:denominators\:are\:equal,\:combine\:the\:fractions}:\quad \frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c}$

$=\frac{-212\cdot \:53-291}{53}$

$=\frac{-11527}{53}$ ∵ $-212\cdot \:53-291=-11527$

$\mathrm{Apply\:the\:fraction\:rule}:\quad \frac{-a}{b}=-\frac{a}{b}$

$=-\frac{11527}{53}$

Therefore,

$\left(\left(\frac{27}{53}\right)-6\right)+\left(53\left(-\left(2^2\right)\right)\right)=-\frac{11527}{53}\quad \left(\mathrm{Decimal:\quad }\:-217.49056\dots \right)$

Keywords: algebraic expression

Learn more about solving algebraic expression from brainly.com/question/4687406

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8 0

4 years ago

Find the solutions to the equation 102x 11 = (x 6)2 – 2. Which values are approximate solutions to the equation? Select two answ

otez555 [7]

You can try finding the roots of the given quadratic equation to get to the solution of the equation.

There are two solutions to the given quadratic equation

$x = 0.202, x = 113.798$

<h3>How to find the roots of a quadratic equation?</h3>

Suppose that the given quadratic equation is $ax^2 + bx +c = 0$

Then its roots are given as:

$x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

<h3>How to find the solution to the given equation?</h3>

First we will convert it in the aforesaid standard form.

$102x + 11 = (x-6)^2 - 2\\102x + 11 + 2 = x^2 + 36 - 12x\\0 = x^2 -114x + 23\\x^2 -114x + 23 = 0\\$

Thus, we have

a = 1. b = -114, c = 23

Using the formula for getting the roots of a quadratic equation,

$x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}\\\\x = \dfrac{114 \pm \sqrt{114^2 - 92}}{2} \\\\ x = 0.202 (\text{used "-" sign})\\\\x = 113.798 ( used "+" sign})$