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3 years ago

each day last week, Ms. wilson walked3/4 mile. what is the total distance, in miles, that Ms. Wilson walked in 4 days?

2 answers:

6 0

Hey there! :D

Since she walks 3/4 a mile a day, we can multiply that by 4.

3/4= .75

4*.75= 3

She walked 3 miles in 4 days.

I hope this helps!
~kaikers

ad-work [718]3 years ago

4 0

The answer is 3 because 3/4 * 4 =3
———————————————————
3/4 * 4/1
3 times 4 equals 12 and 4 times 1 is 4 which means the answer is at this point 12/4 which can be simplified to 3 because 4 goes into 12, 3 times.

Hope this helps:)

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HELP WITH THIS EZ QUESTION

9966 [12]

Answer:

Step-by-step explanation:

$3+75t-16t^{2} =h\\\\$

we evaluate when t=2

$3+75(2)-16(2)^2=h\\\\3+150-64=h\\\\h=89$

so it's 89 feet

7 0

3 years ago

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What is radians converted to degrees? If necessary, round your answer to the nearest degree.

Tomtit [17]

To convert radians to degrees, make use of the fact that pi radians equals one half circle, or 180º.

This means that if you divide radians by pi, the answer is the number of half circles. Multiplying this by 180º will tell you the answer in degrees.

So, to convert radians to degrees, multiply by 180/pi, like this:

Degrees= radians X 180/pi

3 0

2 years ago

Place 3 numbers between 12 and 60 to make a sequence of 5 numbers with a common difference

Vilka [71]

The 3 numbers which should be placed between 12 and 60 are; 24, 36 and 48.

<h3>What three numbers should be placed between 12 and 60?</h3>

It follows from the task content that the first and last numbers are; 12 and 60 respectively.

Hence, since the difference between 12 and 60 is 48 and there are 4 transitions between the two numbers to have 5 total numbers,

It follows that the common difference is; 48/4 = 12 and the three numbers required are; 24,36 and 48.

Read more on common difference;

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5 0

2 years ago

Find a power series for the function, centered at c, and determine the interval of convergence. f(x) = 9 3x + 2 , c = 6

san4es73 [151]

Answer:

$\frac{9}{3x + 2} = 1 - \frac{1}{3}(x - \frac{7}{3}) + \frac{1}{9}(x - \frac{7}{3})^2 - \frac{1}{27}(x - \frac{7}{3})^3 ........$

The interval of convergence is: $(-\frac{2}{3},\frac{16}{3})$

Step-by-step explanation:

Given

$f(x)= \frac{9}{3x+ 2}$

$c = 6$

The geometric series centered at c is of the form:

$\frac{a}{1 - (r - c)} = \sum\limits^{\infty}_{n=0}a(r - c)^n, |r - c| < 1.$

Where:

$a \to$ first term

$r - c \to$ common ratio

We have to write

$f(x)= \frac{9}{3x+ 2}$

In the following form:

$\frac{a}{1 - r}$

So, we have:

$f(x)= \frac{9}{3x+ 2}$

Rewrite as:

$f(x) = \frac{9}{3x - 18 + 18 +2}$

$f(x) = \frac{9}{3x - 18 + 20}$

Factorize

$f(x) = \frac{1}{\frac{1}{9}(3x + 2)}$

Open bracket

$f(x) = \frac{1}{\frac{1}{3}x + \frac{2}{9}}$

Rewrite as:

$f(x) = \frac{1}{1- 1 + \frac{1}{3}x + \frac{2}{9}}$

Collect like terms

$f(x) = \frac{1}{1 + \frac{1}{3}x + \frac{2}{9}- 1}$

Take LCM

$f(x) = \frac{1}{1 + \frac{1}{3}x + \frac{2-9}{9}}$

$f(x) = \frac{1}{1 + \frac{1}{3}x - \frac{7}{9}}$

So, we have:

$f(x) = \frac{1}{1 -(- \frac{1}{3}x + \frac{7}{9})}$

By comparison with: $\frac{a}{1 - r}$

$a = 1$

$r = -\frac{1}{3}x + \frac{7}{9}$

$r = -\frac{1}{3}(x - \frac{7}{3})$

At c = 6, we have:

$r = -\frac{1}{3}(x - \frac{7}{3}+6-6)$

Take LCM

$r = -\frac{1}{3}(x + \frac{-7+18}{3}+6-6)$

r = -\frac{1}{3}(x + \frac{11}{3}+6-6)

So, the power series becomes:

$\frac{9}{3x + 2} = \sum\limits^{\infty}_{n=0}ar^n$

Substitute 1 for a

$\frac{9}{3x + 2} = \sum\limits^{\infty}_{n=0}1*r^n$

$\frac{9}{3x + 2} = \sum\limits^{\infty}_{n=0}r^n$

Substitute the expression for r

$\frac{9}{3x + 2} = \sum\limits^{\infty}_{n=0}(-\frac{1}{3}(x - \frac{7}{3}))^n$

Expand

$\frac{9}{3x + 2} = \sum\limits^{\infty}_{n=0}[(-\frac{1}{3})^n* (x - \frac{7}{3})^n]$

Further expand:

$\frac{9}{3x + 2} = 1 - \frac{1}{3}(x - \frac{7}{3}) + \frac{1}{9}(x - \frac{7}{3})^2 - \frac{1}{27}(x - \frac{7}{3})^3 ................$

The power series converges when:

$\frac{1}{3}|x - \frac{7}{3}| < 1$

Multiply both sides by 3

$|x - \frac{7}{3}|$

Expand the absolute inequality

$-3 < x - \frac{7}{3}$

Solve for x

$\frac{7}{3} -3 < x$

Take LCM

$\frac{7-9}{3} < x$

$-\frac{2}{3} < x$

The interval of convergence is: $(-\frac{2}{3},\frac{16}{3})$

6 0

2 years ago

Let the shortest side of a triangle be x. The longest side of a triangle is six

siniylev [52]

I think you guys are using a^2 + b^2 = c^2

the equation would be

x^2 + 2x^2 = 6^2

3 0

3 years ago

Read 2 more answers