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KIM [24]
2 years ago
10

martha and debbie each have 17 ribbons. they buy 1 package with 8 ribbons in it. how many ribbons do they have now?

Mathematics
2 answers:
goblinko [34]2 years ago
8 0
They have 25 ribbons in all
Mice21 [21]2 years ago
4 0
They have...
If they EACH have 17 and there are 2 of them 17 * 2 = 34 if the package has 8 ribbons then you would add 34 and 8 which equals 42

they have 42 ribbons
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2 years ago
Identify the diameter of the circular base created by folding the figure into a right cone. HELP ASAP PLEASE!!
Akimi4 [234]

let's notice something, we have a circle with a radius of 12 and one 90° sector is cut off, so only three 90° sectors of the circle are left shaded, so namely the cone will be using 3/4 of that circle.

think of it as, this shaded area is some piece of paper, and you need to pull it upwards and have the cutoff edges meet, and when that happens, you'll end up with a cone-shaped paper cup, and pour in some punch.

now, once we have pulled up the center of the circle to make our paper cup, there will be a circular base, its diameter not going to be 24, it'll be less, but whatever that base is, we know that is going to have the same circumference as those in the shaded area.  Well, what is the circumference of that shaded area?

\bf \textit{circumference of a circle}\\\\ C=2\pi r~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ r=12 \end{cases}\implies C=2\pi 12\implies C=24\pi \implies \stackrel{\textit{three quarters of it}}{24\pi \cdot \cfrac{3}{4}} \\\\\\ 6\pi \cdot 3\implies 18\pi

well then, the circumference of that circle at the bottom will be 18π, so, what is the diameter of a circle with a circumferenc of 18π?

\bf \textit{circumference of a circle}\\\\ C=2\pi r~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ C=18\pi \end{cases}\implies 18\pi =2\pi r\implies \cfrac{18\pi }{2\pi }=r\implies 9=r \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill \stackrel{\textit{diameter is twice the radius}}{d=18}~\hfill

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8 0
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What is the partial fraction decomposition of StartFraction 8 x + 19 Over (x + 8) (x minus 1) EndFraction? StartFraction 3 Over
hoa [83]

The partial fraction decomposition is \frac{8x + 19}{(x + 8)(x - 1)} = \frac{5}{x + 8} + \frac{3}{x - 1}

<h3>How to determine the decomposition?</h3>

The fraction is given as:

\frac{8x + 19}{(x + 8)(x - 1)}

Split the fraction as follows:

\frac{8x + 19}{(x + 8)(x - 1)} = \frac{A}{x + 8} + \frac{B}{x - 1}

Take the LCM

\frac{8x + 19}{(x + 8)(x - 1)} = \frac{Ax -A + Bx + 8B}{(x + 8)(x -1)}

Cancel the common factors

8x + 19 = Ax - A + Bx + 8B

By comparison, we have:

Ax + Bx = 8x

-A + 8B = 19

This gives

A + B = 8

-A + 8B = 19

Add both equations

9B = 27

Divide by 9

B = 3

Substitute B = 3 in A + B = 8

A + 3 = 8

Solve for A

A = 5

So, we have:

\frac{8x + 19}{(x + 8)(x - 1)} = \frac{5}{x + 8} + \frac{3}{x - 1}

Hence, the partial fraction decomposition is \frac{8x + 19}{(x + 8)(x - 1)} = \frac{5}{x + 8} + \frac{3}{x - 1}

Read more about partial fraction at:

brainly.com/question/12783868

#SPJ1

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2 years ago
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