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BigorU [14]
3 years ago
9

4y + 3q - 5q^3 PLS ANSWER ASAP

Mathematics
1 answer:
Ilya [14]3 years ago
6 0

Answer:

4y+q(3-5q^2)

Step-by-step explanation:

common taking from 3q-5q^3

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Inga [223]

Answer:

A = 3

B = -5

C = 2

Step-by-step explanation:

The way to format a quadratic equation is: ax^2 + bx + c, so the first step to solving this is to format it in the right way, where the x^2 comes first, the x second, and the number alone.

After formatting, your equation should look like this: 3x^2 - 5x + 2

From here, you can see the the <em>a </em> is 3, the <em>b</em> is -5, and the <em>c </em>is 2

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3 years ago
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SVETLANKA909090 [29]

Answer:

r  = 6

Step-by-step explanation:

To get the value of r we will use the pythagoras theorem a shown;

PR² = QR² + PQ²

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Expand

16+8r+r² = 64 + r²

16 + 8r = 64

8r = 64 - 16

8r = 48

r = 48/8

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Hence the value of r is 6

7 0
3 years ago
A rope of length 18 feet is arranged in the shape of a sector of a circle with central angle O radians, as shown in the
creativ13 [48]

Answer:

A(\theta)=\frac{162 \theta}{(\theta+2)^2}

Step-by-step explanation:

The picture of the question in the attached figure

step 1

Let

r ---> the radius of the sector

s ---> the arc length of sector

Find the radius r

we know that

2r+s=18

s=r \theta

2r+r \theta=18

solve for r

r=\frac{18}{2+\theta}

step 2

Find the value of s

s=r \theta

substitute the value of r

s=\frac{18}{2+\theta}\theta

step 3

we know that

The area of complete circle is equal to

A=\pi r^{2}

The complete circle subtends a central angle of 2π radians

so

using proportion find the area of the sector by a central angle of angle theta

Let

A ---> the area of sector with central angle theta

\frac{\pi r^{2} }{2\pi}=\frac{A}{\theta} \\\\A=\frac{r^2\theta}{2}

substitute the value of r

A=\frac{(\frac{18}{2+\theta})^2\theta}{2}

A=\frac{162 \theta}{(\theta+2)^2}

Convert to function notation

A(\theta)=\frac{162 \theta}{(\theta+2)^2}

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Rachel used 9 cups of flour and 4 teaspoons of sugar.
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Answer:

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Step-by-step explanation:

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