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Kruka [31]
3 years ago
13

Find a which length of arc WW on c

Mathematics
1 answer:
adoni [48]3 years ago
3 0
€ = 7/9
7/9÷2 = .389
C = 2pi•r = 2•pi•18 = 36pi
WV = 7/9 (36pi) = .389 (36)pi = 14pi
So C) 14pi is the answer
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11/98 x 2/99 = ???????
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11•2=22
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4851 can be divided by 11. So we can divide the top and bottom by 11.
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3 years ago
Sara is working on a Geometry problem in her Algebra class. The problem requires Sara to use the two quadrilaterals below to ans
zloy xaker [14]
Part A:

Given a square with sides 6 and x + 4. Also, given a rectangle with sides 2 and 3x + 4

The perimeter of the square is given by 4(x + 4) = 4x + 16

The area of the rectangle is given by 2(2) + 2(3x + 4) = 4 + 6x + 8 = 6x + 12

For the perimeters to be the same

4x + 16 = 6x + 12
4x - 6x = 12 - 16
-2x = -4
x = -4 / -2 = 2

The value of x that makes the <span>perimeters of the quadrilaterals the same is 2.



Part B:

The area of the square is given by

Area=(x+4)^2=x^2+8x+16

The area of the rectangle is given by 2(3x + 4) = 6x + 8

For the areas to be the same

x^2+8x+16=6x+8 \\  \\ \Rightarrow x^2+8x-6x+16-8=0 \\  \\ \Rightarrow x^2+2x+8=0 \\  \\ \Rightarrow x= \frac{-2\pm\sqrt{2^2-4(8)}}{2}  \\  \\ = \frac{-2\pm\sqrt{4-32}}{2} = \frac{-2\pm\sqrt{-28}}{2}  \\  \\ = \frac{-2\pm2i\sqrt{7}}{2} =-1\pm i\sqrt{7}

Thus, there is no real value of x for which the area of the quadrilaterals will be the same.
</span>
7 0
3 years ago
How do I solve 5+a&gt;5
Ksenya-84 [330]

Answer:

5a>5 is =

Step-by-step explanation:

they are equal to each other

7 0
3 years ago
Math question plz help
Svetach [21]

Answer:

Yes they are.

Because:

As said in the rules you have to keep the bases and add the exponents.

7 0
3 years ago
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Line FG goes through the points (4,9) and (1,3). Which equation represents a line that is perpendicular to FG and passes through
stiks02 [169]

Answer:

x + 2y= 2

Step-by-step explanation:

Given

Points:

F = (4,9)

G = (1,3)

Required

Determine the equation of line that is perpendicular to the given points and that pass through (2,0)

First, we need to determine the slope, m of FG

m = \frac{y_2 - y_1}{x_2 - x_1}

Where

F = (4,9) --- (x_1,y_1)

G = (1,3) --- (x_2,y_2)

m = \frac{3 - 9}{1 - 4}

m = \frac{- 6}{- 3}

m =2

The question says the line is perpendicular to FG.

Next, we determine the slope (m2) of the perpendicular line using:

m_2 = -\frac{1}{m}

m_2 = -\frac{1}{2}

The equation of the line is then calculated as:

y - y_1 = m_2(x - x_1)

Where

m_2 = -\frac{1}{2}

(x_1,y_1) = (2,0)

y - 0 = -\frac{1}{2}(x - 2)

y  = -\frac{1}{2}(x - 2)

y  = -\frac{1}{2}x + 1

Multiply through by 2

2y = -x + 2

Add x to both sides

x + 2y= -x +x+ 2

x + 2y= 2

Hence, the line of the equation is x + 2y= 2

8 0
3 years ago
Read 2 more answers
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