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mina [271]
3 years ago
15

What is the length of PR?

Mathematics
1 answer:
GuDViN [60]3 years ago
5 0

Answer:

you have to provide a picture or something as well

Step-by-step explanation:

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Mutiply 4y-4w^8-3y^5w^3
sukhopar [10]

Answer:

48y⁶w¹¹

Step-by-step explanation:

[-3y⁵][4y] → -12y⁶

×

[w³][-4w⁸] → -4w¹¹

______

48y⁶w¹¹

I am joyous to assist you anytime.

8 0
3 years ago
The interior angles formed by the sides of a quadrilateral have measures that sum to 360°.
vova2212 [387]

add all the angles and make them equal to 360

88 + 108 + (3x-6) + 2x = 360

196 + 5x -6 = 360

190 + 5x = 360

5x = 170

x = 34 = answer

4 0
3 years ago
Can someone please help me with this?
jonny [76]
I need to know if you where are you gonna is the day you
4 0
3 years ago
If x=4cos(2t) and y=4sin(2t), what is the parameter of these equations
kobusy [5.1K]

Answer:

y=√(4+x)(4−x)

Step-by-step explanation:

7 0
3 years ago
The area of the shaded segment is 100cm^2. Calculate the value of r.
Reil [10]
Hello, 

The formula for finding the area of a circular region is: A=  \frac{ \alpha *r^{2} }{2}

then:
A_{1} = \frac{80*r^{2} }{2}

With the two radius it is formed an isosceles triangle, so, we must obtain its area, but first we obtain the height and the base.

cos(40)= \frac{h}{r}  \\  \\ h= r*cos(40)\\ \\ \\ sen(40)= \frac{b}{r} \\ \\ b=r*sen(40)

Now we can find its area:
A_{2}=2* \frac{b*h}{2}  \\  \\ A_{2}= [r*sen(40)][r*cos(40)]\\  \\A_{2}= r^{2}*sen(40)*cos(40)

The subtraction of the two areas is 100cm^2, then:

A_{1}-A_{2}=100cm^{2} \\ (40*r^{2})-(r^{2}*sen(40)*cos(40) )=100cm^{2} \\ 39.51r^{2}=100cm^{2} \\ r^{2}=2.53cm^{2} \\ r=1.59cm

Answer: r= 1.59cm
7 0
3 years ago
Read 2 more answers
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