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sweet-ann [11.9K]
2 years ago
15

Simplify the fraction below​

Mathematics
1 answer:
jasenka [17]2 years ago
7 0

Answer:

5/28

Step-by-step explanation:

First you multiply across,

<u>5 · 2 = 10</u>

8 · 7 = 56

Then after multiplying, you will get the fraction, 10/56.

You can simplify that into 5/28.

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Help ............. ​
NARA [144]

What kinda of math is this so i can answer..

:)

5 0
3 years ago
10 – 2v = -5v – 50 what is V
Virty [35]
I think it’s v=-20 i showed you my work if that helps :)

4 0
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
Any portion of the circumference of a circle is called a(n)
Alex777 [14]
Any portion of the circumference of a circle is called a radius
5 0
3 years ago
Given z1 = -3
trasher [3.6K]

Answer:

See explanation

Step-by-step explanation:

The given complex number are:

z_1=-3\sqrt{3}+3i

and

z_2=6\cos 150\degree+6i\sin 150\degree

When we rewrite z_1=-3\sqrt{3}+3i in complex form, we obtain;

z_1=r(\cos \theta+i\sin \theta)

where

r=\sqrt{(-3\sqrt{3})^2+3^2 }=\sqrt{36}=6

and

\theta=tan^{-1}(\frac{y}{x})

\implies \theta=tan^{-1}(\frac{-3\sqrt{3}}{3})=150\degree

Hence,

z_1=6(\cos 150\degree+i\sin 150\degree)

z_1=6\cos 150\degree+6i\sin 150\degree

Hence

z_1=z_2

8 0
3 years ago
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