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3 years ago

What’s the area of the shape

Mathematics

2 answers:

Ugo [173]3 years ago

5 0

Answer:

the answer is 37.68

i hope that helps

Yuki888 [10]3 years ago

3 0

The answer is 37.68 because you first find the area of the circle which is pie r^2 or 50.24 then you find the fourth of that which is 12.56 so you subtract 12.56 from 50.24 to get the answer of 37.68

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Hii can someone help with this pls :)

BlackZzzverrR [31]

Answer:

$R(x,y) = (0.8,-1.2)$

Step-by-step explanation:

The coordinates of QS are:

$Q(x_1,y_1) = (-2,-6)$

$S(x_2,y_2) = (5,6)$

The ratio is represented as:

$R_{m:n} = 2:3$

Required

Find the coordinates of R

The coordinates of R is calculated by:

$R(x,y) = (\frac{mx_2 + nx_1}{m+n},\frac{my_2 + ny_1}{m+n})$

Substitute values for m, n, x's and y's

$R(x,y) = (\frac{2 * 5 + 3 * -2}{2+3},\frac{2*6 + 3*-6}{2+3})$

$R(x,y) = (\frac{4}{5},\frac{-6}{5})$

$R(x,y) = (0.8,-1.2)$

None of the options is correct

6 0

2 years ago

Pls help ASAP!!!!!!

Marrrta [24]

${6x}^{3} - 4x {}^{2} + 4x - 8$

answer is

$6x {}^{2} + 8x + \frac{52}{a = - 2}$

8 0

3 years ago

Explain number 2 & 3 please

VLD [36.1K]

1. 3x - 2y = 0 => y = 3/2x
4x + 2y = 14<=> 4x + 2*(3/2x) = 14<=> 7x = 14<=> x = 2 & y = 3
2.3p + q = 7=> q = 7 - 3p
2p - 2q = -6<=> 2p - 2*(7-3p) = -6<=> 2p - 14 + 6p = -6<=> 8p = -6 + 14 = 8<=> p = 1 & q = 4
3.3x - 2y = 1=> x = (1+2y)/3
8x + 3y = 2<=> 8*(1+2y)/3 + 3y = 2<=> 8*(1+2y)/3 - 2 = -3y<=> 3*(8*(1+2y)/3 - 2) = -3*(3y)<=> 8*(1+2y) - 6 = -9y<=> 8 + 16y - 6 = -9y<=> 2 = -25y<=> y = -2/25 & x = 7/25

7 0

3 years ago

The triangle T has vertices at (-2, 1), (2, 1) and (0,-1). (It might be an idea to

Firdavs [7]

Rewrite the boundary lines y = -1 - x and y = x - 1 as functions of y :

y = -1 - x ==> x = -1 - y

y = x - 1 ==> x = 1 + y

So if we let x range between these two lines, we need to let y vary between the point where these lines intersect, and the line y = 1.

This means the area is given by the integral,

$\displaystyle\iint_T\mathrm dA=\int_{-1}^1\int_{-1-y}^{1+y}\mathrm dx\,\mathrm dy$

The integral with respect to x is trivial:

$\displaystyle\int_{-1}^1\int_{-1-y}^{1+y}\mathrm dx\,\mathrm dy=\int_{-1}^1x\bigg|_{-1-y}^{1+y}\,\mathrm dy=\int_{-1}^1(1+y)-(-1-y)\,\mathrm dy=2\int_{-1}^1(1+y)\,\mathrm dy$

For the remaining integral, integrate term-by-term to get

$\displaystyle2\int_{-1}^1(1+y)\,\mathrm dy=2\left(y+\frac{y^2}2\right)\bigg|_{-1}^1=2\left(1+\frac12\right)-2\left(-1+\frac12\right)=\boxed{4}$

Alternatively, the triangle can be said to have a base of length 4 (the distance from (-2, 1) to (2, 1)) and a height of length 2 (the distance from the line y = 1 and (0, -1)), so its area is 1/2*4*2 = 4.

6 0

3 years ago

Which number is a factor of the prime factorization of 60? 3, 15, 9, or 6

Anton [14]

Answer:3

Step-by-step explanation:

4 0

3 years ago