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Butoxors [25]
3 years ago
10

How do you calculate viewing area

Mathematics
1 answer:
Sauron [17]3 years ago
8 0

Answer:

The way to calculate viewing area would be base multiplied by height, but if you would be more specific I would provide a more in detail answer.

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Show that f(x)=πx-2 and f-1(x)=(x+2/π) are inverse functions of one another
iVinArrow [24]
They are not inverse functions of one another. For f(x) and g(x) to be inverses of each other, f(g(x)) = x. In this case, pi(x+2/pi)-2 = pi* x, which is not x.
5 0
4 years ago
Use elimination to solve the system of equations. <br> x+y=−3 <br> 2x−3y=29
attashe74 [19]

Answer:

(4,-7)

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
X=y-4 and 2x+4y=10 help me ​
Murljashka [212]

Answer:

Step-by-step explanation:

1. x = y-4

0 = y - 4 -x

4 = y - x

2. 2x+4y = 10

x+4y = 5

x+y = 5/4

6 0
3 years ago
Solve for x. Assume that lines which appear to be diameters are actually diameters​
Dafna1 [17]

Answer: number 7. 7

Step-by-step explanation:

Simplifying

7X + 1 + -5X = 15

Reorder the terms:

1 + 7X + -5X = 15

Combine like terms: 7X + -5X = 2X

1 + 2X = 15

Solving

1 + 2X = 15

Solving for variable 'X'.

Move all terms containing X to the left, all other terms to the right.

Add '-1' to each side of the equation.

1 + -1 + 2X = 15 + -1

Combine like terms: 1 + -1 = 0

0 + 2X = 15 + -1

2X = 15 + -1

Combine like terms: 15 + -1 = 14

2X = 14

Divide each side by '2'.

X = 7

Simplifying

X = 7

4 0
3 years ago
Find an equation of the sphere with center (2, −10, 3) and radius 5. $$25=(x−2)2+(y−(−10))2+(z−3)2 use an equation to describe i
lana66690 [7]

In the x,y plane, we have z=0 everywhere. So in the equation of the sphere, we have

25=(x-2)^2+(y+10)^2+(-3)^2\implies(x-2)^2+(y+10)^2=16=4^2

which is a circle centered at (2, -10, 0) of radius 4.

In the x,z plane, we have y=0, which gives

25=(x-2)^2+10^2+(z-3)^2\implies(x-2)^2+(z-3)^2=-75

But any squared real quantity is positive, so there is no intersection between the sphere and this plane.

In the y,z plane, x=0, so

25=(-2)^2+(y+10)^2+(z-3)^2\implies(y+10)^2+(z-3)^2=21=(\sqrt{21})^2

which is a circle centered at (0, -10, 3) of radius \sqrt{21}.

3 0
3 years ago
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