Answer:
x=-2
Step-by-step explanation:
4+8x=-12
-4 -4
8x = -16
/8 /8
x = -2
R=16
J=32
20+40(which is 20×2)=60
20-4=16
16×2=32
Using the pairs (1,32) and (2,64) the rate of speed is (64-32 / 2-1 = 32 feet per second.
To find the rate of decent you would multiply the rate by time:
The equation is y = 32x
B. Replace x with 15 and solve for y:
Y = 32(15)
Y = 480
The rate of speed is 480 feet per second.
a. Length of the fence around the field = perimeter of quarter circle = 892.7 ft.
b. The area of the outfield is about 39,584 sq. ft..
<h3>What is the Perimeter of a Quarter Circle?</h3>
Perimeter of circle = 2πr
Perimeter of a quarter circle = 2r + 1/4(2πr).
a. The length of the fence around the field = perimeter of the quarter circle fence
= 2r + 1/4(2πr).
r = 250 ft
Plug in the value
The length of the fence around the field = 2(250) + 1/4(2 × π × 250)
= 892.7 ft.
b. Size of the outfield = area of the full field (quarter circle) - area of the infield (cicle)
= 1/4(πR²) - πr²
R = radius of the full field = 250 ft
r = radius of the infield = 110/2 = 55 ft
Plug in the values
Size of the outfield = 1/4(π × 250²) - π × 55²
= 49,087 - 9,503
= 39,584 sq. ft.
Learn more about perimeter of quarter circle on:
brainly.com/question/15976233
Answer:
x = 2 sqrt(3) or x = -2 sqrt(3) or x = sqrt(3) or x = -sqrt(3)
Step-by-step explanation:
Solve for x:
-8 + x^2 + (x^2 - 8)^2 = 20
Expand out terms of the left hand side:
x^4 - 15 x^2 + 56 = 20
Subtract 20 from both sides:
x^4 - 15 x^2 + 36 = 0
Substitute y = x^2:
y^2 - 15 y + 36 = 0
The left hand side factors into a product with two terms:
(y - 12) (y - 3) = 0
Split into two equations:
y - 12 = 0 or y - 3 = 0
Add 12 to both sides:
y = 12 or y - 3 = 0
Substitute back for y = x^2:
x^2 = 12 or y - 3 = 0
Take the square root of both sides:
x = 2 sqrt(3) or x = -2 sqrt(3) or y - 3 = 0
Add 3 to both sides:
x = 2 sqrt(3) or x = -2 sqrt(3) or y = 3
Substitute back for y = x^2:
x = 2 sqrt(3) or x = -2 sqrt(3) or x^2 = 3
Take the square root of both sides:
Answer: x = 2 sqrt(3) or x = -2 sqrt(3) or x = sqrt(3) or x = -sqrt(3)
