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kvv77 [185]
3 years ago
6

Complete this correctly and I will give you brainliest

Mathematics
1 answer:
Arada [10]3 years ago
8 0
Part a: c. 96 square feet

part b: 60 inches

part c: 36 square feet


explanation: I multiplied 4 and 5 because area = l•w meaning the garden was originally 20 square feet. I then doubled the making it 8 and added two feet to the length making it 7 feet. Since area equals l•w i multiplied 8 and 7 getting 56 square feet. Since the question asked, “how many square feet larger,” I subtracted the new area, 56 square feet, by the old one, 20 square feet, and got 36 square feet.
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Pedro solves the quadratic equation x^2+6x –14 using the quadratic formula. In which step did Pedro make an error?
Nutka1998 [239]

9514 1404 393

Answer:

  correct choice is marked

Step-by-step explanation:

The error is in step 2. The square root cannot be distributed to a sum.

3 0
3 years ago
Its math help me out?
Hoochie [10]

Its the first one

--------------------

7 0
2 years ago
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Do you guys know how to do this I am struggling and need help?
Oksanka [162]

Answer:

i think its 234

Step-by-step explanation:

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3 years ago
PLEASE HELP MEEEEEEE PLSSS
Novosadov [1.4K]

Answer:

use the desmo calculator

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3 years ago
Let R be the region bounded by the following curves. Use the shell method to find the volume of the solid generated when R is re
Tomtit [17]

Answer:

<u>Volume = 1.535</u>

<u />

Step-by-step explanation:

The region R is bounded by the equations:

y = √sin⁻¹x

y = √(π/2)

y = √(π/3)

x = 0

R is revolved around the x-axis so we will need f(y) for finding out the volume. We need to make x the subject of the equation and then replace it with f(y).

f(x) = √sin⁻¹x

y = √sin⁻¹x

Squaring both sides we get:

y² = sin⁻¹x

x = sin (y²)

f(y) = sin (y²)

Using the Shell Method to find the volume of the solid when R is revolved around the x-axis:

V = 2\pi \int\limits^a_b {f(y)} \, dy

The limits a and b are the equations y = √(π/2) and y = √(π/3) which bound the region R. So, a = √(π/2) and b = √(π/3).

V = 2π \int\limits^\sqrt{\frac{\pi }{2}}_\sqrt{\frac{\pi }{3} }   sin (y²) dy

Integrating sin (y²) dy, we get:

-cos(y²)/2y

So,

V = 2π [-cos(y²)/2y] with limits √(π/2) and √(π/3)

V = 2π [(-cos(√(π/2) ²)/2*√(π/2)] - [(-cos(√(π/3) ²)/2*√(π/3)]

V = 2π [(-cos(π/2)/ 2√(π/2)) - ((-cos(π/3)/ 2√(π/3))]

V = 2π [ 0 - (-0.5/2.0466)]

V = 2π (0.2443)

V = 1.53499 ≅ 1.535

3 0
3 years ago
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