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Anna35 [415]
3 years ago
6

Which function grows

Mathematics
1 answer:
mojhsa [17]3 years ago
4 0
Y = 8x + 29 is the answer for your question
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What the answer is and how?
sladkih [1.3K]

Answer:

PerimeterFigure= pi*DIameter/2= 1.5*pi ft.

Step-by-step explanation:

The perimeter is the length of the figure, since we are dealing with an semi-circle, we can use the perimeter equation for a full circle and divide it by half.

Perimeter= 2pi*R= pi*Diameter.

Since Our figure is a semi circle, as we say earlier, we now will divide the above result by half.

PerimeterFigure= pi*DIameter/2= 1.5*pi ft.

3 0
2 years ago
I don’t know how to do this. can someone please help me :)
Kaylis [27]
The surface area of a cylinder is the sum of the area of the two ends and the lateral area.
.. surface area = 2*(end area) +(lateral area)
.. = 2*(π*r^2) +(2π*r*h)
.. = 2π*r*(r +h)
.. = 2*3.14*(8 in)*(8 in +8 in)
.. = 3.14*256 in^2
.. ≈ 803.8 in^2

The 3rd selection is appropriate.
8 0
3 years ago
There are ___ questions on the ACT math test,<br> А. 75<br> В. 100<br> С. 80<br> D. 60
koban [17]

The answer is: D) 60

5 0
3 years ago
Explain how you could calculate the surface area of a square pyramid.
Damm [24]
Find the area of the base of the pyramid, then find the area of each side then add the areas. hope this help :)
6 0
3 years ago
Read 2 more answers
Solve this query plzzz​
Olin [163]

Step-by-step explanation:

\frac{\sin \:4A}{\cos \:2A}  \times \frac{1 - \cos \:2A}{1 - \cos\:4A}   = \tan \:A \\  \\ LHS =  \frac{\sin \:4A}{\cos \:2A}  \times \frac{1 - \cos \:2A}{1 - \cos\:4A}   \\  \\  =  \frac{2\sin \:2A.\cos \:2A}{\cos \:2A}   \times  \frac{1 - (2 { \cos}^{2}A - 1) }{1 - (2 { \cos}^{2}2A - 1) } \\  \\  = 2\sin \:2A   \times  \frac{1 - 2 { \cos}^{2}A  +  1}{1 - 2 { \cos}^{2}2A  + 1 } \\  \\  = 2\sin \:2A   \times  \frac{2- 2 { \cos}^{2}A  }{2 - 2 { \cos}^{2}2A   } \\  \\   = 2\sin \:2A   \times  \frac{2(1 - { \cos}^{2}A)  }{2 (1-  { \cos}^{2}2A)   } \\  \\   = 2\sin \:2A   \times  \frac{1 - { \cos}^{2}A}{1-  { \cos}^{2}2A   } \\  \\     = 2\sin \:2A   \times  \frac{ { \sin}^{2}A}{{ \sin}^{2}2A   } \\  \\    = 2  \times  \frac{ { \sin}^{2}A}{{ \sin}2A   } \\  \\   = 2  \times  \frac{ { \sin}^{2}A}{{ 2\sin}A. \cos \:   A } \\  \\   = \frac{ { \sin}A}{ \cos \:   A }  \\  \\  = tan \: A \\  \\  = RHS \\

7 0
3 years ago
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