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Korvikt [17]
3 years ago
13

Can you please put the words in s sentence please

Mathematics
1 answer:
yarga [219]3 years ago
3 0
El chico de la foto es mi novio
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A bakery offers a small circular cake with a diameter of 8 inches. It also offers a large circular cake with a diameter of 24 in
Elanso [62]
We must find the area of each cake's top.

Formula for area of a circle:

A =  \pi  r^{2} where r is the radius.

<span>small cake:
</span>Plug in 4 for r because the radius is 4. (They give us the diameter, which is 8, and the radius is half that)

A =  \pi  4^{2}

A = 16π

<span>big cake:
</span>Repeat the process: 
Plug in 12 for r because the radius is 12. (They give us the diameter, which is 24, and the radius is half that)

A = \pi 12^{2}

A = 144π

So our two radii are 144π and 16π. 
The large cake's top is not 3 times the area of the small cake's.

This makes sense because you are squaring the radius, which makes the fact that the larger cake's diameter is triple the smaller cake diameter irrelevant. 

Hope this helped! ^-^
7 0
3 years ago
Read 2 more answers
Please answer this question now
fenix001 [56]

Answer:

44 degrees

Step-by-step explanation:

Since MN is a tangent, it forms a right angle wheng it intersects line MP since MP is the diameter. So, 90+46 is 136 and since there are 180 degrees in a traingle, 180-136 is 44.

Hope this is helpful! :)

6 0
3 years ago
Read 2 more answers
Solve the equations to find the number and type of solutions. The equation 8 − 4x = 0 has real solution(s).
frez [133]

Answer:

The number of solution is, 1 and the type of solution is, Integer solution.

Step-by-step explanation:

Given the equation:

8-4x=0

Add 4x to both sides we have;

8 = 4x

Divide both sides by 4 we have;

2 = x

or

x = 2

Solution for this equation is, 2

Therefore, The number of solution is, 1 and the type of solution is, Integer solution.

5 0
3 years ago
Read 2 more answers
Help with cal 2 exercise involving integrals?
zhannawk [14.2K]

\displaystyle\int\frac{\sqrt{x^2+1}}x\,\mathrm dx=\int\frac{\sqrt{\tan^2B+1}}{\tan B}\sec^2B\,\mathrm dB

\tan^2B+1=\sec^2B

\sqrt{\sec^2B}=\sec B (provided that \sec B>0)

Then

\dfrac{\sec B}{\tan B}\cdot\sec^2B=\dfrac{\sec^2B}{\tan B}\cdot\sec B

(just moving around a factor of \sec B)

\dfrac{\sec B}{\tan B}\cdot\sec^2B=\dfrac{\sec^2B}{\tan^2B}\cdot\sec B\tan B

(multiply by \dfrac{\tan B}{\tan B})

8 0
3 years ago
Given g(x)=-x-2, Find g(1).
lawyer [7]
G(x)=-1-2

g(x)=-3

i’m pretty sure that’s the answer
4 0
3 years ago
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