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Nadya [2.5K]
2 years ago
6

X 12. Find the volume of the composite solid. Use 3.14 for . 5.5 ft Find the volume of the cone using the formula. ​

Mathematics
1 answer:
svetlana [45]2 years ago
6 0

Answer: 74.61 ft^3

Step-by-step explanation:

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Mhanifa please help! Solve for the unknown in each triangle. Round to the nearest tenth.
Scilla [17]

Answer:

(D)

<u>Use the law of sines:</u>

  • 13 / sin x = 12 / sin 67°
  • sin x = 13 sin 67° / 12
  • sin x = 0.997
  • x = arcsin (0.997)
  • x = 85.6°

(E)

<u>Find the third angle:</u>

  • 180° - (48° + 61°) = 71°

<u>Use the las of sines:</u>

  • x / sin 61° = 21 / sin 71°
  • x = 21 sin 61° / sin 71°
  • x = 19.4 cm
8 0
2 years ago
First try was incorrect
gizmo_the_mogwai [7]

Answer:

48%

Step-by-step explanation:

P (not getting a speckled rock) = 12/18 = 2/3

P (not getting a green rock) = 13/18

2/3 * 13/18 = 26/54 = 13/27 ≈ 48%

6 0
2 years ago
Use a graph in a (-2π, 2π, π/2) by (-3, 3, 1) viewing rectangle to complete the identity.
yaroslaw [1]

First, notice that:

2\tan (\frac{x}{2})=2\cdot(\pm\sqrt[]{\frac{1-cosx}{1+\cos x})}

And in the denominator we have:

1+\tan ^2(\frac{x}{2})=1+\frac{1-\cos x}{1+\cos x}=\frac{1+cosx+1-\cos x}{1+cosx}=\frac{2}{1+\cos x}

then, we have on the original expression:

\begin{gathered} \frac{2\tan(\frac{x}{2})}{1+\tan^2(\frac{x}{2})}=\frac{2\cdot\pm\sqrt[]{\frac{1-\cos x}{1+cosx}}}{\frac{2}{1+\cos x}}=\frac{2\cdot(\pm\sqrt[]{1-cosx})\cdot(1+\cos x)}{2\cdot(\sqrt[]{1+cosx})} \\ =(\sqrt[]{1-\cos x})\cdot(\sqrt[]{1+\cos x})=\sqrt[]{(1-\cos x)(1+\cos x)} \\ =\sqrt[]{1-\cos^2x}=\sqrt[]{\sin^2x}=\sin x \end{gathered}

therefore, the identity equals to sinx

8 0
1 year ago
Question 1 (1 point)
ratelena [41]
I have no idea I’m sorry
4 0
3 years ago
Which coefficient matrix represents a system of linear equations that has a unique solution ?
finlep [7]

Answer:

Option C

Step-by-step explanation:

We are given a coefficient matrix along and not the solution matrix

Since solution matrix is not given we cannot check for infinity solutions.

But we can check whether coefficient matrix is 0 or not

If coefficient matrix is zero, the system is inconsistent and hence no solution.

Option A)

|A|=\left[\begin{array}{ccc}4&2&6\\2&1&3\\-2&3&-4\end{array}\right] =0

since II row is a multiple of I row

Hence no solution or infinite

OPtion B

|B|=\left[\begin{array}{ccc}2&0&-2\\-7&1&5\\4&-2&0\end{array}\right] \\=2(10)-2(10)=0

Hence no solution or infinite

Option C

\left[\begin{array}{ccc}6&0&-2\\-2&0&6\\1&-2&0\end{array}\right] \\=2(36-2)=68

Hence there will be a unique solution

Option D

\left[\begin{array}{ccc}5&10&5\\4&1&4\\-1&-2&-1\end{array}\right] \\=2(10)-2(10)=0=0

(since I row is -5 times III row)

Hence there will be no or infinite solution

Option C is the correct answer



4 0
3 years ago
Read 2 more answers
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