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gtnhenbr [62]
3 years ago
11

An Olympic diver jumps off of a platform that is 10 meters high. The diver goes 2 meters underwater after a dive. How far is it

from the starting point at the top of the platform to the depth under water?​
Mathematics
1 answer:
grandymaker [24]3 years ago
4 0

Answer:

12 meters far.

Step-by-step explanation:

The question translates to 10+2. The word problem states that the height of the platform is 10 meters. That means we can say that he'd have already traveled 10 meters down when he dived. And the problem also states that the diver goes two meters under the water, meaning that he had traveled 2 meters, in addition to the 10 meters. And so, we add the 10 and 2 to get 12 meters in total.

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Well...  if she has 8 yards, and a yard equals to 18 inches. So.. I would divide 8 and 1/4. After you do the reciprocal, you would get 32 pieces.
7 0
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Which description matches the graph of the inequality y ≥ |x + 2| – 3? a shaded region above a solid boundary line a shaded regi
Alex777 [14]
The given inequality is y ≥ |x + 2| -3.

This inequality may be written two ways:
(a) y ≥ x + 2 - 3
    or
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(b) y ≥ -x -2 - 3
    or
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A graph of the inequality is shown below. The shaded region satisfies the inequality.

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6 0
3 years ago
Given that F(x) = x 2 + 2, evaluate F(1) + F(5) <br> which is the answer? 38 30 15
Valentin [98]

your answer is 30

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6 0
3 years ago
What is the mean of 18,24,17,21,24,16,29,18
timurjin [86]

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7 0
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Read 2 more answers
Solve the triangle A = 2 B = 9 C =8
VARVARA [1.3K]

Answer:

\begin{gathered} A=\text{ 12}\degree \\ B=\text{ 114}\degree \\ C=54\degree \end{gathered}

Step-by-step explanation:

To calculate the angles of the given triangle, we can use the law of cosines:

\begin{gathered} \cos (C)=\frac{a^2+b^2-c^2}{2ab} \\ \cos (A)=\frac{b^2+c^2-a^2}{2bc} \\ \cos (B)=\frac{c^2+a^2-b^2}{2ca} \end{gathered}

Then, given the sides a=2, b=9, and c=8.

\begin{gathered} \cos (A)=\frac{9^2+8^2-2^2}{2\cdot9\cdot8} \\ \cos (A)=\frac{141}{144} \\ A=\cos ^{-1}(\frac{141}{144}) \\ A=11.7 \\ \text{ Rounding to the nearest degree:} \\ A=12º \end{gathered}

For B:

\begin{gathered} \cos (B)=\frac{8^2+2^2-9^2}{2\cdot8\cdot2} \\ \cos (B)=\frac{13}{32} \\ B=\cos ^{-1}(\frac{13}{32}) \\ B=113.9\degree \\ \text{Rounding:} \\ B=114\degree \end{gathered}\begin{gathered} \cos (C)=\frac{2^2+9^2-8^2}{2\cdot2\cdot9} \\ \cos (C)=\frac{21}{36} \\ C=\cos ^{-1}(\frac{21}{36}) \\ C=54.3 \\ \text{Rounding:} \\ C=\text{ 54}\degree \end{gathered}

3 0
1 year ago
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