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3 years ago

Which number completes the Pythagorean triple: 12,16____? 18 20 22 24

Mathematics

2 answers:

solmaris [256]3 years ago

8 0

Twenty...............

snow_lady [41]3 years ago

6 0

The first thing that would probably come to mind when solving this problem would be to plug in the numbers into the equations. Like this 12^2 + 16^2 = x^2 (X = to the number that you plug in from the answer bank.). This is correct thinking so you can try this:

144 + 256 = 400.

So your square has to be equal to 400.

And it most common knowledge that the square root of 400 is equal to 20.

So your answer is 20.

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Answer:

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3 years ago

To the nearest millimeter, a cell phone is 86 mm long and 40 mm wide. What is the ratio of the width to the length?

MAXImum [283]

Answer:

20 : 43

Step-by-step explanation:

width : length

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2 years ago

A donut store has 11 different types of donuts. You can only buy a bag of 3 of them, where each donut has to be of a different t

MakcuM [25]

Answer:

$165$ .

Step-by-step explanation:

Since repetition isn't allowed, there would be $11$ choices for the first donut, $(11 - 1) = 10$ choices for the second donut, and $(11 - 2) = 9$ choices for the third donut. If the order in which donuts are placed in the bag matters, there would be $11 \times 10 \times 9$ unique ways to choose a bag of these donuts.

In practice, donuts in the bag are mixed, and the ordering of donuts doesn't matter. The same way of counting would then count every possible mix of three donuts type $3 \times 2 \times 1 = 6$ times.

For example, if a bag includes donut of type $x$ , $y$ , and $z$ , the count $11 \times 10 \times 9$ would include the following $3 \times 2 \times 1$ arrangements:

$xyz$ .
$xzy$ .
$yxz$ .
$yzx$ .
$zxy$ .
$zyx$ .

Thus, when the order of donuts in the bag doesn't matter, it would be necessary to divide the count $11 \times 10 \times 9$ by $3 \times 2 \times 1 = 6$ to find the actual number of donut combinations:

$\begin{aligned} \frac{11 \times 10 \times 9}{3 \times 2 \times 1} = 165\end{aligned}$ .

Using combinatorics notations, the answer to this question is the same as the number of ways to choose an unordered set of $3$ objects from a set of $11$ distinct objects:

$\begin{aligned}\begin{pmatrix}11 \\ 3\end{pmatrix} &= \frac{11 !}{(11 - 3)! \times 3 !} \\ &= \frac{11 !}{8 ! \times 3 !} \\ &= \frac{11 \times 10 \times 9}{3 \times 2 \times 1} = 165\end{aligned}$ .