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3 years ago

What is the length of the hypotenuse

Mathematics

2 answers:

Afina-wow [57]3 years ago

8 0

Answer:

Its C

If u didn't feel like reading.

Step-by-step explanation:

nika2105 [10]3 years ago

5 0

<u>Given</u>:

Given that the length of the two legs of the right triangle are 28 m and 45 m.

We need to determine the length of the hypotenuse.

<u>Length of the hypotenuse:</u>

The length of the hypotenuse can be determined using the Pythagorean theorem.

Thus, we have;

$hyp^2=28^2+45^2$

Simplifying, we get;

$hyp^2=784+2025$

$hyp^2=2809$

Taking square root on both sides, we get;

$hyp=\sqrt{2809}$

$hyp=53 \ m$

Thus, the length of the hypotenuse is 53 m

Hence, Option C is the correct answer.

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Chords AC and DB intersect at point E. Find the value of x.

Elina [12.6K]

Answer:

the answer would be 9 so option B

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3 years ago

A basketball player that shoots 80% from the free throw line attempts two free throws. The notation for conditional probability

taurus [48]

Answer:

P(made 2nd attempt|made 1st attempt)=P(made 2nd attempt)

Step-by-step explanation:

Here given that a basketball player that shoots 80% from the free throw line attempts two free throws.

If x is the no of shoots he makes (say) then we find that each throw is independent of the other.

In other words, because he made successful first attempt, his chances for second attempt will not change

Prob for success in each attempt remains the same as 0.80

Hence I throw is independent of II throw.

When A and B are independent,then we have

P(A/B) = P(A)

Hence answer is

P(made 2nd attempt|made 1st attempt)=P(made 2nd attempt)

8 0

3 years ago

Chandra runs 10 miles. After every two miles, she checks the time. The scatter plot shows her results.

masya89 [10]

Answer: 7.5

Step-by-step explanation:

Per 2 mile she ran 15m

Per one it’s 7.5m

3 0

3 years ago

What is 5.316 - 1.942 btw (show ur work) :)

erastovalidia [21]

Answer:

3.374

Step-by-step explanation:

$\mathrm{Write\:the\:numbers\:one\:under\:the\:other,\:line\:up\:the\:decimal\:points.}$

$\mathrm{Add\:trailing\:zeroes\:so\:the\:numbers\:have\:the\:same\:length.}$

$\begin{matrix}\:\:&5&.&3&1&6\\ -&1&.&9&4&2\end{matrix}$

$\mathrm{Subtract\:each\:column\:of\:digits,\:starting\:from\:the\:right\:and\:working\:left}$

$\mathrm{In\:the\:bolded\:column,\:subtract\:the\:second\:digit\:from\:the\:first}:\quad \:6-2=4\\$

$\frac{\begin{matrix}\:\:&5&.&3&1&\textbf{6}\\ -&1&.&9&4&\textbf{2}\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\:\:&\:\:&\textbf{4}\end{matrix}}$

$\frac{\begin{matrix}\:\:&5&.&3&\textbf{1}&6\\ -&1&.&9&\textbf{4}&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\:\:&\textbf{\:\:}&4\end{matrix}}$

$\frac{\begin{matrix}\:\:&5&.&\textbf{3}&1&6\\ -&1&.&\textbf{9}&4&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\textbf{\:\:}&\:\:&4\end{matrix}}$

$\frac{\begin{matrix}\:\:&\textbf{4}&\:\:&10&\:\:&\:\:\\ \:\:&\textbf{\linethrough{5}}&.&3&1&6\\ -&\textbf{1}&.&9&4&2\end{matrix}}{\begin{matrix}\:\:&\textbf{\:\:}&\:\:&\:\:&\:\:&4\end{matrix}}\\$

$\frac{\begin{matrix}\:\:&4&\:\:&\textbf{13}&\:\:&\:\:\\ \:\:&\linethrough{5}&.&\textbf{\linethrough{3}}&1&6\\ -&1&.&\textbf{9}&4&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\textbf{\:\:}&\:\:&4\end{matrix}}$

$\frac{\begin{matrix}\:\:&4&\:\:&\textbf{12}&10&\:\:\\ \:\:&\linethrough{5}&.&\textbf{\linethrough{13}}&1&6\\ -&1&.&\textbf{9}&4&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\textbf{\:\:}&\:\:&4\end{matrix}}$

$\frac{\begin{matrix}\:\:&4&\:\:&12&\textbf{11}&\:\:\\ \:\:&\linethrough{5}&.&\linethrough{13}&\textbf{\linethrough{1}}&6\\ -&1&.&9&\textbf{4}&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\:\:&\textbf{\:\:}&4\end{matrix}}$

$\frac{\begin{matrix}\:\:&4&\:\:&\textbf{12}&11&\:\:\\ \:\:&\linethrough{5}&.&\textbf{\linethrough{13}}&\linethrough{1}&6\\ -&1&.&\textbf{9}&4&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\textbf{3}&7&4\end{matrix}}$

$\frac{\begin{matrix}\:\:&4&\textbf{\:\:}&12&11&\:\:\\ \:\:&\linethrough{5}&\textbf{.}&\linethrough{13}&\linethrough{1}&6\\ -&1&\textbf{.}&9&4&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\textbf{.}&3&7&4\end{matrix}}$

$\frac{\begin{matrix}\:\:&\textbf{4}&\:\:&12&11&\:\:\\ \:\:&\textbf{\linethrough{5}}&.&\linethrough{13}&\linethrough{1}&6\\ -&\textbf{1}&.&9&4&2\end{matrix}}{\begin{matrix}\:\:&\textbf{3}&.&3&7&4\end{matrix}}$

=3.374

6 0

2 years ago

If AB = 10, AC = 6, and BC = 6, find AD:

JulsSmile [24]

AD= 5

Happy 4th of july :)

9 0

4 years ago