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3 years ago

on friday,718 peoole went to school play.on saturday,822 people went.about how many oeople saw the play altogether

Mathematics

1 answer:

IrinaVladis [17]3 years ago

3 0

About 1540 people saw the play

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10 points, doing this one again because I got a wrong answer last time but regardless thank you for trying to help

Alik [6]

Hey ! there

Answer:

Value of missing side i.e. TE is 12 feet

Step-by-step explanation:

In this question we are provided with a right angle triangle having TS - 35 ft and SE - 37 ft . And we are asked to find the missing side that is TE using Pythagorean Theorem .

Pythagorean Theorem : -

According to Pythagorean Theorem sum of squares of perpendicular and base is equal to square of hypotenuse in a right angle triangle i.e.

H² = P² + B²

Where ,

H refers to Hypotenuse

P refers to Perpendicular

B refers to Base

Solution : -

In the given triangle ,

Base = TE

Perpendicular = TS ( 35 feet )

Hypotenuse = SE ( 37 feet )

Now applying Pythagorean Theorem :

$\quad \longmapsto \qquad \:SE {}^{2} = TS {}^{2} + TE {}^{2}$

Substituting values :

$\quad \longmapsto \qquad \:37 {}^{2} = 35 {}^{2} + TE {}^{2}$

Simplifying it ,

$\quad \longmapsto \qquad \:1369 = 1225 + TE {}^{2}$

Subtracting 1225 on both sides :

$\quad \longmapsto \qquad \:1369 - 1225 = \cancel{1225} + TE {}^{2} - \cancel{1225}$

We get ,

$\quad \longmapsto \qquad \:144 = TE {}^{2}$

Applying square root to both sides :

$\quad \longmapsto \qquad \ \sqrt{ 144} = \sqrt{TE {}^{2}}$

We get ,

$\quad \longmapsto \qquad \: \red{\underline{\boxed{\frak{TE = 12 \: feet}}}} \quad \bigstar$

Henceforth , value of missing side is 12 feet .

Verifying : -

Now we are verifying our answer using Pythagorean Theorem . We know that according to Pythagorean Theorem ,

SE² = TS² + TE²

Substituting value of SE , TS and TE :

37² = 35² + 12²

1369 = 1225 + 144

1369 = 1369

L.H.S = R.H.S

Hence , Verified .

Therefore , our answer is correct .

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6 0

2 years ago

Read 2 more answers

A Geometric series has a first term of 4 and a common ratio of 6. what formula is equivalent to the partial sum of the first n t

IRISSAK [1]

$\bf \qquad \qquad \textit{sum of a finite geometric sequence}
\\\\
S_n=\sum\limits_{i=1}^{n}\ a_1\cdot r^{i-1}\implies S_n=a_1\left( \cfrac{1-r^n}{1-r} \right)\quad 
\begin{cases}
n=n^{th}\ term\\
a_1=\textit{first term's value}\\
r=\textit{common ratio}\\
----------\\
a_1=4\\
r=6
\end{cases}
\\\\\\
S_n=4\left( \cfrac{1-6^n}{1-6} \right)\implies S_n=4\left( \cfrac{1-6^n}{-5} \right)$