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andreyandreev [35.5K]
3 years ago
8

A triangle has one side that measures 12 inches and another side that measures 33 inches. Which are possible side lengths of the

third side?
Mathematics
1 answer:
In-s [12.5K]3 years ago
5 0

Answer:

Greater than 21

Less than 45

Step-by-step explanation:

You add and subtract the two numbers you are given.

33 - 12 = 21

The third side has to be bigger than 21.

33 + 12 = 45

The third side has to be less than 45.

It can be written in one math sentence:

21 < x < 45

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Please help I need just 4 Correct for answers to pass
zlopas [31]

Answer:

Hey!

Your answer is 14m (2sf)

Step-by-step explanation:

<u>WE HAVE TO USE PYTHAGORAS!</u> ⇒ a²+b²=c² THEN √c = (unknown side)

Here: 12² + 8²= 208

Now SQUARE-ROOT 208! ⇒ √208

ANSWER: 14.42220510185596

The answers there are rounded up so we round this number to get 14m!

B IS YOUR OPTION!

<h2><u></u></h2><h2><u>I HOPE THIS HELPED YOU!</u></h2>
5 0
3 years ago
Suppose theta is an angle in the standard position whose terminal side is in quadrant 4 and cot theta = -6/7. find the exact val
zimovet [89]

First off, let's notice that the angle is in the IV Quadrant, where sine is negative and the cosine is positive, likewise the opposite and adjacent angles respectively.

Also let's bear in mind that the hypotenuse is never negative, since it's simply just a radius unit.

\bf cot(\theta )=\cfrac{\stackrel{adjacent}{6}}{\stackrel{opposite}{-7}}\qquad \impliedby \textit{let's find the \underline{hypotenuse}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies c=\sqrt{a^2+b^2} \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ c=\sqrt{6^2+(-7)^2}\implies c=\sqrt{36+49}\implies c=\sqrt{85} \\\\[-0.35em] ~\dotfill

\bf tan(\theta)=\cfrac{\stackrel{opposite}{-7}}{\stackrel{adjacent}{6}} ~\hfill csc(\theta)=\cfrac{\stackrel{hypotenuse}{\sqrt{85}}}{\stackrel{opposite}{-7}} ~\hfill sec(\theta)=\cfrac{\stackrel{hypotenuse}{\sqrt{85}}}{\stackrel{adjacent}{6}} \\\\\\ sin(\theta)=\cfrac{\stackrel{opposite}{-7}}{\stackrel{hypotenuse}{\sqrt{85}}}\implies \stackrel{\textit{and rationalizing the denominator}}{sin(\theta)=\cfrac{-7}{\sqrt{85}}\cdot \cfrac{\sqrt{85}}{\sqrt{85}}\implies sin(\theta)=-\cfrac{7\sqrt{85}}{85}}

\bf cos(\theta)=\cfrac{\stackrel{adjacent}{6}}{\stackrel{hypotenuse}{\sqrt{85}}}\implies \stackrel{\textit{and rationalizing the denominator}}{cos(\theta)=\cfrac{6}{\sqrt{85}}\cdot \cfrac{\sqrt{85}}{\sqrt{85}}\implies cos(\theta)=\cfrac{6\sqrt{85}}{85}}

6 0
3 years ago
A cleaners recommend 1 1/2 cup of cleaner to 12 cups of water what is the ratio of cleaner to water in simplest form
Evgen [1.6K]

Answer:

8 cups of water : 1 cup of cleaner

Step-by-step explanation:

To find the ratio, divide one value by the other. 12÷1½=8, so the ratio is 8 cups of water : 1 cup of cleaner.

6 0
4 years ago
What is the anwser to this?
Elza [17]
Which one is it the top one or the bottom one?
6 0
3 years ago
Read 2 more answers
33. Suppose that the scores on a statewide standardized test are normally distributed with a mean of 78 and a standard deviation
tino4ka555 [31]

Given the scores on a statewide standardized test are normally distributed

Mean = μ = 78

Standard deviation = σ = 3

Normalize the data using the z-score by using the following formula and chart:

z=\frac{x-\mu}{\sigma}

Estimate the percentage of scores of the following cases:

(a) between 75 and 81

so, the z-score for the given numbers will be:

\begin{gathered} 75\rightarrow z=\frac{75-78}{3}=\frac{-3}{3}=-1 \\ 81\rightarrow z=\frac{81-78}{3}=\frac{3}{3}=1 \end{gathered}

As shown, the percentage when (-1 < z < 1) = 68%

(b) above 87

87\rightarrow z=\frac{87-78}{3}=\frac{9}{3}=3

The percentage when (z > 3) = 0.5%

(c) below 72

72\rightarrow z=\frac{72-78}{3}=\frac{-6}{3}=-2

The percentage when (z < -2) = 0.5 + 2 = 2.5%

(d) between 75 and 84

\begin{gathered} 75\rightarrow z=\frac{75-78}{3}=-\frac{3}{3}=-1 \\ 84\rightarrow z=\frac{84-78}{3}=\frac{6}{3}=2 \end{gathered}

The percentage when ( -1 < z < 2 ) = 68 + 13.5 = 81.5%

3 0
1 year ago
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