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Sav [38]
2 years ago
15

For the equation: 3x+1=2x+7 what are the 1 st and 2 nd opposite operations?

Mathematics
1 answer:
qaws [65]2 years ago
4 0

Answer:

Step-by-step explanation:

3x +1 = 2x +7  subtract 1 from both sides ( first)

3x +1 -1 = 2x + 7 -1

3x = 2x + 6 subtract 2x brom both sides (second)

3x -2x = 2x -2x +6

5x = 6

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A bag contains 4 white, 3 black, and 6 green balls. Balls are picked at random. Explain why the events picking a white ball and
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Answer: D


Step-by-step explanation:

D is the correct answer because first of all, there are 13 balls total. So the denominator would have to be 13. Also, white is the second largest number of balls in the set, so it is pretty likely that you will pick it. Although the most common picked would be green since it has the most balls.

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Which soap deal has the lowest price per ounce?
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It’s 5.75 for 14 oz. It’s answer is 0.41 which is the lowest out of all of them.
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What is the answer to this question
Anon25 [30]

Answer:

2118/12681

Step-by-step explanation:

Number of students absent in high school/Total number of students in high school

2118/12681

4 0
3 years ago
Find the exact value of tan (arcsin (two fifths)). For full credit, explain your reasoning.
Hitman42 [59]
\bf sin^{-1}(some\ value)=\theta \impliedby \textit{this simply means}
\\\\\\
sin(\theta )=some\ value\qquad \textit{now, also bear in mind that}
\\\\\\
sin(\theta)=\cfrac{opposite}{hypotenuse}\qquad 
\qquad 
% tangent
tan(\theta)=\cfrac{opposite}{adjacent}\\\\
-------------------------------\\\\

\bf sin^{-1}\left( \frac{2}{5} \right)=\theta \impliedby \textit{this simply means that}
\\\\\\
sin(\theta )=\cfrac{2}{5}\cfrac{\leftarrow opposite}{\leftarrow hypotenuse}\qquad \textit{now let's find the adjacent side}
\\\\\\
\textit{using the pythagorean theorem}\\\\
c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a\qquad 
\begin{cases}
c=hypotenuse\\
a=adjacent\\
b=opposite\\
\end{cases}

\bf \pm \sqrt{5^2-2^2}=a\implies \pm\sqrt{21}=a
\\\\\\
\textit{we don't know if it's +/-, so we'll assume is the + one}\quad \sqrt{21}=a\\\\
-------------------------------\\\\
tan(\theta)=\cfrac{opposite}{adjacent}\qquad \qquad tan(\theta)=\cfrac{2}{\sqrt{21}}
\\\\\\
\textit{and now, let's rationalize the denominator}
\\\\\\
\cfrac{2}{\sqrt{21}}\cdot \cfrac{\sqrt{21}}{\sqrt{21}}\implies \cfrac{2\sqrt{21}}{(\sqrt{21})^2}\implies \cfrac{2\sqrt{21}}{21}

7 0
3 years ago
Trigonometry, can anybody help??
WINSTONCH [101]

Answer:

angle B: 39

AC: 7.288

AB: 11.58

(note: you cut off the part about rounding so make sure that it's rounded correctly before you put in your answer)

Step-by-step explanation:

To solve this we will use SOH, CAH, TOA

we have the angle and the one opposite to it which means we can use either SOH or TOA

let's use TOA

tan(51)=(9/x)

x= 7.288

We can now use pahtagaryous theroem to solve for the hyptonouse

we have

9²+7.288²=C²

C=11.58

Finally, to find angle B we will recall that the angles of a triangle must add to 180.

51+90+a=180

a=39

5 0
2 years ago
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