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BartSMP [9]
3 years ago
14

Help please with question 5-8, show the solution

Mathematics
1 answer:
ludmilkaskok [199]3 years ago
8 0

For each of these problems, remember SOH-CAH-TOA.

Sine = opposite/hypotenuse

Cosine = adjacent/hypotenuse

Tangent = opposite/adjacent

5) Here we are looking for the cosine of the 30 degree angle. Cosine uses the adjacent side to the angle over the hypotenuse. Therefore, cos(30) = 43/50.

6) We have an unknown side length, of which is adjacent to 22 degrees, and the length of the hypotenuse. Since we know the adjacent side and the hypotenuse, we should use Cosine. Therefore, our equation to find the missing side length is cos(22) = x / 15.

7) When finding an angle, we always use the inverse of the trigonometry function we originally used. Therefore, if sin(A) = 12/15, then the inverse of that would be sin^-1 (12/15) = A.

8) We are again using an inverse trigonometry function here. We know the hypotenuse, as well as the side adjacent to the angle. Therefore, we should use the inverse cosine function. Using the inverse cosine function gives us cos^-1 (9/13) = 46 degrees.

Hope this helps!

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the interest earned on an account varies directly with the balance in the account. if an account with a balance of $200 earns $1
sp2606 [1]

\bf \qquad \qquad \textit{direct proportional variation} \\\\ \textit{\underline{y} varies directly with \underline{x}}\qquad \qquad y=kx\impliedby \begin{array}{llll} k=constant\ of\\ \qquad variation \end{array} \\\\[-0.35em] \rule{34em}{0.25pt}

\bf \stackrel{\textit{\underline{I}nterest varies with \underline{b}alance}}{I=kb}\qquad \qquad \textit{we know that } \begin{cases} b=200\\ I=12.5 \end{cases} \\\\\\ 12.5=k200 \implies \cfrac{12.5}{200}=k\implies \cfrac{~~\frac{125}{10}~~}{\frac{200}{1}}=k\implies \cfrac{125}{10}\cdot \cfrac{1}{200}=k \\\\\\ \cfrac{1}{16}=k \qquad \qquad \textit{therefore}\qquad \boxed{I=\cfrac{1}{16}b} \\\\\\ \textit{when b = 500, what is \underline{I}?}\qquad I=\cfrac{1}{16}(500)\implies I=\cfrac{125}{4}\implies I=31.25

3 0
3 years ago
Which system of inequalities is represented by the graph?
alexgriva [62]

Answer:

\left\{\begin{array}{l}y\ge -x-6\\ \\ y\le 2x+4\end{array}\right.

Step-by-step explanation:

The diagram shows shaded region which boundary is bounded with two lines.

<u>Equation of the 1st line:</u>

This line passes through the points (-6,0) and (0,-6), so its equation is

\dfrac{x-(-6)}{0-(-6)}=\dfrac{y-0}{-6-0}\\ \\\dfrac{x+6}{6}=\dfrac{y}{-6}\\ \\y=-x-6

<u>Equation of the 2nd line:</u>

This line passes through the points (-2,0) and (0,4), so its equation is

\dfrac{x-(-2)}{0-(-2)}=\dfrac{y-0}{4-0}\\ \\\dfrac{x+2}{2}=\dfrac{y}{4}\\ \\y=2x+4

The origin belongs to the shaded region, so it must satisfy both inequalities, so

-0-6=-6\le 0\Rightarrow correct inequality is -x-6\le y

2\cdot 0+4=4\ge 0\Rightarrow correct inequality is 2x+4\ge y

Note that signsare with notion "or equal to" because lines are solid.

Answer:

\left\{\begin{array}{l}-x-6\le y\\ \\2x+4\ge y\end{array}\right.\Rightarrow \left\{\begin{array}{l}y\ge -x-6\\ \\ y\le 2x+4\end{array}\right.

7 0
3 years ago
What is the expression (a^2+b^2)^2 is equivalent to
netineya [11]
<span>(a^2+b^2)^2
since (a+b)^2=a^2+2ab+b^2, substitute a with a^2 and b with b^2, which yields:
a^4+2(a^2*b^2)+b^4</span>
7 0
3 years ago
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21+19=40;19+21=40 a commutative property or associative property
Nikolay [14]

Answer:

it is a commutative property

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3 years ago
Which of the following are integer solutions to the inequality below?
PtichkaEL [24]

Answer:

0.5

Step-by-step explanation:

3 0
3 years ago
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