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Gelneren [198K]
3 years ago
12

Use properties to find the sum or product of 93+(68+7)

Mathematics
1 answer:
zaharov [31]3 years ago
7 0
93 + (68 + 7)
93 + 75
168
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I have to use trigonometric identities to solve. But I’m having trouble finding the values of cos A and sin B. Can anyone help m
katrin [286]

let's notice something, angles α and β are both in the I Quadrant, and on the first quadrant the x-coordinate/cosine and y-coordinate/sine are both positive.

\bf \textit{Sum and Difference Identities} \\\\ cos(\alpha - \beta)= cos(\alpha)cos(\beta) + sin(\alpha)sin(\beta) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ sin(\alpha)=\cfrac{\stackrel{opposite}{15}}{\stackrel{hypotenuse}{17}}\impliedby \textit{let's find the \underline{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{17^2-15^2}=a\implies \pm\sqrt{64}=a\implies \pm 8 = a\implies \stackrel{I~Quadrant}{\boxed{+8=a}} \\\\[-0.35em] ~\dotfill\\\\ cos(\beta)=\cfrac{\stackrel{adjacent}{3}}{\stackrel{hypotenuse}{5}}\impliedby \textit{let's find the \underline{opposite side}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \pm\sqrt{c^2-a^2}=b \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases}

\bf \pm\sqrt{5^2-3^2}=b\implies \pm\sqrt{16}=b\implies \pm 4=b\implies \stackrel{\textit{I~Quadrant}}{\boxed{+4=b}} \\\\[-0.35em] ~\dotfill

\bf cos(\alpha - \beta)=\stackrel{cos(\alpha)}{\left( \cfrac{8}{17} \right)}\stackrel{cos(\beta)}{\left( \cfrac{3}{5} \right)}+\stackrel{sin(\alpha)}{\left( \cfrac{15}{17} \right)}\stackrel{sin(\beta)}{\left( \cfrac{4}{5} \right)}\implies cos(\alpha - \beta)=\cfrac{24}{85}+\cfrac{60}{85} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill cos(\alpha - \beta)=\cfrac{84}{85}~\hfill

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3 years ago
Find the missing side of each triangle.
Bogdan [553]
They answer is 12.t obeys
6 0
3 years ago
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If you deposit $750 into a bank account that pays 1.75% interest compounded continuously, how much will be in the account after
lisabon 2012 [21]

is it compounded yearly? if so,

CI = p(1+ r/100)^n

= 750 (1+ 1.75/100)^5

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5 0
3 years ago
How to create a pattern with the rule n 3?
SCORPION-xisa [38]
If you copied the "n 3" part, it's very likely that your job was to create a pattern with either the rule n^3 or n*3.

In the case of the former, we can start with the initial number of 1 and increase by 1.
In that way, using the rule n^3 would create this pattern of numbers: 1, 8, 27, 64, and so on. Or stated in another way 1*1*1, 2*2*2, 3*3*3, 4*4*4 ...

In the case of the latter, we can start with the initial number of 1 and increase it by 1. 
In this way, using the rule of n*3 would create this pattern of numbers: 3, 6, 9, 12, 15, 18 and o son. Or stated in another way 1*3, 2*3, 3*3, 4*3, 5*3 ...
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3 years ago
Given (x – 7)2 = 36, select the values of x.
SVEN [57.7K]

Answer:

x = 25

Step-by-step explanation:

Solve by isolating x:
(x-7)2 = 36
x-7 = 18
x = 25

3 0
1 year ago
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