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denpristay [2]
3 years ago
9

Need help as soon as possible​

Mathematics
2 answers:
choli [55]3 years ago
7 0

Answer:

Sin(F)  = 12/13

Cos(F)  = 5/13

Tan(F) = 12/5

Step-by-step explanation:

Sin = Opp./Hypo, Cos = Adj./Hypo. and Tan = Opp./Adj.

So

Sin(F) = Opp./Hypo = 12/13

Cos(F) = Adj./Hypo = 5/13

Tan(F) = Opp./Adj. = 12/5

Crank3 years ago
3 0

Answer:

see explanation

Step-by-step explanation:

sinF = \frac{opposite}{hypotenuse} = \frac{HG}{HF} = \frac{12}{13}

------------------------------------------------------

cosF = \frac{adjacent}{hypotenuse} = \frac{FG}{HF} = \frac{5}{13}

-------------------------------------------------------

tanF = \frac{opposite}{adjacent} = \frac{HG}{FG} = \frac{12}{5}

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. Write the exponential function modeling an initial population of 67,000 and a relative growth rate of 1.7% per year. What popu
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Answer:

In 37 years the population will be 125,012

Step-by-step explanation:

Giving the following information:

Initial population (PV)= 67,000

Growth rate (g)= 1.7%= 0.017

Number of years (n)= 37

<u>First, we will establish the future value (FV) formula:</u>

FV= PV*(1+i)^n

FV= 67,000*(1.017^n)

<u>Now, for 37 years:</u>

FV= 67,000*(1.017^37)

FV= 125,012

In 37 years the population will be 125,012

6 0
2 years ago
3)<br> Evaluate the expression 2x-5 when x = 3.
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Answer:

1

Step-by-step explanation:

Plug in the x to the original equation given.

2(3)-5

2 times 3 is 6 so then,

6-5 =1

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2 years ago
A business had sales last year of $139,526,000. Write the value of the 9 digit in this number. _________________________________
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7 0
3 years ago
Use the substitution u = tan(x) to evaluate the following. int_0^(pi/6) (text(tan) ^2 x text( sec) ^4 x) text( ) dx
Rudiy27
If we use the substitution u = \tan x, then du = \sec^2 {x}\ dx. If you try substituting just u and du into the integrand, though, you'll notice that there's a \sec^2x left over that we have to deal with.

To get rid of this problem, use the identity \tan^2 x + 1 = \sec^2 x and substitute in the left side of the identity for the extra \sec^2x, as shown:

\int\limits^{\pi/6}_0 {tan^2 x \ sec^4 x} \, dx
\int\limits^{\pi/6}_0 {tan^2 x \ (tan^2 x + 1) \ sec^2 x} \, dx

From there, we can substitute in u and du, and then evaluate:

\int\limits^{\pi/6}_0 {tan^2 x \ (tan^2 x + 1) \ sec^2 x} \, dx
\int\limits^{\frac{1}{\sqrt{3}}}_0 {u^2(u^2 + 1)} \, du
\int\limits^{\frac{1}{\sqrt{3}}}_0 {u^4 + u^2} \, du
= \left.\frac{u^5}{5} + \frac{u^3}{3}\right|_0^\frac{1}{\sqrt{3}}
= (\frac{(\frac{1}{\sqrt{3}})^5}{5} + \frac{(\frac{1}{\sqrt{3}})^3}{3}) - (\frac{(0)^5}{5} + \frac{(0)^3}{3})
= \frac{1}{45\sqrt{3}} + \frac{1}{9\sqrt{3}} = \frac{6}{45\sqrt{3}} = \bf \frac{2}{15\sqrt{3}}


8 0
3 years ago
A savings account that pays interest every 6 months is said to have _______ interest period. A. Quarterly B. Monthly C. Daily D.
AveGali [126]
The answer is Semi-annual
4 0
3 years ago
Read 2 more answers
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