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Akimi4 [234]
2 years ago
15

Which trig function would you use?

Mathematics
2 answers:
Vinil7 [7]2 years ago
8 0

Answer:

tangent.

Step-by-step explanation:

with reference angle 43°

perpendicular (p) = a

base (b) = 11

So here we use

tangent = p / b

Hope it will help :)

vovikov84 [41]2 years ago
7 0
I did not think was doing a good job
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3.5<br><br> 11.5<br><br> 28<br><br> 33<br> please help
IgorC [24]

Answer:

28

Step-by-step explanation:

First, find the ratio between the triangles. To do this, divide any side of the big triangle by corresponding side in the small triangle. I chose 35. So, 35/10 = 3.5. 3.5 is the ratio between the triangles. Multiply 3.5 by 8 to get 28, the length of the missing side. :)

4 0
3 years ago
Linda takes out a loan for $4,000 she will repay the loan in 4 years Linda will pay $560 in interest how much is Linda's loan pa
Schach [20]

Answer:

$1,280 every year and $106.66 every month.

Step-by-step explanation:

You would do 4,000 by 4. Which would be 1,000. Then you would divide 560 by 4. After, you add both of your products together and you get 1,280. Obviously there are 12 months in a year, so you would divide your sum by 12. Then you would get 106.66. Basically, that is how you would get the answer.

I hope this helps!

(If you have the kindness in your heart to give me brainliest, I would appreciate it.)

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3 years ago
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Step-by-step explanation:

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3 years ago
The graph represents data from an experiment.
grandymaker [24]

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3 years ago
If $8700 is invested at 3% annual simple interest, how much should be invested at 6% annual simple interest so that the total ye
True [87]
First off, see  how much 8700 as principal, yields at 3% APR
that is \bf \qquad \textit{Simple Interest Earned}\\\\&#10;I = Prt\qquad &#10;\begin{cases}&#10;I=\textit{interest earned}\\&#10;P=\textit{original amount deposited}\to& \$8700\\&#10;r=rate\to 3\%\to \frac{3}{100}\to &0.03\\&#10;t=years\to &1&#10;\end{cases}

it will yield some amount

subtract that amount from 393
the difference is how much the yield will be on the 6% investment
so

\bf \qquad \textit{Simple Interest Earned}\\\\&#10;I = Prt\quad &#10;\begin{cases}&#10;I=\textit{interest earned}\\&#10;P=\textit{original amount deposited}\to& \$8700\\&#10;r=rate\to 3\%\to \frac{3}{100}\to &0.03\\&#10;t=years\to &1&#10;\end{cases}&#10;\\\\\\&#10;\implies \boxed{?}\\\\&#10;-----------------------------\\\\&#10;\textit{how much to invest at 6\%?}&#10;\\\\\\&#10;

\bf \qquad \textit{Simple Interest Earned}\\\\&#10;(393-\boxed{?}) = Prt\quad &#10;\begin{cases}&#10;I=\textit{interest earned}\\&#10;P=\textit{original amount deposited}\to& \$\\&#10;r=rate\to 6\%\to \frac{6}{100}\to &0.06\\&#10;t=years\to &1&#10;\end{cases}&#10;\\\\\\&#10;\textit{solve for "P", to see how much should the Principal be}\\\\&#10;\textit{keep in mind that }P+\boxed{?}=393\leftarrow \textit{both yields added}
6 0
3 years ago
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