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DanielleElmas [232]
3 years ago
15

Help ASAP with this question.

Mathematics
1 answer:
Sergeu [11.5K]3 years ago
7 0
X = 3

It all simplifies to this 
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A project on Kickstarter.com was aiming to raise $15000 for a precision coffee press. They ended up with 714 supporters, raising
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For this, we can set up a proportion.
Their aim was 15000, so if they raised 100% of their goal, they wouldve raised exactly 15000.
They raised 557%, so how much did they make?

Proportion:

$15000 100%
$x 557%

Our expression would be x=$15000*557% divided by 100%, which would be $83550, so... a lot of money.

(we did an “x” formation with the proportion, ie “connected” x with 100, and “connected” 15k and 557. The number connecting to x is what we divide the equation by, and the other two are multiplied and then divided by the first number connecting to x. Otherwise we couldve calculated 557% of 15000 and gotten the same result. Whichever way is easier for you :) )
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Which simplified expression represents the area of the parallelogram
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Just multiply the sum of the numbers then divide
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2 years ago
Find the exact value of cos(sin^-1(-5/13))
son4ous [18]

bearing in mind that the hypotenuse is never negative, since it's just a distance unit, so if an angle has a sine ratio of -(5/13) the negative must be the numerator, namely -5/13.

\bf cos\left[ sin^{-1}\left( -\cfrac{5}{13} \right) \right] \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{then we can say that}~\hfill }{sin^{-1}\left( -\cfrac{5}{13} \right)\implies \theta }\qquad \qquad \stackrel{\textit{therefore then}~\hfill }{sin(\theta )=\cfrac{\stackrel{opposite}{-5}}{\stackrel{hypotenuse}{13}}}\impliedby \textit{let's find the \underline{adjacent}}

\bf \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} \\\\\\ \pm\sqrt{13^2-(-5)^2}=a\implies \pm\sqrt{144}=a\implies \pm 12=a \\\\[-0.35em] ~\dotfill\\\\ cos\left[ sin^{-1}\left( -\cfrac{5}{13} \right) \right]\implies cos(\theta )=\cfrac{\stackrel{adjacent}{\pm 12}}{13}

le's bear in mind that the sine is negative on both the III and IV Quadrants, so both angles are feasible for this sine and therefore, for the III Quadrant we'd have a negative cosine, and for the IV Quadrant we'd have a positive cosine.

8 0
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