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Mathematics

2 answers:

Tomtit [17]3 years ago

7 0

Con qué necesitas ayuda?

forsale [732]3 years ago

5 0

En que te puedo ayudar??

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Solve 2x+y=7 for y what is y ?

masha68 [24]

You cant solve y but you can form an equation in terms of y
which is y=7-2x

if you were given another equation that has the term y and x than you can solve it.also that is a simultaneous equation.

Hope this helps,dont forget to put this as the brainliest answer if it does to help me out too. xD

6 0

3 years ago

Read 2 more answers

Please help with this

Murljashka [212]

Answer:

i think its b

Step-by-step explanation:

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3 years ago

2) Shane puts aside $40 per week for his monthly car payment. He earns $320 per week.

ExtremeBDS [4]

Answer:

12.5%

Step-by-step explanation:

320 divided by 40 = 8

100 divided by 8 = 12.5

100% is 320 dollars, which means that 40 dollars is 1/8 of 320. And 1/8 of 100 is 12.5%

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3 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.

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2 years ago

NEED HELP PLEASE!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

vodomira [7]

Y=5/2x-20
There’s answer

6 0

3 years ago