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kvv77 [185]
3 years ago
7

WHENS MY BDAY I SAID THIS

Mathematics
1 answer:
Alja [10]3 years ago
8 0

Answer:

reeee whenever it was happy bday ily by :)

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Write the following equations in slope-intercept form.5x+y=30
Maurinko [17]

Answer:

y=-5x+30

Step-by-step explanation:

I HOPE THIS HELPS

BRAINLIEST PLSSS I NEED IT SO BAD

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3 years ago
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4- 5i= 8- 9i<br> 32i- 24= 7i+ 26
o-na [289]

Answer:

Sovle for I and go step by step and how your teacher has showed you i dont know how your teacher has showed you

Step-by-step explanation:

3 0
3 years ago
F(x) =2x-5 and g(x) =x+52 find f(g(x)
irga5000 [103]
If f(x) = 2x - 5 and g(x) = x + 52, then f(g(x)) can be deduced by placing g(x) in the spot of x in the f(x) equation as follows:

f(g(x)) = 2(g(x)) - 5

Since we know g(x) = x + 52, let's plug it in:

f(g(x)) = 2(x + 52) - 5
f(g(x)) = 2x + 104 - 5
f(g(x)) = 2x + 99
4 0
3 years ago
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What temperature are<br> 22.7 degrees colder than 96.5 degrees?
Gwar [14]

Answer:

73.8º

Step-by-step explanation:

96.5-22.7=

73.8º

4 0
2 years ago
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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
3 years ago
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