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andre [41]
3 years ago
5

MATH HELP PLEASE ASAP

Mathematics
1 answer:
HACTEHA [7]3 years ago
3 0
The answers is for that one is D
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2. A paving company is paving the rectangular student parking lot at Hamilton High School. The
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Answer:

15p2 - p - 28

Step-by-step explanation:

4 0
3 years ago
Help -8(9n-10) -20 a. 90n+54 b. -72n+60 c. -17n – 80 d. -160n
Anna [14]

Answer:

1)  -8(9n-10) - 20

-72n + 80 - 20

-7n + 60

Can't factor

2) 90n+54

18(5n+3)

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-12(6n-5)

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5) -160n

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3 0
3 years ago
Read 2 more answers
Andrew has 212 coins, all of
Igoryamba

Answer:

158 quarters

54 dimes

Step-by-step explanation:

Let d = no. of dimes

     q = no. of quarters

10d = value of all the dimes since each dime is worth 10 cents

and 25q = value of all the quarters since each quarter is worth 25 cents

$44.90 = 4490 cents

  10d + 25q = 4490              d + q = 212

<u> - 10d - 10q = -2120</u>

             15q = 2370

                 q = 158 quarters   d + 158 = 212

                                                        d = 54 dimes

8 0
3 years ago
Find the squares of (x+3)<br>​
d1i1m1o1n [39]

Answer:

{x}^{2}  + 6x + 9 \\

Step-by-step explanation:

{(x + 3)}^{2}  \\ (x + 3)(x + 3) \\ x(x + 3) + 3(x + 3) \\  {x}^{2}  + 3x + 3x + 9 \\  {x}^{2}  + 6x + 9 \\

5 0
3 years ago
Given sin x = -4/5 and x is in quadrant 3, what is the value of tan x/2
love history [14]

bearing in mind that, on the III Quadrant, sine as well as cosine are both negative, and that hypotenuse is never negative, so, if the sine is -4/5, the negative number must be the numerator, so sin(x) = (-4)/5.


\bf sin(x)=\cfrac{\stackrel{opposite}{-4}}{\stackrel{hypotenuse}{5}}\impliedby \textit{let's find the \underline{adjacent}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \sqrt{c^2-b^2}=a \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{5^2-(-4)^2}=a\implies \pm\sqrt{9}=a\implies \pm 3=a \\\\\\ \stackrel{III~Quadrant}{-3=a}~\hfill cos(x)=\cfrac{\stackrel{adjacent}{-3}}{\stackrel{hypotenuse}{5}} \\\\[-0.35em] ~\dotfill

\bf tan\left(\cfrac{\theta}{2}\right)= \begin{cases} \pm \sqrt{\cfrac{1-cos(\theta)}{1+cos(\theta)}} \\\\ \cfrac{sin(\theta)}{1+cos(\theta)}\qquad \leftarrow \textit{let's use this one} \\\\ \cfrac{1-cos(\theta)}{sin(\theta)} \end{cases} \\\\[-0.35em] ~\dotfill

\bf tan\left( \cfrac{x}{2} \right)=\cfrac{~~\frac{-4}{5}~~}{1-\frac{3}{5}}\implies tan\left( \cfrac{x}{2} \right)=\cfrac{~~\frac{-4}{5}~~}{\frac{2}{5}}\implies tan\left( \cfrac{x}{2} \right)=\cfrac{-4}{5}\cdot \cfrac{5}{2} \\\\\\ tan\left( \cfrac{x}{2} \right)=\cfrac{-4}{2}\cdot \cfrac{5}{5}\implies tan\left( \cfrac{x}{2} \right)=-2

4 0
4 years ago
Read 2 more answers
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