PLEASE HELP ME

Mathematics

1 answer:

Marianna [84]4 years ago

7 0

So 70 divided by 2 is 35 so 35 minutes

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Write a function rule for the table

jeka57 [31]

The correct answer is B.

Find the slope using the formula m= y2-y1 / x2-x1. You will get 11.50/2, which is p = $5.75 per hour.

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3 years ago

Please answer this your award will be 20 points for the first answer and brainliest

Anna [14]

Step-by-step explanation:

pool = 1600 sq.cm

1600 sq.sm = 50cm × 32cm

AB = 50cm

the area of the park of A'B' :

50cm×4=200 sq.cm

5 0

3 years ago

NEED URGENT HELP

AnnyKZ [126]

Answer:

32.5

Step-by-step explanation:

If CED is 65, then AEB is 65, therefore we can calculate that CEA is 115 because 180 - 65 = 115. Then we do 180 - 115 = 65 which is the sum of angles ACE and CAE so 65 / 2 = 32.5 which is CAE.

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3 years ago

Gretta is 1 1/2 meters tall. Which of the following is equivalent to 1 1/2 meters?

yan [13]

Answer:

100 million

Step-by-step explanation:

7 0

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.

8 0

3 years ago