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denis23 [38]
3 years ago
10

42/-8 Divide Please answer asap

Mathematics
2 answers:
krok68 [10]3 years ago
5 0
It's -5.25 because 5*8 is 40 and 6*8 is 48.
AleksAgata [21]3 years ago
4 0
The answer is -5.25 and I know I am right
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Can someone please help me ..
Naddika [18.5K]
(0+1+2+3+4) /5 = 2
Hope this helps!
7 0
3 years ago
Read 2 more answers
Find the missing side lengths. Leave your answers as radicals in simplest form.
tankabanditka [31]

Answer:

A.

x = \frac{7\sqrt{6}{3}

y = \frac{7\sqrt{6}}{3}

Step-by-step explanation:

Reference angle = 60°

Opposite = \frac{7\sqrt{2}}{2}

Hypotenuse = x

Adjacent = y

✔️To find x, apply the trigonometric function SOH:

Sin 60° = Opp/Hyp

sin 60° = \frac{\frac{7\sqrt{2}}{2}}{x}

\frac{\sqrt{3}}{2} = \frac{\frac{7\sqrt{2}}{2}}{x} (sin 60 = √3/2)

\frac{\sqrt{3}}{2} = \frac{7\sqrt{2}}{2}*\frac{1}{x}

\frac{\sqrt{3}}{2} = \frac{7\sqrt{2}}{2x}

Cross multiply

\sqrt{3}*2x = 7\sqrt{2}*2

2\sqrt{3}*x = 14\sqrt{2}

Divide both sides by 2

\sqrt{3}*x = 7\sqrt{2}

Divide both sides by √3

x = \frac{7\sqrt{2}}{\sqrt{3}}

Rationalize

x = \frac{7\sqrt{2}*\sqrt{3}}{\sqrt{3}*\sqrt{3}}

x = \frac{7\sqrt{6}{3}

✔️To find y, apply the trigonometric function TOA:

Tan 60° = Opp/Adjacent

Tan 60° = \frac{\frac{7\sqrt{2}}{2}}{y}

\sqrt{3} = \frac{\frac{7\sqrt{2}}{2}}{y} (tan 60 = √3)

\sqrt{3} = \frac{7\sqrt{2}}{2}*\frac{1}{y}

\sqrt{3} = \frac{7\sqrt{2}}{2y}

Cross multiply

\sqrt{3}*2y = 7\sqrt{2}

2\sqrt{3}*y = 7\sqrt{2}

Divide both sides by √3

y = \frac{7\sqrt{2}}{\sqrt{3}}

Rationalize

y = \frac{7\sqrt{2}*\sqrt{3}}{\sqrt{3}*\sqrt{3}}

y = \frac{7\sqrt{6}}{3}

4 0
3 years ago
Yolanda went for a drive in her new car. She drove
GenaCL600 [577]
Her speed was 62.5 miles per hour.
5 0
3 years ago
se necesitan 247,8kg de hielo para elaborar un iglu que necesita 48bloques de hielocada uno de4,6kg cuanto hielo se necedita par
MrMuchimi
No mas hablas Espanol
5 0
4 years ago
Can someone help please
kicyunya [14]

Answer:

14.4

Step-by-step explanation:

It makes the most sense.

4 0
3 years ago
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