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alexandr1967 [171]
2 years ago
5

What is (-m)⁻³n if m = 2 and n = -24?

Mathematics
1 answer:
Galina-37 [17]2 years ago
4 0

Answer:

-3

Step-by-step explanation:

(-2)^-3 x (-24)

(-2)^3 becomes 1/(-2)^3 in order to make the negative exponent a positive one.

then, you do 1/-8 (the -8 is the (-2)^3 simplified) x -24

1/-8 x (24) = 24/-8 = -3.

Hope this helps! :)

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Answer:

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Step-by-step explanation:

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1. (2 - 4) / (-2 - 0) = -2 / -2 = 1

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3 years ago
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What is the value of 8+ h² when h = 3?
MrMuchimi

Answer:

17

Step-by-step explanation:

8 + h^2

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4 0
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If Ryan pays his car insurance for the year in full, he will get a credit of $28. If he chooses to pay a monthly premium, he wil
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The one year-plan would have a credit, so it would have a positive sign. In the monthly plan, there is a high risk of being late in paying the bills. That's why a fine of $10 is given for every month that you are late. If you are not time conscious and you end up being late every month, it would give you a negative balance.
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2 years ago
An aeroplane X whose average speed is 50°km/hr leaves kano airport at 7.00am and travels for 2 hours on a bearing 050°. It then
Zigmanuir [339]

Answer:

(a)123 km/hr

(b)39 degrees

Step-by-step explanation:

Plane X with an average speed of 50km/hr travels for 2 hours from P (Kano Airport) to point Q in the diagram.

Distance = Speed X Time

Therefore: PQ =50km/hr X 2 hr =100 km

It moves from Point Q at 9.00 am and arrives at the airstrip A by 11.30am.

Distance, QA=50km/hr X 2.5 hr =125 km

Using alternate angles in the diagram:

\angle Q=110^\circ

(a)First, we calculate the distance traveled, PA by plane Y.

Using Cosine rule

q^2=p^2+a^2-2pa\cos Q\\q^2=100^2+125^2-2(100)(125)\cos 110^\circ\\q^2=34175.50\\q=184.87$ km

SInce aeroplane Y leaves kano airport at 10.00am and arrives at 11.30am

Time taken =1.5 hour

Therefore:

Average Speed of Y

=184.87 \div 1.5\\=123.25$ km/hr\\\approx 123$ km/hr (correct to three significant figures)

(b)Flight Direction of Y

Using Law of Sines

\dfrac{p}{\sin P} =\dfrac{q}{\sin Q}\\\dfrac{125}{\sin P} =\dfrac{184.87}{\sin 110}\\123 \times \sin P=125 \times \sin 110\\\sin P=(125 \times \sin 110) \div 184.87\\P=\arcsin [(125 \times \sin 110) \div 184.87]\\P=39^\circ $ (to the nearest degree)

The direction of flight Y to the nearest degree is 39 degrees.

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3 years ago
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