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3 years ago

Simplify the expression. (−5a5b5)2(a3b3)5

Mathematics

1 answer:

Murrr4er [49]3 years ago

4 0

The answer with exponents would be -50a^8 b^8

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67.5 into a fraction

andrey2020 [161]

Answer:

67 1/2

Step-by-step explanation:

67 5/10

67 1/2

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2 years ago

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Can someone please help me to solve this? ASAP Please with explanation, no spam

snow_lady [41]

Answer:

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3 years ago

A right triangle has legs that are 21 in. and 25 in. long. What is the area of the triangle?

vfiekz [6]

Answer:

The area of the right triangle is 262.5

Step-by-step explanation:

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2 years ago

6.) -2x + y = -1 X Y -4 -3 -2 1 0

MrRa [10]

Answer:

Step-by-step explanation:

i ASSUME THE QUESTION IS TO DETERMINE Y?

Rearrange the equation to y = 2x - 1 and solve for each value of x

X Y

-4 -9

-3 -7

-2 -5

1 1

0 -1

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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.

7 0

3 years ago