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Margaret [11]
3 years ago
11

A box is 1 m high, 2.5 m long, and 1.5 m wide, what is its volume?

Mathematics
2 answers:
pochemuha3 years ago
6 0

Answer:

3.75

Step-by-step explanation:

<em>v = lbh  \\ 2.5 \times 1.5 \times 1 \\  = 3.75</em>

Margaret [11]3 years ago
4 0

Answer:

3.75

Step-by-step explanation:

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Clair has250 pennies some pennies are in a box the rest are in her bankthere are more then 100 pennies in each place how many pe
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In each place there must be more 100 pennies, but as the total 250 pennies, there are several ways that the money is distributed.

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3 years ago
Read 2 more answers
24. A farmstand collected $350 from sales on Wednesday. This amount was $75
larisa [96]

Answer:

$100

Step-by-step explanation:

Given:

  • Wednesday sales = $350

If the amount collected from sales on Wednesday was <u>$75 greater</u> than the amount collected from sales on Tuesday, then Tuesday's sales were $75 <u>less</u> than Wednesday's sales:

  • Tuesday sales = $350 - $75 = $275

If the amount collected from sales on Wednesday was <u>two times as great</u> as the amount collected from sales on Monday, then Monday's sales were <u>half</u> Wednesday's sales:

  • Monday sales = $350 ÷ 2 = $175

To calculate how much more the farmstand collected from sales on Tuesday than it collected on Monday, <u>subtract</u> the Monday sales from the Tuesday sales:

  • $275 - $175 = $100
8 0
2 years ago
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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
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