All categories

3 years ago

Please help me idk this

Mathematics

1 answer:

madreJ [45]3 years ago

8 0

Answer:

660in

Step-by-step explanation:

Rectangle:

A=l*w

A=20*30=60

Triangle:

A=b*h/2

A=40*30/2=600

600+60=660

You might be interested in

A rectangular prism measures 225 inches by 315 inches by 2 inch.

Bas_tet [7]

Answer:

No I don't agree with Priya

The reason why I don't agree with Priya is because 25 Inch side blocks are BIGGER than 15 inch side blocks. That means that it will take less amount of 25 inch side blocks to fill up the volume of the rectangular prism rather than 15 inch side blocks.

(I'm sorry, I don't know what was your previous question, so I can't answer that part. If you comment it below, I will do it!)

5 0

2 years ago

S(x)= 22x-8 and f(x)= -5. What is s(x) x f(x)?

zloy xaker [14]

-110x+40
You need to use the distributive property for this problem
You multiply 22x times -5 (which is -110x) and -8 times -5 (which is 40) to get -110x+40

6 0

3 years ago

How long is a elaphants head

sergejj [24]

Answer: Depends on the elephant ;)

7 0

3 years ago

Write (5.64) x 10-7 as a decimal.

Cloud [144]

Answer:

16.92

Step-by-step explanation:

5.64 x 3 = 16.92

<em>Hope that helps!</em>

3 0

3 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