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2 years ago

Find the area of the shaded portions. Round answers to the tenths place if necessary

Mathematics

1 answer:

Sladkaya [172]2 years ago

6 0

the answer is 3!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

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Describe a time when someone around you made a decision based on authority of someone else?

dalvyx [7]

There was a time that a friend of mine got to a decision because the leader of the group said so (it was a group project). Although it was clear that he did not want to comply to the decision, he made it because it was what the leader said so.

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3 years ago

Read 2 more answers

What is the area of a triangle with vertices (1,0) (5,0) (3,4)

Aleksandr-060686 [28]

Answer:

The area of the triangle is of 8 units of area.

Step-by-step explanation:

Answer:

The area of the triangle is of 21 units of area.

Step-by-step explanation:

The area of a triangle with three vertices $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ is given by the determinant of the following matrix:

$A = \pm 0.5 \left|\begin{array}{ccc}x_1&y_1&1\\x_2&y_2&1\\1_3&y_3&1\end{array}\right|$

In this question:

Vertices (1,0) (5,0) (3,4). So

$A = \pm 0.5 \left|\begin{array}{ccc}1&0&1\\5&0&1\\3&4&1\end{array}\right|$

$A = \pm 0.5(1*0*1+0*1*3+1*5*4-1*0*3-0*5*1-1*1*4)$

$A = \pm 0.5*(20-4)$

$A = \pm 0.5*16 = 8$

The area of the triangle is of 8 units of area.

3 0

3 years ago

Assume you have a poker chip set containing blue, red, and white chips, all of the same size. You place 30 blue chips, 20 red ch

Zigmanuir [339]

Answer:

0.3

Step-by-step explanation:

30/(30+20+50) = 30/100 = 0.3

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3 years ago

Jim bought 3 cds at a cost of $14.99 each what will he pay including 6% sales tax

Charra [1.4K]

14.99 *3 = 44.97

44.97*1.06 =47.6682, round off to $47.67 with tax

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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.

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3 years ago