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AnnZ [28]
3 years ago
6

Please answer this question in two minutes

Mathematics
1 answer:
EastWind [94]3 years ago
7 0

Answer:

The other point is  R(19, 4)

Step-by-step explanation:

\frac{20+x}{2}=19.5\\\\\Rightarrow x = 19\\\\\frac{y+15}{2}=9.5\\\\\Rightarrow y=4

Best Regards!

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5. trey made pasta in the Shape of a square ; He draws 2 lines to cut the pasta into smaller pieces The smaller pieces have L I
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Wouldn't it be. 8 pieces because you cut each sides into two pieces so like this. —|—. You'll end up having four square pieces, but count the pieces that creates an L shape.


best wishes good luck

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PLEASE ANSWER.<br> ill mark most brainliest.
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the answer is C

Step-by-step explanation:

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What is the percent of change from 28 to 70
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Your answer will be 150 percent
3 0
2 years ago
Shobos mother present age is six times shobos present age. Shobos age five from now will be one third of his mother's present ag
r-ruslan [8.4K]

Answer:

Step-by-step explanation:

From the problem statement, we can setup the following equations:

M = 6S

S + 5 = \frac{1}{3}M

where S is the current age of Shobos and M is the current age of his mother.

Substituting the first equation into the second will allow us to solve for S:

S + 5 = \frac{1}{3}M

S + 5 = \frac{1}{3}(6S)

S + 5 = 2S

S = 5

Substituting this value into the first equation gives us the age of the mother:

M = 6S

M = 6(5)

M = 30

5 0
3 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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