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3 years ago

In a class of 34 students, 19 of them are girls.

Mathematics

1 answer:

cluponka [151]3 years ago

8 0

Answer:

55.8

Step-by-step explanation:

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The water level at your dock is 5 ft above sea level. The tide goes out and the water decreases by 8 ft. What integer represents

Tanya [424]

Answer:

-3

Step-by-step explanation:

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3 years ago

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Jake set up a study schedule. The plan called for him to study 1/4 hour, 5/8 hour and 1 hour on Monday Tuesday and Wednesday in

Ludmilka [50]

Each day his study time is increasing. Find a common denominator, in this case 8

1/4 = 2/8
5/8 = 5/8
1 = 8/8

So he went from

2/8 to 5/8 to 8/8 which is an increase of 3/8 each day. If this pattern continues Thursday would be 11/8 and Friday 14/8.

14/8 = 1 6/8 = 1 3/4 hours

(an hour and 45 minutes)

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3 years ago

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if understand this correctly

Step-by-step explanation:

it would be half a meter

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2 years ago

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What segment is congruent to AC?<br> A. BD<br> B. BE<br> C. CE<br> D. none

Phoenix [80]

You can find the segment congruent to AC by finding another segment with the same length. So first, you need to find the length of AC.

C - A = AC
0 - (-6) = AC Cancel out the double negative
0 + 6 = AC
6 = AC

Now, find another segment that also has a length of 6.

D - B = BD
2 - (-2) = BD Cancel out the double negative
2 + 2 = BD
4 = BD
4 ≠ 6

E - B = BE
4 - (-2) = BE Cancel out the double negative
4 + 2 = BE
6 = BE
6 = 6

So, the segment congruent to AC is B. BE .

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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