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abruzzese [7]
2 years ago
6

Write y = x2 − 4x − 1 in vertex form. y = (x + 2)2 − 5 y = (x + 2)2 + 5 y = (x − 2)2 − 5 y = (x − 2)2 + 5

Mathematics
1 answer:
marta [7]2 years ago
4 0

Answer:

.

Step-by-step explanation:

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The width of a rectangle measures (3.8b-1.5) centimeters, and it's length measures (4.3b-7.2) centimeters. Which expression repr
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-17.4+16.2b

Step-by-step explanation:

I just guessed and it was right

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2 years ago
Can anyone help me with my homework
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GFBH is the code

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Giving right answer a branleist
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yes, (8×10) - (8×3)

hope this helps :)

6 0
2 years ago
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4) Allison practices her violin for at least 12 hours per week. She practices for three fourths of an hour each
zaharov [31]

Answer:

12

Step-by-step explanation:

So, she practices 12 hrs per week, and already has practiced 3 hrs this week.

This means she has 9 hrs left to practice.

3/4 of an hour is one session, so to find out the number of sessions she needs, you have to do (9 hours)/(one session)

which is basically 9 / ( 3/4)

\frac{9}{3/4}=\frac{9}{1} / \frac{3}{4}  = \frac{9}{1} * \frac{4}{3} = \frac{36}{3}  = 12

She needs 12 more sessions.

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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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