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3 years ago

Graph the function. Identify the vertex and axis of symmetry f(x) = –x2 – x + 2

Mathematics

1 answer:

sergij07 [2.7K]3 years ago

5 0

I hope this is right (-5) + (-4) + (7);
= -9 + 7;
= -2

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A carpet remnant is shaped like a triangle. It has a height of 32 inches and a base of 8 inches. Write an equation to represent

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128 square inches.

The area of a triangle is 1/2bh, where b is the base and h is the height. Substitute the values in.

1/2 * 8 * 32

Now, just solve.

4 * 32

128 square inches

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3 years ago

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Sin(5x) cos(3x)-1/2sin(8x) = sin(x) cos(x)

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Answer:

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Step-by-step explanation:

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3 years ago

For f(x) = x3 – x2, which gives an output of 48? O f(-8) O fl-4) O f(4) O f(8)

Misha Larkins [42]

Answer:

f(4)

Step-by-step explanation:

The output is the value of f(x) for a given value of x ( the input )

Substitute the given values of x into f(x) to determine the output, that is

f(- 8) = (- 8)³ - (- 8)² = - 512 - 64 = - 576 ≠ 48

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f(4) = 4³ - 4² = 64 - 16 = 48

Thus f(4) gives an output of 48

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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

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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

Please help me.. I really need it, and I will give you brainlyest:)))

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Answer:

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so how you ahsbdgkjsgdejgwqjda is how you ajsdbguyaweguidasdh to the question

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