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3 years ago

If f(x)=4x^2+1 and g(x) = x^2-5 fins (f-g)(x)

Mathematics

1 answer:

arlik [135]3 years ago

5 0

Answer:

Given:

f(x) = 4x² + 1

g(x) = x² - 5

Then

(f + g)(x) = 4x² + 1 + x² - 5

= 5x² - 4

Answer: 5x² - 4

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So I don’t really understand this so yeh pls help *pic included* by the way this is volume of cubes and cuboids and the question

faust18 [17]

Answer:

A2: 4*4*4 = 64 cm³; 6*6*6 = 216 cm³; etc...

B2: 7*4.5*4 = 126 cm³; 10.5*6*2 = 126 cm³; etc...

Step-by-step explanation:

So in A2 you have cubes, so l=b=h, and only one has to be given, called the "length of side".

In B2 they are different, hence there are 3 rows with l, b and h. But still you multiply them l x b x h

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2 years ago

What is the area of the sector with a central angle of 97 degrees and a diameter of 10cm

vampirchik [111]

Answer:

21. 162 (rounded to nearest thousandth)

Step-by-step explanation:

Area of a sector: (degree/360) (pi*radius^2)

degree given= 97

radius= diameter/2 = 5

(97/360) (pi*5^2)

(97/360) (pi*25) = 21.16211718

4 0

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)$