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3 years ago

8x + 11 = 2(4x - 7) + 25 Step by step please!

Mathematics

1 answer:

Dimas [21]3 years ago

3 0

Step-by-step explanation:

8x + 11=2(4x-7) + 25

8x + 11 = 8x - 14 + 25

8x + 11=8x- 14 + 14 +25 + 11

8x-8x + 11 -11 = 14 + 25 + 11

8x-8x =14 + 25 + 11

X = 50

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Pilar says that two linear systems below have the same solution. (see picture) Is she correct? Explain.

wariber [46]

The first set:

3x + 2y = 2 ---1)
5x + 4y = 6 ---2)

From 1), multiply all by 2, 6x + 4y = 4 ---3)

3) - 2),

6x + 4y - (5x + 4y) = 6 - 4
6x + 4y - 5x - 4y = 2
x = 2

Sub in x = 2 into 1),

3(2) + 2y = 2
2y = -4
y = -2

(2 , -2)

The second set:

3x + 2y = 2 ---1)
11x + 8y = 10 ---2)

From 1), multiply all by 4, 12x + 8y = 8 ---3)

3) - 2),

12x + 8y - (11x + 8y) = 8 - 10
12x + 8y - 11x - 8y = -2
x = -2

From this x value alone, we can tell that these two linear systems do NOT have the same solution as they meet at different coordinates.

Hope this helped! Ask me if there's any working from here that you don't understand! :)

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Answer:

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Answer:

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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.

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