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Solnce55 [7]
3 years ago
7

25 POINTS!!! Simplify 3x - 4 - 5x = 6 + 4x + 2

Mathematics
1 answer:
Maksim231197 [3]3 years ago
6 0
3x-5x-4x=6+2+4
-6x=12
x=-2

Hope this can help.
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You are the administrator of an annual essay contest scholarship fund. This year a $90,000 college scholarship is being divided
Vinil7 [7]

Answer:

Runner-up: $15,000

Winner: $75,000

Step-by-step explanation:

Let's create an equation for this problem.

Let x=the amount the runner-up receives

Let 5x=the amount the winner receives

Then,

x+5x=90,000

6x=90,000

x=15,000

5x=75,000

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3 years ago
What is the greatest common factor of the polynomial below 12x^2-9x
wlad13 [49]

Answer:

the greatest common factor of this is 3

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The diameter of a Frisbee is 12 inches. what's the area of the Frisbee?
xenn [34]
A= πr² = π × 6² = "113.1" is the area of the frisbee. 

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7 0
3 years ago
Samantha just opened a new restaurant. She earned $550 on the first day and $750 on the third day that she was open for business
mr Goodwill [35]

Answer:

Step-by-step explanation:

If Samantha's earnings continue to increase at the same rate, this means that her earning is increasing arithmetically.

If she earned $550 in the first day, we can say the first term is $550

If she earned $750 on the third day, we can say the third term is $750

For us to know by how much her money is increasing, we need to find the common difference d formed by the sequence

550, x , 750

T1 = 550

T2 = x

T3 = 750

Common difference d = T2-T1 = T3-T2

x - 550 = 750 - x = d

Let's calculate the second term first i.e x

Since x - 550 = 750 - x = d

x - 550 = 750 - x

Collect like terms

x+x = 750+550

2x = 1300

x = 1300/2

x = 650

d = T2-T1

d = x - T1

d = 650-550

d = 100

Hence her money keeps increasing by $100 each day

8 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)

The direction of flight Y to the nearest degree is 39 degrees.

7 0
3 years ago
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