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Dafna1 [17]
1 year ago
11

Michael has saved $40. He got a summer job mowing lawns, and gets $9 for every lawn he mows. How many lawns does Michael need to

mow to save exactly $121, the cost of a concert ticket?
Mathematics
1 answer:
Shkiper50 [21]1 year ago
3 0

Answer:

9

Step-by-step explanation:

9 × 9 = 81

81 + 40 = 121

121 is the amount of money you want so you just have to mow 9 lawns to get your exact amount of money you need.

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This doesn’t make sense explain plz
AnnZ [28]

Its simple, graph them.

For the first equation, go on the Y-Axis (the vertical one) and go to 4. Then from there go up 1, right 6.

For the second equation go on the Y-Axis (the vertical one) and plot a point at 1 (aka 0,1) Now you go up 1, right 3.

When you see an equation like y=2x+3, the 3 represents the point (0,3) as when the x is 0, y=3. Just plug the numbers in. And as for the "2x" 2 is the slope. Slope is always rise/run or up, then right. So if its 2 your slope is 2/1, rise 2, over 1. If it "-2x" is your slope then all you have to do is go down 2, right 1.

I hope this cleared up your confusion, brainliest/heart would help.

5 0
2 years ago
Explain:<br><br><br><br><br><br><br><br><br><br> 10 to the 3rd power
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10x10x10 which is 1000 have a great day
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2 years ago
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Does the graph represent a function?
saveliy_v [14]

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Step-by-step explanation:

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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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