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3 years ago

HELP ME PLLEEAASSEEE

Mathematics

1 answer:

matrenka [14]3 years ago

6 0

Answer:

The relationship is that it might goes by 2

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What the answer to <br>-2v+8 (1+2v)=-90

dexar [7]

Answer:

v=2-2root7, v=2+2root7

Step-by-step explanation:

-2v+8(1+2v)=-90

-2-4v^2+8+16v=-90

6-4v^2+16v=-90

-4v^2+16v=-96

v=2-2root7, v=2+2root7

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3 years ago

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4) From a graphed linear line, how do you find the<br> Slope:<br> Y-intercept:

Ad libitum [116K]

Rise over Run

Put the amount you go up or down on the Y-axis over how much you go right or the x-axis

the Y-intercept is found simply by looking where the graph crosses the y, if the equation is written instead of graphed, graph it using a graphing calculator, then use the information I gave you previously : )

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3 years ago

The adults of a certain type of insect have a mean length of 0.6 inch. The students in a science class measured 10 insects of th

mixer [17]

Answer:

Step-by-step explanation:

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3 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

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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3 years ago

PLEASE HELP ME WITH GEO TEST!

pantera1 [17]

Answer:

The first one is ur answer

Step-by-step explanation:

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