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2 years ago

Jesse collects data on his math scores on the last 12 tests and the number of hours he spent completing his math homework

Mathematics

1 answer:

oksian1 [2.3K]2 years ago

8 0

Answer:14

Step-by-step explanation:7+7=14

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The radius of Earth is 6,378.1 km, which is 18,983.9 km less than the radius of Uranus. Find the radius of Uranus.

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Uranus has an equatorial radius of 25,559 km and a radius of 24,973 km at the poles.

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2 years ago

Slope is 0 and (0,5) is on the line<br> What is standard form please!

stellarik [79]

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3 years ago

3) If it takes 3/29 of the coal for the train to go 1 mile, how much of the coal would it take to go 6.3 miles?

nevsk [136]

Answer:

18.9/29 or 65.17% of the coal.

Step-by-step explanation:

3/29 x 6.3 = 18.9/29

18.9/29 = 0.6517 x 100 = 65.17

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3 years ago

A center has a center at X(4, -3) and a point on the circle is W(8, 0). What is the location of Y on the diameter WY?

igor_vitrenko [27]

Answer: (0, -6)

Step-by-step explanation:

The center is the midpoint of the diameter, meaning that the coordinates of Y must be (0, -6).

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2 years ago

Find the exact value of each trigonometric function for the given angle θ.

Kay [80]

Answer:

$\sin (240^\circ)=-\dfrac{\sqrt{3}}{2},\cos (240^\circ)=-\dfrac{1}{2},\tan (240^\circ)=\sqrt{3},\cot (240^\circ)=\dfrac{1}{\sqrt{3}},\sec (240^\circ)=-2,\csc (240^\circ)=\dfrac{2}{\sqrt{3}}.$

Step-by-step explanation:

The given angle is 240 degrees.

We need to find the exact value of each trigonometric function for the given angle θ.

Since $\theta=240$ , it means θ lies in 3rd quadrant. In 3d quadrant only tan and cot are positive.

$\sin (240^\circ)=\sin (180^\circ+60^\circ)=-\sin (60^\circ)=-\dfrac{\sqrt{3}}{2}$

$\cos (240^\circ)=\cos (180^\circ+60^\circ)=-\cos (60^\circ)=-\dfrac{1}{2}$

$\tan (240^\circ)=\tan (180^\circ+60^\circ)=\tan (60^\circ)=\sqrt{3}$

$\cot (240^\circ)=\cot (180^\circ+60^\circ)=\cot (60^\circ)=\dfrac{1}{\sqrt{3}}$

$\sec (240^\circ)=\sec (180^\circ+60^\circ)=-\sec (60^\circ)=-2$

$\csc (240^\circ)=\csc (180^\circ+60^\circ)=-\csc (60^\circ)=-\dfrac{2}{\sqrt{3}}$

Therefore, $\sin (240^\circ)=-\dfrac{\sqrt{3}}{2},\cos (240^\circ)=-\dfrac{1}{2},\tan (240^\circ)=\sqrt{3},\cot (240^\circ)=\dfrac{1}{\sqrt{3}},\sec (240^\circ)=-2,\csc (240^\circ)=-\dfrac{2}{\sqrt{3}}.$

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2 years ago