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3 years ago

Write an equation of the line in slope-intercept form that passes through the point (3, 8) with slope 4?

1 answer:

8 0

Slope intercept form:
Y = mx + b
m = slope, b = y intercept
Given the slope of 4
Y = 4x + b
Plug in the point (3,8)
8 = 4(3) + b
8 = 12 + b, b = -4
Y intercept = -4

Solution: y = 4x - 4

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The height of a lighthouse in Pensacola, Florida, is 150 feet. Standing next to the statue, Natalie casts a 1.1-foot shadow. She

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30 feet

Step-by-step explanation:

Since Natalie is standing right next to the Penthouse, we can establish the ration between her height and the shadow she casts, which is;

1.1 : 5.5

The two are related by a factor of 1/5

We can use this factor to establish the shadow cast by the penthouse;

1/5 * 150

= 30

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3 years ago

11 divided by 1/3 (fraction form)

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11 / 1/3
11 * 3/1
33/1

Fraction Form: 33/1
Simplified Form: 33

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3 years ago

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How many millilitres in a centimeter

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Answer:1 millilitres

Step-by-step explanation:

One millilitres equals one centimeters

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3 years ago

seagull is flying at 125 feet above sea level. A dolphin is swimming 23 feet below sea level. What is the distance in feet betwe

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Answer:

102 feet.

Step-by-step explanation:

-23ft + 125ft = 102 ft

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3 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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3 years ago