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2 years ago

Find the missing angle.

Mathematics

2 answers:

OLEGan [10]2 years ago

8 0

Answer:

x= 127 degrees

Step-by-step explanation:

use the vertical angles theorem to say that x=90*+37*. So, x=127 degrees.

MrRa [10]2 years ago

5 0

Answer:

127

Step-by-step explanation:

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F(y) = 6y – 2 and g(y) = 3y2 + 7y – 8, find f (y) times g(y)

tamaranim1 [39]

Answer:12

Step-by-step explanation:

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2 years ago

I NEED HELP ASAPPP!!!! It’s urgentttttt!!!!

Akimi4 [234]

Answer: (x - 2)² + (y - 14)² = 1

Step-by-step explanation:

Concept:

Here, we need to know the idea of the circle formula.

Circle formula: (x - h)² + (y - k)² = r²

Center = (h, k)

Radius = r

If you are still confused, please refer to the attachment below for a graphical explanation.

Solve:

Center = (2, 14)

Radius = 1

Given formula

(x - h)² + (y - k)² = r²

Substitute the value into the formula

(x - 2)² + (y - 14)² = (1)²

Simplify

(x - 2)² + (y - 14)² = 1

Hope this helps!! :)

Please let me know if you have any questions

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3 years ago

What is the total surface area of the square pyranid?

monitta

Answer:

The general formula for the total surface area of a regular pyramid is T. S. A. =12pl+B where p represents the perimeter of the base, l the slant height and B the area of the base.

8 0

3 years ago

Read 2 more answers

Solve for L: P = 2L + 2W

natulia [17]

$P = 2L + 2W\\2L=P-2W\\L=\dfrac{P-2W}{2}$

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

8 0

2 years ago