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podryga [215]
3 years ago
13

Whats the answer to the problem in the picture

Mathematics
1 answer:
vesna_86 [32]3 years ago
5 0

Answer:

2/6

1/6

-------

1/6

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Circumference is 2πr, and the large cake has a radius 3 times the radius of the small cake. So, the circumference of the large cake will be 3 times that of the small cake.
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On the day a coffee shop first opened, it had 140 customers. 65% of the customers ordered a large coffee.
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140/100 * 65 = 91 people
Divide 140 by 100 to find 1 percent then multiply by 65 to find 65 percent.
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The set of all months that contain less than 31 years
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The set of all months that contains less than 31 days is; {February, April, June, September, November}.

<h3>What is the set of all months that contain less than 31 days?</h3>

It follows from the task content that the set whose elements are to be determined is characterized by containing less than 31 days.

On this note, it follows that the elements of the set are; February (28/29 days), April (30 days), September (30 days), June (30 days) and November (30 days).

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2 years ago
Somebody please try and help me with #9
Maurinko [17]
First, you subtract the equations. (3x^2-2x+5) - (x^2+3x-2) so you get 3x^2 - 2x + 5 - x^2 - 3x + 2. This leads you to 2x^2 - 5x + 7. If you multiply that by 1/2 x^2, you get x^4 - 5/2(x^3) + 7/2(x^2)
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3 years ago
the point (3,-6) lies on the terminal side of angle 0. Find the exact value of the six trigonometric functions of 0. Sin cos tan
Shkiper50 [21]

ANSWER

See explanation

EXPLANATION

The point (3,-6) is the fourth quadrant.

In this quadrant only the cosine ratio and the secant ratio are positive.

The remaining four trigonometric ratios are negative.

The diagram is shown in the attachment.

We use the Pythagoras Theorem to find the hypotenuse, h.

{h}^{2}  =  {6}^{2}  +  {3}^{2}

{h}^{2}  = 36 + 9

{h}^{2}  = 45

h =  \sqrt{45}

h = 3 \sqrt{5}

\sin( \theta)  =   -  \frac{opp}{hyp}

\sin( \theta)  =   -  \frac{6}{3 \sqrt{5} }

\sin( \theta)  =   -  \frac{2 \sqrt{5} }{ 5 }

\sin( \theta)  =   -  \frac{2}{ \sqrt{5} }

\ cos(\theta)  =   \frac{adj}{hyp}

\ cos(\theta)  =   \frac{3}{3 \sqrt{5} }

\ cos(\theta)  =   \frac{1}{ \sqrt{5} }

\ cos(\theta)  =   \frac{ \sqrt{5} }{5}

\tan(\theta)= -  \frac{opp}{hyp}

\tan(\theta)= -  \frac{6}{3}  =  - 2

\csc(\theta)= -  \frac{hyp}{opp}

\csc(\theta)= -  \frac{3 \sqrt{5} }{6}  =  -   \frac{\sqrt{5} }{2}

\sec( \theta)  =   \frac{hyp}{adj}

\sec( \theta)  = \frac{3 \sqrt{5} }{3}  =  \sqrt{5}

\cot( \theta)  =  -  \frac{adj}{opp}

\cot( \theta)  =   - \frac{3}{6}  =  - \frac{1}{2}

6 0
3 years ago
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