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laila [671]
3 years ago
15

I need help with this, can someone help me?

Mathematics
1 answer:
Alinara [238K]3 years ago
5 0
The answer is C cause if you rotate it the shape looks different but if it had rotational symmetry it could be rotated less than 360 degrees and still look the same
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What is the area??? I don't understand I have tried so hard.
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8 * 5.7 = 45.6


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1 represent 3 days , 2 represent 6 days , 3 represent 9 days , and 4 represent 12 days
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3 years ago
On a sunny day, Larry notices that his shadow is 24 inches long. If it is 74 inches from the top of his head to the end of the s
kicyunya [14]

Answer:

70 inches

Step-by-step explanation:

Larry forms a right triangle with the ground (this is one leg of a right triangle) and he is 24 inches from his shadow (this is the second leg).

74 inches is the hypotenuse of the right triangle

use pythagorean theorem:  a^2 + b^2 = c^2

let h = Larry's height

h^2 + 24^2 = 74^2

h^2 + 576 = 5476

h^2 = 4900

h = sq rt 4900

h = 70

5 0
2 years ago
Verify that:
Lelu [443]

Answer:

See Below.

Step-by-step explanation:

Problem 1)

We want to verify that:

\displaystyle \left(\cos(x)\right)\left(\cot(x)\right)=\csc(x)-\sin(x)

Note that cot(x) = cos(x) / sin(x). Hence:

\displaystyle \left(\cos(x)\right)\left(\frac{\cos(x)}{\sin(x)}\right)=\csc(x)-\sin(x)

Multiply:

\displaystyle \frac{\cos^2(x)}{\sin(x)}=\csc(x)-\sin(x)

Recall that Pythagorean Identity: sin²(x) + cos²(x) = 1 or cos²(x) = 1 - sin²(x). Substitute:

\displaystyle \frac{1-\sin^2(x)}{\sin(x)}=\csc(x)-\sin(x)

Split:

\displaystyle \frac{1}{\sin(x)}-\frac{\sin^2(x)}{\sin(x)}=\csc(x)-\sin(x)

Simplify:

\csc(x)-\sin(x)=\csc(x)-\sin(x)

Problem 2)

We want to verify that:

\displaystyle (\csc(x)-\cot(x))^2=\frac{1-\cos(x)}{1+\cos(x)}

Square:

\displaystyle \csc^2(x)-2\csc(x)\cot(x)+\cot^2(x)=\frac{1-\cos(x)}{1+\cos(x)}

Convert csc(x) to 1 / sin(x) and cot(x) to cos(x) / sin(x). Thus:

\displaystyle \frac{1}{\sin^2(x)}-\frac{2\cos(x)}{\sin^2(x)}+\frac{\cos^2(x)}{\sin^2(x)}=\frac{1-\cos(x)}{1+\cos(x)}

Factor out the sin²(x) from the denominator:

\displaystyle \frac{1}{\sin^2(x)}\left(1-2\cos(x)+\cos^2(x)\right)=\frac{1-\cos(x)}{1+\cos(x)}

Factor (perfect square trinomial):

\displaystyle \frac{1}{\sin^2(x)}\left((\cos(x)-1)^2\right)=\frac{1-\cos(x)}{1+\cos(x)}

Using the Pythagorean Identity, we know that sin²(x) = 1 - cos²(x). Hence:

\displaystyle \frac{(\cos(x)-1)^2}{1-\cos^2(x)}=\frac{1-\cos(x)}{1+\cos(x)}

Factor (difference of two squares):

\displaystyle \frac{(\cos(x)-1)^2}{(1-\cos(x))(1+\cos(x))}=\frac{1-\cos(x)}{1+\cos(x)}

Factor out a negative from the first factor in the denominator:

\displaystyle \frac{(\cos(x)-1)^2}{-(\cos(x)-1)(1+\cos(x))}=\frac{1-\cos(x)}{1+\cos(x)}

Cancel:

\displaystyle \frac{\cos(x)-1}{-(1+\cos(x))}=\frac{1-\cos(x)}{1+\cos(x)}

Distribute the negative into the numerator. Therefore:

\displaystyle \frac{1-\cos(x)}{1+\cos(x)}=\displaystyle \frac{1-\cos(x)}{1+\cos(x)}

3 0
3 years ago
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