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2 years ago

Answer 11-12 it’s ok if your don’t no one of them. Please do at least one. Thanks.

Mathematics

1 answer:

Naddika [18.5K]2 years ago

3 0

[11]

Equation: 180° - 75°

Measure: 105°

A straight line has an angle measurement of 180°. Since we have GFB, 75°, we can find EFB this way.

[12]

Equation: 90° - 40°

Measure: 50°

The little box in the angle shows that it is equal to 90°. Since they give us HGB, 40°, we can solve for EFB using this information.

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17.69
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3 years ago

Which two numbers below have the least common multiple of 48?

Tems11 [23]

Answer:

D. 3 and 16

Step-by-step explanation:

A common multiple is the result of multiplication of two numbers

3x16= 48

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2 years ago

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Sophie and Marcos are each saving for new bicycles. So far, Sophie has $6 saved and can earn $5 per hour walking dogs. Marcos ha

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It will take Marco and Sophie 3 hours to get the same amount of money. The amount of money they will have will be $21.

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2 years ago

Considering only the values of β for which sinβtanβsecβcotβ is defined, which of the following expressions is equivalent to sinβ

-Dominant- [34]

Answer:

$\tan(\beta)$

Step-by-step explanation:

For many of these identities, it is helpful to convert everything to sine and cosine, see what cancels, and then work to build out to something. If you have options that you're building toward, aim toward one of them.

${\tan(\theta)}={\dfrac{\sin(\theta)}{\cos(\theta)}$ and ${\sec(\theta)}={\dfrac{1}{\cos(\theta)}$

Recall the following reciprocal identity:

$\cot(\theta)=\dfrac{1}{\tan(\theta)}=\dfrac{1}{ \left ( \dfrac{\sin(\theta)}{\cos(\theta)} \right )} =\dfrac{\cos(\theta)}{\sin(\theta)}$

So, the original expression can be written in terms of only sines and cosines:

$\sin(\beta)\tan(\beta)\sec(\beta)\cot(\beta)$

$\sin(\beta) * \dfrac{\sin(\beta) }{\cos(\beta) } * \dfrac{1 }{\cos(\beta) } * \dfrac{\cos(\beta) } {\sin(\beta) }$

$\sin(\beta) * \dfrac{\sin(\beta) \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{---}}{\cos(\beta) \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{---}} * \dfrac{1 }{\cos(\beta) } * \dfrac{\cos(\beta) \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{---}} {\sin(\beta) \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{---}}$

$\sin(\beta) *\dfrac{1 }{\cos(\beta) }$

$\dfrac{\sin(\beta)}{\cos(\beta) }$

Working toward one of the answers provided, this is the tangent function.

The one caveat is that the original expression also was undefined for values of beta that caused the sine function to be zero, whereas this simplified function is only undefined for values of beta where the cosine is equal to zero. However, the questions states that we are only considering values for which the original expression is defined, so, excluding those values of beta, the original expression is equivalent to $\tan(\beta)$ .

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2 years ago

Complete the table to show the steps for combining like terms

Fantom [35]

Answer:

check the illustration I provided

Step-by-step explanation:

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2 years ago