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2 years ago

10 POINTS!!! PLease help !!!!!

Mathematics

1 answer:

NemiM [27]2 years ago

3 0

It’s only 5 points but I’ll answer the first one he makes 10$ an hour so if they want you to Wright it in a equation it’s would be y=10x+0

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Answer this and I’ll mark you the Brainlest

Elena-2011 [213]

Answer:

D. 88

Step-by-step explanation:

if you add the given numbers up (equals 100) them subtract it from 540(the degrees of a Pentagon) you get 440. then divide that by 5 (number of xs left) you get 88. You can plug in 88 for x and check but I already did, I hope this helps you lemme know if it does :)

3 0

3 years ago

Read 2 more answers

What is the value of t? 13t - 8.7 = 82.3

Dennis_Churaev [7]

Answer:

Step-by-step explanation:

13t = 82.3 + 8.7

13t = 91

t = 91/13 = 7

5 0

3 years ago

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Please help me finish these for summer school :)

ValentinkaMS [17]

13.45 will be the answer

since we are finding the hypotenuse, the formula is a^2 + b^2 = c^2

this will then be 10^2 + 9^2 = c^2

181=c^2

c=13.45 :)

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2 years ago

Read 2 more answers

Which is defined using the undefined terms point and line

____ [38]

Answer:

It's a ray because it has a point and one line that goes on in a direction.

Step-by-step explanation:

5 0

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 $\tan(\beta)$ .

8 0

1 year ago