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Alex777 [14]
3 years ago
10

Find a​ point-slope form for the line with slope

Mathematics
1 answer:
Katen [24]3 years ago
7 0

Answer:

It is Third quardant in Graph

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What is <br> 162divide[6 multiply (7-4) to the second power]<br> help please
Setler [38]

Answer:

please mark me brainlist

Step-by-step explanation:

i need it

3 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
2 years ago
10. What are the solutions to ?<br> F. -3 and 7 =<br> G. -7 and 3<br> H. -7 and -3<br> J. 3 and 7
Masja [62]
What do you mean what are the solutions to?
I don’t understand the question?
3 0
3 years ago
Read 2 more answers
2x+5y=20. , -2x+20y=-90. one solution. infinity many ir no solution. use elimination to find your answer​
scoundrel [369]
2x+5y=20
X=10-2.5y

-2(10-2.5y)+20y=-90
-20+5y+20y=-90
25y=-70
Y=2.8

2(2.8)+5y=20

5.6+5y=20
5y=14.4
Y=2.88

CHECK:
2(2.8)+5(2.88)=20

5 0
3 years ago
Ayuda necesito resolver este problema con procedimiento ;)
Paraphin [41]

x^3-2x^2+x-1 is one of the prime factors of the polynomial

<h3>How to factor the expression?</h3>

The question implies that we determine one of the prime factors of the polynomial.

The polynomial is given as:

x^8 - 3x^6 + x^4 - 2x^3 - 1

Expand the polynomial by adding 0's in the form of +a - a

x^8 - 3x^6 + x^4 - 2x^3 - 1 = x^8 -2x^7 + 2x^7 - 4x^6 +x^6 + 2x^5 -2x^5- 3x^4 + 4x^4 + 2x^3 -6x^3+2x^3- x^2  -3x^2 +4x^2-2x+2x-1

Rearrange the terms

x^8 - 3x^6 + x^4 - 2x^3 - 1 = x^8 -2x^7 + 2x^5 - 3x^4 + 2x^3 - x^2 + 2x^7 - 4x^6 + 4x^4 -6x^3+4x^2-2x+x^6-2x^5+2x^3-3x^2+2x-1

Factorize the expression

x^8 - 3x^6 + x^4 - 2x^3 - 1 = x^2(x^6-2x^5+2x^3-3x^2+2x-1) + 2x(x^6-2x^5+2x^3-3x^2+2x-1) + 1(x^6-2x^5+2x^3-3x^2+2x-1)

Factor out x^6-2x^5+2x^3-3x^2+2x-1

x^8 - 3x^6 + x^4 - 2x^3 - 1 = (x^2+2x + 1)(x^6-2x^5+2x^3-3x^2+2x-1)

Express x^2 + 2x + 1 as a perfect square

x^8 - 3x^6 + x^4 - 2x^3 - 1 = (x+1)^2(x^6-2x^5+2x^3-3x^2+2x-1)

Expand the polynomial by adding 0's in the form of +a - a

x^8 - 3x^6 + x^4 - 2x^3 - 1 = (x+1)^2(x^6- 2x^5+x^4-x^4-x^3 +x^3-2x^3-x^2 -2x^2 +x+x - 1)

Rearrange the terms

x^8 - 3x^6 + x^4 - 2x^3 - 1 = (x+1)^2(x^6- 2x^5+x^4-x^3-x^4-2x^3-x^2+x+x^3-2x^2 +x - 1)

Factorize the expression

x^8 - 3x^6 + x^4 - 2x^3 - 1 = (x+1)^2(x^3(x^3-2x^2+x-1) -x(x^3-2x^2+x-1)+1(x^3-2x^2+x-1))

Factor out x^3-2x^2+x-1

x^8 - 3x^6 + x^4 - 2x^3 - 1 = (x+1)^2(x^3 -x+1)(x^3-2x^2+x-1)

One of the factors of the above polynomial is x^3-2x^2+x-1.

This is the same as the option (c)

Hence, x^3-2x^2+x-1 is one of the prime factors of the polynomial

Read more about polynomials at:

brainly.com/question/4142886

#SPJ1

4 0
2 years ago
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