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Elodia [21]
3 years ago
7

Is y-4=2(x-2) and 4x-2y=5 perpendicular

Mathematics
1 answer:
djyliett [7]3 years ago
7 0

Answer:

Therefore,

Slopes are Equal, Hence the Lines are not Perpendicular.

Step-by-step explanation:

Given:

y-4=2(x-2)=2x=-4\\\\y=2x .......Equation ( 1 )

4x-2y=5\\2y=4x-5\\\\y=2x-\dfrac{5}{2}  .......Equation ( 2 )

Comparing with,  

y=mx+c ......General Slope point form  

Where m =slope  

We get  

Slope = m1 = 2   For Equation ( 1 )

Slope = m2 = 2   For Equation ( 2 )

Hence,

m1=m2[=2

Therefore, Slopes are Equal Hence the Lines are Parallel not Perpendicular.

For Perpendicular we require

m1\times m2=-1

Therefore,

Slopes are Equal, Hence the Lines are not Perpendicular.

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Step-by-step explanation: Theirs your help!

5 0
3 years ago
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(x^2-2)^2-10(x^2-2)+21=0 find one value of x that is a solution to the equation
Sloan [31]

Answer:

x=3; x=-3; x=√5; x=-√5

Step-by-step explanation:

6 0
3 years ago
Let f (x) = 1/x<br> and g(x) = x² – 3x. What<br> two numbers are not in the domain of fºg?
iren [92.7K]

For this case we have the following equations:

f (x) = \frac {1} {x}\\g (x) = x ^ 2-3x

We must find (f_ {o} g) (x):

By definition of composition of functions we have to:

(f_ {o} g) (x) = f (g (x))

So:

(f_ {o} g) (x) = \frac {1} {x ^ 2-3x}

We must find the domain of f (g (x)). The domain will be given by the values for which the function is defined, that is, when the denominator is nonzero.

x ^ 2-3x = 0\\x (x-3) = 0

So, the roots are:

x_ {1} = 0\\x_ {2} = 3

The domain is given by all real numbers except 0 and 3.

Answer:

x other than 0 and 3

8 0
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Answer:

linear function is a function whose graph is a line. Use the graph to write a linear function that relates y to x. The points lie on a line. Find the slope and y-intercept of the line.

Step-by-step explanation:

8 0
2 years ago
Answer quickly please
KIM [24]

The correct answer is C.


You can tell this by factoring the equation to get the zeros. To start, pull out the greatest common factor.


f(x) = x^4 + x^3 - 2x^2


Since each term has at least x^2, we can factor it out.


f(x) = x^2(x^2 + x - 2)


Now we can factor the inside by looking for factors of the constant, which is 2, that add up to the coefficient of x. 2 and -1 both add up to 1 and multiply to -2. So, we place these two numbers in parenthesis with an x.


f(x) = x^2(x + 2)(x - 1)


Now we can also separate the x^2 into 2 x's.


f(x) = (x)(x)(x + 2)(x - 1)


To find the zeros, we need to set them all equal to 0


x = 0


x = 0


x + 2 = 0

x = -2


x - 1 = 0

x = 1


Since there are two 0's, we know the graph just touches there. Since there are 1 of the other two numbers, we know that it crosses there.

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