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natima [27]
4 years ago
15

Need Answer ASAP

Mathematics
2 answers:
Minchanka [31]4 years ago
5 0

PARALLEL slopes always have the same slope no matter what because they are parallel

jonny [76]4 years ago
5 0

Answer:

y=47x-654

Step-by-step explanation:

To find the line parallel to the given linear equation, we have to use the same slope, because the condition of parallelism is that they must have the same slope.

So, the slope of the given line is m=47, because it's expressed in slope-intercept form, where the coefficient of the variable <em>x </em>is the slope.

Now, we know that they new parallel lines must have a slope equal to 47, and must pass through (14,4). Using this data, we apply the point-slope formula to find the equation of the new line:

y-y_{1}=m(x-x_{1})\\y-4=47(x-14)\\y=47x-658+4\\y=47x-654

The image attached shows the parallelism.

Therefore, the answer is y=47x-654

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3 years ago
The domain of an exponential function is _______
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4 0
3 years ago
Read 2 more answers
What is the quadratic in vertex form that has a vertex of (-1, 5)<br><br> Options in image
dimaraw [331]

Answer:

Third option

y = (x+1) ^ 2 + 5

Step-by-step explanation:

In this case we must write a quadratic equation in the vertex form.

We have the vertice. (-1, 5)

We know that the vertex form for a quadratic equation is:

y = (x-h) ^ 2 + k.

Where (h, k) is the vertex.

So if the vertex is (-1, 5), the equation sought is:

y = (x-(-1)) ^ 2 + 5

y = (x+1)) ^ 2 + 5

Therefore the answer is the Third option.

5 0
3 years ago
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which function has real zeros at x = −10 and x = −6? f(x) = x2 16x 60 f(x) = x2 − 16x 60 f(x) = x2 4x 60 f(x) = x2 − 4x 60
LiRa [457]

Answer:

Option A is correct

The function x^2+16x+60 has real zeroes at x =-10 and x =-6

Explanation:

Given: The real zeroes or roots are x = -10, and x = -6

To find the quadratic function of degree 2.

x^2- (\alpha+\beta)x + \alpha\beta =0 where α,β are real roots.   ....[1]

Here, α= -10  and β= -6

Sum of the roots:

α+β =  -10+(-6) = -10-6 = -16

Product of the roots:

αβ = (-10)(-6)= 60

Substitute these value in equation [1] we have;

x^2-(-16)x+60 = x^2+16x+60

Therefore, the quadratic function for the real roots at x =-10 and x =-6 ;

x^2+16x+60

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3 years ago
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