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3 years ago

Line l has a slope of 1.2. What is the slope of line p that is parallel to line l

Mathematics

1 answer:

Katena32 [7]3 years ago

3 0

The slope would be 1.2. If they are parallel lines, they have the same slope. :)

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Salina spent 100 minutes finishing her homework last night. She spent 23% of her time completing her math homework. Which model

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3 years ago

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6. Let A and B be nxn matrices . Compute (A + B) (A + B). Explain all steps.

Pie

Answer:

(A+B)(A+B)=A.A+B.A+A.B+B.B

Step-by-step explanation:

Given that matrices A and B are nxn matrices

We need to find (A+B)(A+B)

For understanding the multiplication of matrices let'take A is mxn and B is pxq matrices,we can multiple only when n=p,so our Ab matrices will be mxq.

We know that that in matrices AB is not equal to BA.

Now find

(A+B)(A+B)=A.A+B.A+A.B+B.B

So from we can say that (A+B)(A+B) is not equal to A.A+2B.A+B.B because AB is not equal to BA in matrices.

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3 years ago

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2 years ago

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Find the equation, (f(x) = a(x - h)2 + k), for a parabola containing point (2, -1) and having (4, -3) as a vertex. What is the s

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Answer:

$f(x)=\frac{1}{2}x^2-4x+5$

Step-by-step explanation:

A parabola is written in the form

$f(x)=a((x-h)^2+k)$ (1)

where:

$h$ is the x-coordinate of the vertex of the parabola

$ak$ is the y-coordinate of the vertex of the parabola

$a$ is a scale factor

For the parabola in the problem, we know that the vertex has coordinates (4,-3), so we have:

$h=4$ (2)

$ak=-3$

From this last equation, we get that $a=\frac{-3}{k}$ (3)

Substituting (2) and (3) into (1) we get the new expression:

$f(x)=-\frac{3}{k}((x-4)^2+k) = -\frac{3}{k}(x-4)^2 -3$ (4)

We also know that the parabola contains the point (2,-1), so we can substitute

x = 2

f(x) = -1

Into eq.(4) and find the value of k:

$-1=-\frac{3}{k}(2-4)^2-3\\-1=-\frac{3}{k}\cdot 4 -3\\2=-\frac{12}{k}\\k=-\frac{12}{2}=-6$

So we also get:

$a=-\frac{3}{k}=-\frac{3}{-6}=\frac{1}{2}$

So the equation of the parabola is:

$f(x)=\frac{1}{2}((x-4)^2 -6)$ (5)

Now we want to rewrite it in the standard form, i.e. in the form

$f(x)=ax^2+bx+c$

To do that, we simply rewrite (5) expliciting the various terms, we find:

$f(x)=\frac{1}{2}((x^2-8x+16)-6)=\frac{1}{2}(x^2-8x+10)=\frac{1}{2}x^2-4x+5$

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Find the distance between the lines with equations -4x + 7y + 9 = 0 and y = 4/7x - 9

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It’s 6.72 you add the two numbers

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