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salantis [7]
3 years ago
11

How to write y= 2x^2 + 8x +7 in standard (vertex) form

Mathematics
1 answer:
Anni [7]3 years ago
3 0
2x^2+8x+7= \\
2(x^2+4x)+7= \\
2(x^2+4x+4-4)+7= \\
2((x+2)^2-4)+7= \\
2(x+2)^2-8+7= \\
2(x+2)^2-1 \\ \\
\boxed{y=2(x+2)^2-1}
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I need to know how to do this, plz explain-Which function grows at the fastest rate for increasing values of x?
saul85 [17]
If the first function is
.. h(x) = 2^x
it will be the one that grows the fastest. The exponential function with the largest base value will grow faster than any polynomial.

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For khan academy . Need answer immediately
kolbaska11 [484]

It's easy just use y=mx+b form ok?

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If y= 14 when x=8, find y when x=12.
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Triangle PQR is transformed to triangle P'Q'R'. Triangle PQR has vertices P(4, 0), Q(0, −4), and R(−8, −4). Triangle P'Q'R' has
Whitepunk [10]

Answer:

Please find attached the required plot accomplished with an online tool

Part A:

1/4

Part B:

P''(-1, 0),  Q''(0, -1), and R''(2, -1)

Part C:

Triangle PQR is similar to triangle P''Q''R'' but they are not congruent

Step-by-step explanation:

Part A:

Triangle ΔPQR has vertices P(4, 0), Q(0, -4), R(-8, -4)

Triangle ΔP'Q'R' has vertices P'(1, 0), Q'(0, -1), R'(-2, -1)

The dimensions of the sides of the triangle are given by the relation;

l = \sqrt{\left (y_{2}-y_{1}  \right )^{2}+\left (x_{2}-x_{1}  \right )^{2}}

Where;

(x₁, y₁) and (x₂, y₂) are the coordinates on the ends of the segment

For segment PQ, we place (x₁, y₁) = (4, 0) and (x₂, y₂) = (0, -4);

By substitution into the length equation, we get;

The length of segment PQ = 4·√2

The length of segment PR = 4·√10

The length of segment RQ = 8  

The length of segment P'Q' = √2

The length of segment P'R' = √10

The length of segment R'Q' = 2

Therefore, the scale factor of the dilation of ΔPQR to ΔP'Q'R' is 1/4

Part B:

Reflection of (x, y) across the y-axis gives;

(x, y) image after reflection across the y-axis = (-x, y)

The coordinates after reflection of P'(1, 0), Q'(0, -1), R'(-2, -1) across the y-axis is given as follows;

P'(1, 0) image after reflection across the y-axis = P''(-1, 0)

Q'(0, -1) image after reflection across the y-axis = Q''(0, -1)

R'(-2, -1) image after reflection across the y-axis = R''(2, -1)

Part C:

Triangle PQR is similar to triangle P''Q''R'' but they are not congruent as the dimensions of the sides of triangle PQR and P''Q''R'' are not the same.

6 0
4 years ago
Find the equation of the line through (- 4, 5) which is perpendicular to the line y = x/4 - 1
iragen [17]

We conclude that the linear equation perpendicular to y = x/4 - 1 and that passes through (-4, 5) is y = -4*x - 11

<h3>How to find the perpendicular line?</h3>

A general linear equation is of the form:

y = a*x + b

Where a is the slope and b is the y-intercept.

Two lines are perpendicular if the slope of one line is equal to the inverse of the opposite of the other slope.

So, if we have:

y = x/4 - 1

The slope is a = (1/4)

The inverse of the opposite is:

a' = -4

Then the perpendicular line is of the form:

y = -4*x + c

Now we want this to pass through the point (-4, 5), then we must have:

5 = -4*-4 + c

5 = 16 + c

5 - 16 = c = -11

We conclude that the linear equation perpendicular to y = x/4 - 1 and that passes through (-4, 5) is y = -4*x - 11

If you want to learn more about linear equations:

brainly.com/question/1884491

#SPJ1

4 0
2 years ago
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