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2 years ago

Please help me w this an explain how you got it!.

Mathematics

1 answer:

pishuonlain [190]2 years ago

6 0

Answer:

Vertex → (2, 4)

Step-by-step explanation:

Quadratic equation has been given as,

y = -x² + 4x

We rewrite this equation in the form of a function as,

f(x) = - x² + 4x

By comparing this equation with the standard quadratic equation,

y = ax² + bx + c

a = -1 and b = 4

Vertex of the parabola represented by this equation is given by $[-\frac{b}{2a}, f(\frac{-b}{2a})]$

x coordinate = $-\frac{4}{2(-1)}$

= 2

y-coordinate = f(2)

= - (2)² + 4(2)

= -4 + 8

= 4

Therefore, vertex of the given function is (2, 4)

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3 years ago

The sum of 2112 + 2212 + 2312 when divided by 4, leaves no remainder. What will be the remainder when the sum of 2113 + 2213 + 2

trasher [3.6K]

Step-by-step explanation:

2112 + 2212 + 2312 / 4 leaves no remainder, as already the individual numbers could be divided by 4 without remainder.

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3 years ago

Simplify: (7x^2 − 8x − 4) + (−2x^2 + 3x + 5) + (5x^2 − 10x − 8)

kramer

Step 1. Simplify
7x^2 - 8x - 4 - 2x^2 + 3x + 5 + 5x^2 - 10x - 8
Step 2. Collect like terms
(7x^2 - 2x^2 + 5x^2) + (-8x + 3x - 10x) + (-4 + 5 - 8)
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5 0

3 years ago

Without plotting points, let M=(-2,-1), N=(3,1), M'= (0,2), and N'=(5, 4). Without using the distanceformula, show that segments

kramer

Given:

M=(x1, y1)=(-2,-1),

N=(x2, y2)=(3,1),

M'=(x3, y3)= (0,2),

N'=(x4, y4)=(5, 4).

We can prove MN and M'N' have the same length by proving that the points form the vertices of a parallelogram.

For a parallelogram, opposite sides are equal

If we prove that the quadrilateral MNN'M' forms a parallellogram, then MN and M'N' will be the oppposite sides. So, we can prove that MN=M'N'.

To prove MNN'M' is a parallelogram, we have to first prove that two pairs of opposite sides are parallel,

Slope of MN= Slope of M'N'.

Slope of MM'=NN'.

$\begin{gathered} \text{Slope of MN=}\frac{y2-y1}{x2-x1} \\ =\frac{1-(-1)}{3-(-2)} \\ =\frac{2}{5} \\ \text{Slope of M'N'=}\frac{y4-y3}{x4-x3} \\ =\frac{4-2}{5-0} \\ =\frac{2}{5} \end{gathered}$

Hence, slope of MN=Slope of M'N' and therefore, MN parallel to M'N'

$\begin{gathered} \text{Slope of MM'=}\frac{y3-y1}{x3-x1} \\ =\frac{4-(-1)}{5-(-2)} \\ =\frac{3}{2} \\ \text{Slope of NN'=}\frac{y4-y2}{x4-x2} \\ =\frac{4-1}{5-3} \\ =\frac{3}{2} \end{gathered}$

Hence, slope of MM'=Slope of NN' nd therefore, MM' parallel to NN'.

Since both pairs of opposite sides of MNN'M' are parallel, MM'N'N is a parallelogram.

Since the opposite sides are of equal length in a parallelogram, it is proved that segments MN and M'N' have the same length.

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1 year ago

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We have to use the distance formula between the two points provided

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3 years ago