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3 years ago

Find the axis symmetry of x y=4x-6x+3

Mathematics

1 answer:

galina1969 [7]3 years ago

6 0

Answer:

x = 0.75.

Step-by-step explanation:

y = 4x^2 - 6x + 3.

( I have assumed that the first term is 4x^2 not 4x).

Convert to vertex form:

y = 4(x^2 - 1.5x) + 3

y = 4[ (x - 0.75)^2 - 0.5625 ] + 3

y = 4(x - 0.75)^2 - 2.25 + 3

y = 4(x - 0.75)^2 + 0.75.

So the axis of symmetry is x = 0.75.

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How do u compare rate of change and slope help plz

antoniya [11.8K]

Step-by-step explanation:

The slope is described by the the initial and final value of one quantity over the difference between the initial and final of the other

Rate: for points on a Cartesian plane,(x1, y1) and (x2, y2)

the slope, m= (y2 - y1)/(x2 - x1)

The rate of change on the other hand is mostly refers to how much a quantity is changing with time i.e the rate of change of y from y1 to y2 can be expressed as

rate = (y2 - y1)/(t2 - t1)...where t1 and t2 are the various time in y1 and y2 respectively...

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3 years ago

Help get all this points pls

balandron [24]

It would be b.) y=5x+5

8 0

3 years ago

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Solve for XX. Assume XX is a 2×22×2 matrix and II denotes the 2×22×2 identity matrix. Do not use decimal numbers in your answer.

sveticcg [70]

The question is incomplete. The complete question is as follows:

Solve for X. Assume X is a 2x2 matrix and I denotes the 2x2 identity matrix. Do not use decimal numbers in your answer. If there are fractions, leave them unevaluated.

$\left[\begin{array}{cc}2&8\\-6&-9\end{array}\right]$ · X· $\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]$ =I.

First, we have to identify the matrix I. As it was said, the matrix is the identiy matrix, which means

I = $\left[\begin{array}{ccc}1&0\\0&1\end{array}\right]$

So, $\left[\begin{array}{cc}2&8\\-6&-9\end{array}\right]$ · X· $\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]$ = $\left[\begin{array}{ccc}1&0\\0&1\end{array}\right]$

Isolating the X, we have

X· $\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]$ = $\left[\begin{array}{cc}2&8\\-6&-9\end{array}\right]$ - $\left[\begin{array}{ccc}1&0\\0&1\end{array}\right]$

Resolving:

X· $\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]$ = $\left[\begin{array}{ccc}2-1&8-0\\-6-0&-9-1\end{array}\right]$

X· $\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]$ = $\left[\begin{array}{ccc}1&8\\-6&-10\end{array}\right]$

Now, we have a problem similar to A.X=B. To solve it and because we don't divide matrices, we do X=A⁻¹·B. In this case,

X= $\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]$ ⁻¹· $\left[\begin{array}{ccc}1&8\\-6&-10\end{array}\right]$

Now, a matrix with index -1 is called Inverse Matrix and is calculated as: A . A⁻¹ = I.

So,

$\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]$ · $\left[\begin{array}{ccc}a&b\\c&d\end{array}\right]$ = $\left[\begin{array}{ccc}1&0\\0&1\end{array}\right]$

9a - 3b = 1

7a - 6b = 0

9c - 3d = 0

7c - 6d = 1

Resolving these equations, we have a= $\frac{2}{11}$ ; b= $\frac{7}{33}$ ; c= $\frac{-1}{11}$ and d= $\frac{-3}{11}$ . Substituting:

X= $\left[\begin{array}{ccc}\frac{2}{11} &\frac{-1}{11} \\\frac{7}{33}&\frac{-3}{11} \end{array}\right]$ · $\left[\begin{array}{ccc}1&8\\-6&-10\end{array}\right]$

Multiplying the matrices, we have

X= $\left[\begin{array}{ccc}\frac{8}{11} &\frac{26}{11} \\\frac{39}{11}&\frac{198}{11} \end{array}\right]$

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3 years ago

Solve the equation: c + 4c = -90

Liono4ka [1.6K]

Answer:

c= -18

Step-by-step explanation:

c+4c= -90

combine like terms

5c= -90

Divide by 5 on both sides

c= -18

6 0

2 years ago

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You fly a hot air balloon 3.5 miles above the ground. What is the measure of BD⌢, the portion of Earth that you can see? Round y

STatiana [176]

Answer:

$\Delta s \approx 334.545\,m$

Step-by-step explanation:

Let assume that Earth is a sphere, the following trigonometric diagram is constructed and presented below. The central angle is given by this inverse trigonometric equation:

$\theta = 2\cdot \cos^{-1} \left(\frac{r}{r+h} \right)$

$\theta = 2\cdot \cos^{-1}\left(\frac{4000\,mi}{4000\,mi+3.5\,mi} \right)$

$\theta = 4.792^{\circ}$

The distance of the portion of Earth that can be seen is:

$\Delta s = \frac{4.792^{\circ}}{360^{\circ}}\cdot (2\pi)\cdot (4000\,mi)$

$\Delta s \approx 334.545\,m$

7 0

3 years ago