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Masteriza [31]
3 years ago
12

the graph of F(x), shown below in pink, has the same shape as the graph of G(x) = x^2, shown in gray. Which of the following is

the equation for F(x)?

Mathematics
2 answers:
rusak2 [61]3 years ago
7 0
The education is (x-h)^2 + k ands the vertex is (h,k)
valina [46]3 years ago
3 0

Answer:

The correct option is B.

Step-by-step explanation:

The given function is

G(x)=x^2

The gray curve represent the graph of function G(x)

The vertex form of the parabola is

F(x)=a(x-h)^2+k

Where, (h,k) is the vertex and a is stretch factor.

From the given graph is clear that the vertex of F(x) is at point (1,-2) and the stretch factor is 1. So, the function F(x) is defined as

F(x)=(x-1)^2+(-2)

F(x)=(x-1)^2-2

Therefore the correct option is B.

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Step-by-step explanation:

6 0
3 years ago
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] =<em>I</em>.

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

<em>I</em> = \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]

6 0
3 years ago
15y^-11/3y^-11, show steps I need help please don't understand how to do it.
tester [92]
One may note, you never quite asked anything per se, ahemm, if you meant simplification.

\bf \left.\qquad \qquad \right.\textit{negative exponents}\\\\&#10;a^{-{ n}} \implies \cfrac{1}{a^{ n}}&#10;\qquad \qquad&#10;\cfrac{1}{a^{ n}}\implies a^{-{ n}}&#10;\qquad \qquad &#10;a^{{{  n}}}\implies \cfrac{1}{a^{-{{  n}}}}\\\\&#10;-------------------------------\\\\&#10;\cfrac{15y^{-11}}{3y^{-11}}\implies \cfrac{15}{3}\cdot \cfrac{y^{-11}}{y^{-11}}\implies 5\cdot y^{-11}\cdot y^{+11}\implies 5\cdot y^{-11+11}&#10;\\\\\\&#10;5y^0 \implies \boxed{5}
8 0
3 years ago
Find the value of a and of b for which<br>(a) the solution of x2 + ax &lt; b is<br>-2&lt;x &lt; 4​
KengaRu [80]

Answer: a = -2

b = 8

Step-by-step explanation:

Given :

x^{2} +ax

re - writing the equation , we have

x^{2} +ax-b

we need to find the value of a and b for which -2<x < 4 , this means that the roots of the quadratic equation are -2<x < 4.

The formula for finding the quadratic equation when the roots are known is :

x^{2} - sum of roots(x) + product of root = 0

sum of roots = -2 + 4 = 2

product of roots = -2 x 4 = -8

substituting into the formula , we have:

x^{2} -2x-8=0 , which could be written in inequality form as

x^{2} -2x-8

comparing with x^{2} +ax-b , it means that :

a = -2

b = 8

3 0
3 years ago
Previous
Natali5045456 [20]

Answer: y = 5d + 7 / c - 3

Step-by-step explanation:

Step 1: Add -3y to both sides

cy - 7 + -3y = 5d + 3y + -3y

= cy - 3y - 7 = 5d

Step 2: Now, add 7 on both sides

cy - 3y - 7 +7 = 5d +7

= cy - 3y = 5d +7

Step 3: Factor out y

y(c - 3) = 5d + 7

Step 4: Divide both sides by c-3

y(c - 3) / c - 3 = 5d+7 / c-3

Your answer for this should be

y = 5d + 7 / c - 3

Hope this helped!

3 0
3 years ago
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