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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)?

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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Mashcka [7]

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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.

$\bf \left.\qquad \qquad \right.\textit{negative exponents}\\\\
a^{-{ n}} \implies \cfrac{1}{a^{ n}}
\qquad \qquad
\cfrac{1}{a^{ n}}\implies a^{-{ n}}
\qquad \qquad 
a^{{{ n}}}\implies \cfrac{1}{a^{-{{ n}}}}\\\\
-------------------------------\\\\
\cfrac{15y^{-11}}{3y^{-11}}\implies \cfrac{15}{3}\cdot \cfrac{y^{-11}}{y^{-11}}\implies 5\cdot y^{-11}\cdot y^{+11}\implies 5\cdot y^{-11+11}
\\\\\\
5y^0 \implies \boxed{5}$

8 0

3 years ago

Find the value of a and of b for which<br>(a) the solution of x2 + ax < b is<br>-2<x < 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

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