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3 years ago

Please help me with this question

Mathematics

1 answer:

gayaneshka [121]3 years ago

5 0

9514 1404 393

Answer:

Step-by-step explanation:

The leading term is of odd degree and negative coefficient. The odd degree tells you the end behaviors will have opposite signs. The negative coefficient tells you the left end behavior will be positive, and the right end behavior will be negative.

$\textbf{D. }\lim\limits_{x\to -\infty}f(x)=\infty\quad\lim\limits_{x\to \infty}f(x)=-\infty$

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Triangle GHJ with vertices G(-2, 4), H(3, 6), and J(3, -2) is dilated by a factor of 1/3 centered at the origin. What are the co

Grace [21]

Answer:

The coordinates of G' are $G'(x,y) = \left(-\frac{2}{3}, \frac{4}{3} \right)$ .

Step-by-step explanation:

From Linear Algebra, we define the dilation of point G by the following expression:

$G'(x,y) = O(x,y) + k\cdot [G(x,y)-O(x,y)]$ (1)

Where:

$O(x,y)$ - Center of dilation, dimensionless.

$k$ - Scale factor, dimensionless.

$G(x,y)$ - Coordinates of point G, dimensionless.

$G'(x,y)$ - Coordinates of point G', dimensionless.

If we know that $O(x,y) = (0,0)$ , $k = \frac{1}{3}$ , $G(x,y) = (-2,4)$ , then the point G' is:

$G'(x,y) = (0,0) + \frac{1}{3}\cdot [(-2,4)-(0,0)]$

$G'(x,y) = \left(-\frac{2}{3}, \frac{4}{3} \right)$

The coordinates of G' are $G'(x,y) = \left(-\frac{2}{3}, \frac{4}{3} \right)$ .

7 0

3 years ago

2/9 dived 1/2 helpppp

kupik [55]

Answer:

.44 or 11/25

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Pls solve this with step by step.

MAVERICK [17]

Step-by-step explanation:

-x+5+6x-7x-14

6x-8x+5-14

-2x-9

3 0

2 years ago

Decimal modle 0.5×1.7

NikAS [45]

You mean the product of 0.5 and 1.7 which is 0.85.

8 0

4 years ago

y = x2 − 6x 8 2y x = 4 the pair of points representing the solution set of this system of equations is___ and ___?

mariarad [96]

Some operators are missing.
If the right equations are y = x^2 -6x +8 and 2y + x =4
This is the solution

2y + x = 4
2y = -x +4
y = -x/2 + 2

y = x^2 -6x + 8 = -x/2 + 2

2x^2 - 12x + 16 = -x + 4
2x^2 -11x +12 = 0

Factor

(x-4)(2x-3)=0

x= 4 and x =3/2

y = -x/2 +2

y = -4/2 + 2 = 0
y = - (3/2)/2 + 2 = -3/4 + 2 = 5/4

Answer (4,0) and (3/2, 5/4)

3 0

3 years ago

Please help me with this question ​

Please help me with this question