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3 years ago

15

Work backward to solve. What is the starting position (x, y)?

1 answer:

levacccp [35]3 years ago

6 0

14,4 because if you are working backwards, you’re going to do the opposite each time and 6,8 was the point after only TWO of the transformations

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What is the range of the function y = x 2?

Leno4ka [110]

Answer:

$\mathrm{Range\:of\:}x^2:\quad \begin{bmatrix}\mathrm{Solution:}\:&\:f\left(x\right)\ge \:0\:\\ \:\mathrm{Interval\:Notation:}&\:[0,\:\infty \:)\end{bmatrix}$

The graph is also attached below.

Step-by-step explanation:

Given the function

$y=x^2$

We know that the range of a function is the set of values of the dependent variable for which a function is defined.

$\mathrm{For\:a\:parabola}\:ax^2+bx+c\:\mathrm{with\:Vertex}\:\left(x_v,\:y_v\right)$

$\mathrm{If}\:a$

$\mathrm{If}\:a>0\:\mathrm{the\:range\:is}\:f\left(x\right)\ge \:y_v$

$a=1,\:\mathrm{Vertex}\:\left(x_v,\:y_v\right)=\left(0,\:0\right)$

$f\left(x\right)\ge \:0$

Thus,

$\mathrm{Range\:of\:}x^2:\quad \begin{bmatrix}\mathrm{Solution:}\:&\:f\left(x\right)\ge \:0\:\\ \:\mathrm{Interval\:Notation:}&\:[0,\:\infty \:)\end{bmatrix}$

The graph is also attached below.

5 0

3 years ago

What is the cube of 3

Sladkaya [172]

Answer:

1.44 or $\sqrt[3]{3}$

Step-by-step explanation:

6 0

3 years ago

PLZ HELP FAST!!

bazaltina [42]

The trapezoid's area is 126 inch².

Step-by-step explanation:

Step 1:

The trapezoid's area is calculated by averaging the base lengths and multiplying it with the trapezoid's height.

The trapezoid's area, $A = \frac{b_{1}+b_{2}}{2} h$ .

Here $b_{1}$ is the lower base and $b_{2}$ is the upper base, height is h.

Step 2:

In the given diagram,

$b_{1} = 12$ inches and $b_{2}= 2+12+2=16$ inches and h = 9 inches.

$A = \frac{b_{1}+b_{2}}{2} h = \frac{12+16}{2} (9) = 14(9)= 126$ inch².

So the given trapezoid's area is 126 inch².

8 0

4 years ago

Kindly answer it thanks

yawa3891 [41]

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5 0

3 years ago

Let the vector \mathbf{v}v have an initial point at (1, -2)(1,−2) and a terminal point at (0, 0)(0,0). Plot the vector \mathbf{v

ASHA 777 [7]

Given that the initial point of the vector is (1,-2) and the termination point of the vector is (0,0).

So, the tail of the vector is at point (1,-2) and the head of the vector is at (0,0). The vector, $\vec{v}$ , has been shown in the figure.

The magnitude of the vector:

$|\vec{v}|=\sqrt{(1-0)^2+(-2-0)^2}=\sqrt{1+4}=\sqrt{5}$

The direction of a vector is the angle made by a vector with the positive direction of the x-axis.

From the figure, $\theta = 180 \degree\tan ^{-1} \left|\frac {0-(-2)}{0-1}\right|$

$\theta = 180 ^{\circ} - \tan^{-1}(2) = 180 ^{\circ} - 63.43^{\circ}$

$\theta =116.57 ^{\circ}$

Hence, the required vector having a magnitude $\sqrt {5}$ and direction $116.57 ^{\circ}$ has been shown in the figure.

4 0

4 years ago