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Kamila [148]
3 years ago
7

The graph of F(x), shown below, has the same shape as the graph of G(x) = x2, but it is shifted to the left 3 units. What is its

equation?

Mathematics
2 answers:
Tems11 [23]3 years ago
6 0

Answer:

f(x) = (x - 3)²

Step-by-step explanation:

If the function given below is g(x) = x²

Then we have to find the function f(x) which is same as g(x) but shifted 3 units left.

As we know in the given equation f(x) = (a + x)²+ k, x defines the shift of the function and and k defines the shift of a graph on y-axis upwards or downwards.

Here the graph is shifted left so the value of x will be (-3) which defines the shifting of 3 left on x-axis.

Therefore the new function will be f(x) = (x - 3)²

Korvikt [17]3 years ago
5 0

Answer:

F(x)=(x+3)^2

Step-by-step explanation:

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. Given that O is the centre of the following circle, find the values of the unknowns.​
alex41 [277]

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To find value of 'a', we will use Pythagoras theorem as the triangle is a right angle triangle,

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Happy to help :)

If u need any help feel free to ask

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7 0
3 years ago
Please help me with this​
denis23 [38]

Answer:

20) \displaystyle [4, 1]

19) \displaystyle [-5, 1]

18) \displaystyle [3, 2]

17) \displaystyle [-2, 1]

16) \displaystyle [7, 6]

15) \displaystyle [-3, 2]

14) \displaystyle [-3, -2]

13) \displaystyle NO\:SOLUTION

12) \displaystyle [-4, -1]

11) \displaystyle [7, -2]

Step-by-step explanation:

20) {−2x - y = −9

{5x - 2y = 18

⅖[5x - 2y = 18]

{−2x - y = −9

{2x - ⅘y = 7⅕ >> New Equation

__________

\displaystyle \frac{-1\frac{4}{5}y}{-1\frac{4}{5}} = \frac{-1\frac{4}{5}}{-1\frac{4}{5}}

\displaystyle y = 1[Plug this back into both equations above to get the x-coordinate of 4]; \displaystyle 4 = x

_______________________________________________

19) {−5x - 8y = 17

{2x - 7y = −17

−⅞[−5x - 8y = 17]

{4⅜x + 7y = −14⅞ >> New Equation

{2x - 7y = −17

_____________

\displaystyle \frac{6\frac{3}{8}x}{6\frac{3}{8}} = \frac{-31\frac{7}{8}}{6\frac{3}{8}}

\displaystyle x = -5[Plug this back into both equations above to get the y-coordinate of 1]; \displaystyle 1 = y

_______________________________________________

18) {−2x + 6y = 6

{−7x + 8y = −5

−¾[−7x + 8y = −5]

{−2x + 6y = 6

{5¼x - 6y = 3¾ >> New Equation

____________

\displaystyle \frac{3\frac{1}{4}x}{3\frac{1}{4}} = \frac{9\frac{3}{4}}{3\frac{1}{4}}

\displaystyle x = 3[Plug this back into both equations above to get the y-coordinate of 2]; \displaystyle 2 = y

_______________________________________________

17) {−3x - 4y = 2

{3x + 3y = −3

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\displaystyle \frac{-y}{-1} = \frac{-1}{-1}

\displaystyle y = 1[Plug this back into both equations above to get the x-coordinate of −2]; \displaystyle -2 = x

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16) {2x + y = 20

{6x - 5y = 12

−⅓[6x - 5y = 12]

{2x + y = 20

{−2x + 1⅔y = −4 >> New Equation

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\displaystyle \frac{2\frac{2}{3}y}{2\frac{2}{3}} = \frac{16}{2\frac{2}{3}}

\displaystyle y = 6[Plug this back into both equations above to get the x-coordinate of 7]; \displaystyle 7 = x

_______________________________________________

15) {6x + 6y = −6

{5x + y = −13

−⅚[6x + 6y = −6]

{−5x - 5y = 5 >> New Equation

{5x + y = −13

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\displaystyle \frac{-4y}{-4} = \frac{-8}{-4}

\displaystyle y = 2[Plug this back into both equations above to get the x-coordinate of −3]; \displaystyle -3 = x

_______________________________________________

14) {−3x + 3y = 3

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{−5x + y = 13

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13) {−3x + 3y = 4

{−x + y = 3

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{x - y = −1⅓ >> New Equation

{−x + y = 3

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\displaystyle 1\frac{2}{3} ≠ 0; NO\:SOLUTION

_______________________________________________

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_______________________________________________

11) {x + 3y = 1

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\displaystyle \frac{-2x}{-2} = \frac{-14}{-2}

\displaystyle x = 7[Plug this back into both equations above to get the y-coordinate of −2]; \displaystyle -2 = y

I am delighted to assist you anytime my friend!

7 0
3 years ago
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