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3 years ago

Write the equation Of a function who parent function, f(x)=x+8 is shifted 2 units to the right

Mathematics

1 answer:

Juli2301 [7.4K]3 years ago

5 0

Answer:

g(x)=x+6

Step-by-step explanation:

we have

f(x)=x+8

If f(x) is shifted 2 units to the right

then

The rule of the translation is

g(x)=f(x-2)

g(x)=(x-2)+8

g(x)=x+6

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10x-2y=18 8x=y solve by substitution

lana [24]

Answer:

x = -1 and y = 18

Step-by-step explanation:

Given the system of equations

y + 8x = 10... 1

2y - 4x = 40 ... 2

From 1; y = 10-8x

Substitute into 2;

2(10-8x) - 4x = 40

20 - 16x - 4x = 40

20 - 20x = 40

-20x = 40 - 20

-20x = 20

x = -1

Recall that y = 10-8x

y = 10 - 8(-1)

y = 10 + 8

y = 18

therefore your answer is x = -1 and y = 18

4 0

2 years ago

Read 2 more answers

What is the solution to StartFraction 5 over 6 EndFraction x minus one-third greater-than 1 and one-third?

Vikentia [17]

Answer:

The correct answer is x > 2.

Step-by-step explanation:

$\dfrac{5}{6} x - \dfrac{1}{3} > 1\dfrac{1}{3}$

An inequality compares two quantities unlike an equality. An inequality is written with either a greater than ( > ) or lower than ( < ) or greater than equal to ( $\geq$ ) or less than equal to ( $\leq$ ) signs. We solve the above given inequality to find the solutions of the unknown x.

$\dfrac{5}{6} x - \dfrac{1}{3} > 1\dfrac{1}{3} \\\dfrac{5}{6} x - \dfrac{1}{3} > \dfrac{4}{3} \\\dfrac{5}{6} x > \dfrac{1}{3} + \dfrac{4}{3} \\\dfrac{5}{6} x > \dfrac{5}{3}\\x > 2$

Firstly we change the right hand side quantity to fraction.

We then transfer the - $\frac{1}{3}$ to the right hand side and add them. The inequality sign does not change as we are simply adding or subtracting terms from both the ends.

Finally we divide both sides with $\frac{6}{5}$ to get the required solution. The inequality sign does not change as we are multiplying both the ends with a positive quantity.