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grandymaker [24]

3 years ago

7

Is g(x)=-2f(x)+4 a vertical shift?

1 answer:

AnnyKZ [126]3 years ago

8 0

Yes. It is a vertical shift up by 4 units.

A vertical shift will be up or down along the y-axis.

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Solve the equation. Round to the nearest hundredth. Show work.

Burka [1]

Answer:

$x=-0.75$

Step-by-step explanation:

The given equation is

$4^{-5x-7}=6^{2x-1}$

We take logarithm of both sides to base 10.

$\log(4^{-5x-7})=\log(6^{2x-1})$

$(-5x-7)\log(4)=(2x-1)\log(6)$

We expand the brackets to get;

$-5x\log(4)-7\log(4)=2x\log(6)-\log(6)$

Group similar terms;

$-7\log(4)+\log(6)=2x\log(6)+5x\log(4)$

$-7\log(4)+\log(6)=(2\log(6)+5\log(4))x$

$\frac{-7\log(4)+\log(6)}{(2\log(6)+5\log(4))}=x$

$x=-0.752478$

To the nearest hundredth.

$x=-0.75$

3 0

3 years ago

Each leg of a triangle measures 12 cm what is the length of the hypotenuse

alexira [117]

The legs are 12 cm each, so the hypotenuse is√(144+144)=12√2☺☺☺☺

6 0

3 years ago

What is the length of the hypotenuse of the triangle when x=15?

Luden [163]

Answer:

114.24°

Step-by-step explanation:

7(15)+6= 105

3(15)= 45, now we plug it in the formula, $a^{2}+b^{2}=c^{2}$

$105^{2}+45^{2} = 13050$

so now we square root it

$\sqrt{13050}$ = 114.24

6 0

3 years ago

Read 2 more answers

Which equation has the solutions x=1+or-square root of 5?

stiv31 [10]

We will proceed to solve each case to determine the solution of the problem.

<u>case a)</u> $x^{2}+2x+4=0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$x^{2}+2x=-4$

Complete the square. Remember to balance the equation by adding the same constants to each side.

$x^{2}+2x+1=-4+1$

$x^{2}+2x+1=-3$

Rewrite as perfect squares

$(x+1)^{2}=-3$

$(x+1)=(+/-)\sqrt{-3}\\(x+1)=(+/-)\sqrt{3}i\\x=-1(+/-)\sqrt{3}i$

therefore

case a) is not the solution of the problem

<u>case b)</u> $x^{2}-2x+4=0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$x^{2}-2x=-4$

Complete the square. Remember to balance the equation by adding the same constants to each side.

$x^{2}-2x+1=-4+1$

$x^{2}-2x+1=-3$

Rewrite as perfect squares

$(x-1)^{2}=-3$

$(x-1)=(+/-)\sqrt{-3}\\(x-1)=(+/-)\sqrt{3}i\\x=1(+/-)\sqrt{3}i$

therefore

case b) is not the solution of the problem

<u>case c)</u> $x^{2}+2x-4=0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$x^{2}+2x=4$

Complete the square. Remember to balance the equation by adding the same constants to each side.

$x^{2}+2x+1=4+1$

$x^{2}+2x+1=5$

Rewrite as perfect squares

$(x+1)^{2}=5$

$(x+1)=(+/-)\sqrt{5}\\x=-1(+/-)\sqrt{5}$

therefore

case c) is not the solution of the problem

<u>case d)</u> $x^{2}-2x-4=0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$x^{2}-2x=4$

Complete the square. Remember to balance the equation by adding the same constants to each side.

$x^{2}-2x+1=4+1$

$x^{2}-2x+1=5$

Rewrite as perfect squares

$(x-1)^{2}=5$

$(x-1)=(+/-)\sqrt{5}\\x=1(+/-)\sqrt{5}$

therefore

case d) is the solution of the problem

therefore

<u>the answer is</u>

$x^{2}-2x-4=0$

5 0

3 years ago

Read 2 more answers

Which value of y makes the inequality 3y^2 + 2(y - 5) >8 true?

AveGali [126]

Answer:

-3

Step-by-step explanation:

7 0

3 years ago