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3 years ago

PLEASE HELP!!!! I WILL GIVE BRAINLIEST!!!!!

Mathematics

2 answers:

Bad White [126]3 years ago

3 0

2^x-3

im taking this test today as well, and i have the answer to this question on the study guide so this is right.

marta [7]3 years ago

3 0

Answer: g(x) = $2^{x-3}$ .

Step-by-step explanation:

We are given graph of f(x) and g(x).

f(x) = $2^x$ .

Let us observe the coordinates of f(x) and g(x).

(0,1) ---> (3,1)

(1,2) ---> (4,2)

(2,4) ---> (5,4)

From the above coordinates we can see than coordinates of g(x) are being shifted 3 units right of the coordinates of f(x).

So, the transformation is being applied f(x-3).

According to rules of transformation, y=f(x-c) function f(x) gets transfer c units right.

Therefore, we need to replace x by x-3 in the exponent of 2 of f function.

Therefore, g(x) = $2^{x-3}$ .

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3 years ago

Can anybody explain how to turn percentages to fractions or fractions to percentages, and diving and dividing decimals and study

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3 years ago

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Is the following statement TRUE or FALSE? Select the best choice from the following to justify your

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3 years ago

Match the circle equations in general form with their corresponding equations in standard form. Not all will be used.

Xelga [282]

The standard form of the equation of a circumference is given by the following expression:

 $(x-h)^{2}+(y-k)^{2}=r^{2} \\ \\ where \ (h, k) \ is \ the \ center \ of \ the \ circumference \ and \ r \ the \ radius$

On the other hand, the general form is given as follows:

 $x^{2}+y^{2}+Dx+Ey+F=0 \\ \\ where: \\ D=-2h, \ E=-2k, \ F=h^{2}+k^{2}-r^{2}$ 

In this way, we can order the mentioned equations as follows:

Equations in Standard Form:

 $\bold{a)} \ (x-6)^{2}+(y-4)^{2}=56 \\ \bold{b)} \ (x-2)^{2} + (y+6)^{2}=60 \\ \bold{c)} \ (x+2)^{2}+(y+3)^{2}=18 \\ \bold{d)} \ (x+1)^{2}+(y-6)^{2}=46$

Equations in General Form:

$\bold{1)} \ x^{2}+y^{2}-4x+12y-20=0 \\ \bold{2)} \ x^{2}+y^{2}+6x-8y-10=0 \\ \bold{3)} \ 3x^{2}+3y^{2}+12x+18y-15=0 \\ \\ If \ we \ divide \ this \ equation \ by \ 3, \ the \ equation \ becomes: \\ x^{2}+y^{2}+4x+6y-5=0 \\ \\ \bold{4)} \ 5x^{2}+5y^{2}-10x+20y-30=0 \\ \\ If \ we \ divide \ this \ equation \ by \ 5, \ the \ equation \ becomes: \\ x^{2}+y^{2}-2x+4y-6=0 \\ \\ \bold{5)} \ 2x^{2}+2y^{2}-24x-16y-8=0 \\ \\ If \ we \ divide \ this \ equation \ by \ 2, \ the \ equation \ becomes: \\ x^{2}+y^{2}-12x-8y-4=0$

$\bold{6)} \ x^{2}+y^{2}+2x-12y$

So let's match each equation:

$\bold{From \ a)} \\ \\ (h,k)=(6,4),\ r=2\sqrt{14} \\ D=-12, \ E=-8 \\ F=-4$

Then, its general form is:

$x^{2}+y^{2}-12x-8y-4=0$

First. a) matches 5) 

$\bold{From \ b)} \\ \\ (h,k)=(2,-6),\ r=2\sqrt{15} \\ D=-4, \ E=12 \\ F=-20$

Then, its general form is:

$x^{2}+y^{2}-4x+12y-20=0$

Second. b) matches 1) 

$\bold{From \ c)} \\ \\ (h,k)=(-2,-3),\ r=3\sqrt{2} \\ D=4, \ E=6 \\ F=-5$

Then, its general form is:

$x^{2}+y^{2}+4x+6y-5=0$

Third. c) matches 3)

$\bold{From \ d)} \\ \\ (h,k)=(-1,6),\ r=\sqrt{46} \\ D=2, \ E=-12, \ F=-9$

Then, its general form is: $x^{2}+y^{2}+2x-12y-9=0$

Fourth. d) matches 6)

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3 years ago

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omeli [17]

Answer:

The answer is (-4/3)

Step-by-step explanation:

3(x − 4) = 12x

Multiply (x – 4) by 3 we get,

3(x – 4) = 12x

3x – 12 = 12x

3x – 12x – 12 + 12 = 12x – 12x + 12

-9x = 12

-9x/(-9) = -12/9

x = -4/3

Thus, The value of x is -4/3

-TheUnknownScientist 72

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2 years ago

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