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2 years ago

Can someone please help me with question 6

Mathematics

1 answer:

olasank [31]2 years ago

7 0

Answer:

5323

Step-by-step explanation:

453

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Write a rule for the linear function in the table.

JulijaS [17]

Answer:

I guess that you want to know the transformations:

We start with:

f(x) = y = 4*x + 3

a)the transformed function is:

f(x) = y = -4*x - 3

So the sign changed.

This means that we go from (x, y) to (x, - y)

This is a reflection over the x-axis which changes the sin of the y component.

b) Now we go to f(x) = 4*x + 3

So the coefficient in the leading term changed.

This is a horizontal contraction:

A horizontal contraction of factor K for the function g(x) is: g(K*x)

In our case, we have:

f(K*x) = 4*(k*x) + 3 = x + 3

4*k*x = x

4*k = 1

k = 1/4

Then the transformation is an horizontal contraction of scale factor 1/4.

6 0

2 years ago

Show that <1+<2+<3=180, if line l and m are parallels cut by two transversals.

Svet_ta [14]

Step-by-step explanation:

Angle 1 2 and 3 are angles that are in a triangle and all the angles in a triangle always add up to 180

so <1 +<2 +<3=180

Hope that helps :)

7 0

3 years ago

Which expression is equivalent to 12x+3y−9x ?

Stels [109]

The answer is 6x

You have to add 12 and 3 which is 15 then you have to do 15 minus 9 then it is 6 then you will put the x there because it had 2 x and 1y so 6x is the answer

4 0

3 years ago

Read 2 more answers

Will give brainliest, thanks, and 5 stars! LOTS OF POINTS! 50 POINTS!

sweet-ann [11.9K]

Answer/Step-by-step explanation:

All of the solutions to the equation 3x^2 - 12 = 0 are x = 12 and x = -2

Answer: False

Explanation:

$3x^2-12+12=0+12$

$3x^2=12$

$\frac{3x^2}{3}=\frac{12}{3}$

$x^2=4$

$\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}$

$x=\sqrt{4},\:x=-\sqrt{4}$

$x=2,\:x=-2$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

There are two unique solutions to the equations (x-3)^2 = 16

Note: Each variable in the matrix can have only one possible value, and this is how you know that this matrix has one unique solution.

Answer: True

Explanation:

$\mathrm{For\:}\left(g\left(x\right)\right)^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}g\left(x\right)=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}$

$x=7,\:x=-1$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Ths solutions for the equation 2(x-3)^3 - 18 = 0 are x = 6 and x = 0

Answer: False

Explanation:

$2\left(x-3\right)^3-18+18=0+18$

$2\left(x-3\right)^3=18$

$\frac{2\left(x-3\right)^3}{2}=\frac{18}{2}$

$\left(x-3\right)^3=9$

$\mathrm{For\:}g^3\left(x\right)=f\left(a\right)\mathrm{\:the\:solutions\:are\:}g\left(x\right)=\sqrt[3]{f\left(a\right)},\:\sqrt[3]{f\left(a\right)}\frac{-1-\sqrt{3}i}{2},\:\sqrt[3]{f\left(a\right)}\frac{-1+\sqrt{3}i}{2}$

$=\sqrt[3]{9}+3,\:x=\frac{6-\sqrt[3]{9}}{2}+i\frac{\sqrt[3]{9}\sqrt{3}}{2},\:x=\frac{6-\sqrt[3]{9}}{2}-i\frac{\sqrt[3]{9}\sqrt{3}}{2}$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~The solutions for the equation 2(x-5)^2-8=0 are x = 7 and x = -7

Answer: False

Explanation:

$2\left(x-5\right)^2-8+8=0+8$

$2\left(x-5\right)^2=8$

$\frac{2\left(x-5\right)^2}{2}=\frac{8}{2}$

$\left(x-5\right)^2=4$

$\mathrm{For\:}\left(g\left(x\right)\right)^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}g\left(x\right)=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}$

$x=7,\:x=3$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

The solutions for the equation (x + 3)^2-25 = -8 are x = 2 and x = -8

Answer: False

Explanation:

$\left(x+3\right)^2-25+25=-8+25$

$\left(x+3\right)^2=17$

$\mathrm{For\:}\left(g\left(x\right)\right)^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}g\left(x\right)=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}$

$x=\sqrt{17}-3,\:x=-\sqrt{17}-3$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

The solutions for the equation 2(2x-1)^2=18 are x = 5 and x = -4

Answer: False

Explanation:

$\frac{2\left(2x-1\right)^2}{2}=\frac{18}{2}$

$\left(2x-1\right)^2=9$

$\mathrm{For\:}\left(g\left(x\right)\right)^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}g\left(x\right)=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}$

$x=2,x=-1$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

The only solution for equation (2x-1)^2-49=0 is x = 4

Answer: False
Explanation:

$\left(2x-1\right)^2-49+49=0+49$

$\left(2x-1\right)^2=49$

$\mathrm{For\:}\left(g\left(x\right)\right)^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}g\left(x\right)=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}$

$x=4,\:x=-3$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

The solutions for the equation 3(x+2)^2 - 3 = 0 are x = -3 and x = -1

Answer: True
Explanation:

$3\left(x+2\right)^2-3+3=0+3$

$3\left(x+2\right)^2=3$

$\frac{3\left(x+2\right)^2}{3}=\frac{3}{3}$

$\left(x+2\right)^2=1$

$\mathrm{For\:}\left(g\left(x\right)\right)^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}g\left(x\right)=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}$

$x=-1,\:x=-3$