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3 years ago

Can somebody help me please? I haven’t gotten help at all.

Mathematics

1 answer:

Elena-2011 [213]3 years ago

4 0

Answer:

f(x) = -3x+12

I would advise you to study this part again :)

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Graph the image of square TUVW after the following sequence of transformations:

GuDViN [60]

Move the shape down by one quadrant and flip it so where “U” was at, “W” is in its place and Vice versa

4 0

2 years ago

one vertex of a triangle is located at (0,5) on a coordinate grid after a transformation the vertex is located at(5,0) which tra

Alik [6]

Answer:

It's a rotation clockwise of 90 degrees about the origin.

Step-by-step explanation:

The given point moves from y = 5 to x = 5 so it has passed through 90 degrees about the origin.

It's a rotation clockwise of 90 degrees about the origin.

6 0

2 years ago

Solve the system of linear equations using elimination. −2x − y = 3 −9x − y = 17

Maurinko [17]

-2×-y=3-9×-y=17

=2×+9×=1y+1y=17+3

11×=2y=20

11=y=20÷2

11y=10

y=10/11

6 0

3 years ago

WILL MARK BRANLIEST the first one plz

ruslelena [56]

Answer:

A,D,E, and F

Step-by-step explanation:

They all equal -1/4.

3 0

2 years ago

Find the product of all real values of r for which 1/2x=r-x/7

Dahasolnce [82]

Answer:

$r = \±\sqrt{14$

$Product = -14$

Step-by-step explanation:

Given

$\frac{1}{2x} = \frac{r - x}{7}$

Required

Find all product of real values that satisfy the equation

$\frac{1}{2x} = \frac{r - x}{7}$

Cross multiply:

$2x(r - x) = 7 * 1$

$2xr - 2x^2 = 7$

Subtract 7 from both sides

$2xr - 2x^2 -7= 7 -7$

$2xr - 2x^2 -7= 0$

Reorder

$- 2x^2+ 2xr -7= 0$

Multiply through by -1

$2x^2 - 2xr +7= 0$

The above represents a quadratic equation and as such could take either of the following conditions.

(1) No real roots:

This possibility does not apply in this case as such, would not be considered.

(2) One real root

This is true if

$b^2 - 4ac = 0$

For a quadratic equation

$ax^2 + bx + c = 0$

By comparison with $2x^2 - 2xr +7= 0$

$a = 2$

$b = -2r$

$c =7$

Substitute these values in $b^2 - 4ac = 0$

$(-2r)^2 - 4 * 2 * 7 = 0$

$4r^2 - 56 = 0$

Add 56 to both sides

$4r^2 - 56 + 56= 0 + 56$

$4r^2 = 56$

Divide through by 4

$r^2 = 14$

Take square roots

$\sqrt{r^2} = \±\sqrt{14$

$r = \±\sqrt{14$

Hence, the possible values of r are:

$\sqrt{14$ or $-\sqrt{14$

and the product is:

$Product = \sqrt{14} * -\sqrt{14}$

$Product = -14$

8 0

2 years ago