All categories

3 years ago

Translate quadrilateral ABCD 5 units to the right and 6 units down.

Mathematics

1 answer:

dezoksy [38]3 years ago

6 0

Answer:

D. A'(1,-7), B'(2,-1), C'8,-2) and D'(7,-5)

Step-by-step explanation:

Probably too late, but I took the quiz

You might be interested in

Can someone please help me ASAP

Sonja [21]

Answer:

x^2 + 5n = 4mx

=> x^2 - 4mx + 5n = 0

D=0 (since one root then roots are equal)

b^2-4ac = 0

(-4m)^2 - 4(1)(5n) = 0

16m^2 - 20n = 0

16m^2 = 20n

8m^2 = 10n

4m^2 = 5n

hope it helps.......

7 0

3 years ago

Find the area of the given triangles.

vovikov84 [41]

154 yd ²

the formula is (b • h) / 2 = A

so plug it in (14 • 22) / 2 = 154

5 0

3 years ago

At what value of x does the graph of the following function F(x) have a

tankabanditka [31]

Given:

The function is

$F(x)=\dfrac{1}{x+2}$

To find:

The vertical asymptote of the given function.

Solution:

Vertical asymptote are the vertical line passes thought the values for which the function is not defined.

To find the vertical asymptote, equate the denominator equal to 0.

We have,

$F(x)=\dfrac{1}{x+2}$

Denominator is (x+2).

$x+2=0$

$x=-2$

The vertical asymptote is $x=-2$ .

Therefore, the correct option is B.

8 0

2 years ago

Read 2 more answers

You have a photo that measures 1 in. wide by 3 in. tall. If you want to enlarge the photo so that the height is 18 in., what wil

kiruha [24]

6in... 3 times 6 is 18, so 1 times 6 is 6

4 0

3 years ago

4TH TIME ASKING THIS!!! Please help me! Someone pleaseeee. I need the correct answers. I don’t want to fail

Alexeev081 [22]

Answer:

The functions are inverses; f(g(x)) = x ⇒ answer D

$h^{-1}(x)=\sqrt{\frac{x+1}{3}}$ ⇒ answer D

Step-by-step explanation:

* Lets explain how to find the inverse of a function

- Let f(x) = y

- Exchange x and y

- Solve to find the new y

- The new y = $f^{-1}(x)$

* Lets use these steps to solve the problems

∵ $f(x)=\sqrt{x-3}$

∵ f(x) = y

∴ $y=\sqrt{x-3}$

- Exchange x and y

∴ $x=\sqrt{y-3}$

- Square the two sides

∴ x² = y - 3

- Add 3 to both sides

∴ x² + 3 = y

- Change y by $f^{-1}(x)$

∴ $f^{-1}(x)=x^{2}+3$

∵ g(x) = x² + 3

∴ $f^{-1}(x)=g(x)$

∴ The functions are inverses to each other

* Now lets find f(g(x))

- To find f(g(x)) substitute x in f(x) by g(x)

∵ $f(x)=\sqrt{x-3}$

∵ g(x) = x² + 3

∴ $f(g(x))=\sqrt{(x^{2}+3)-3}=\sqrt{x^{2}+3-3}=\sqrt{x^{2}}=x$

∴ f(g(x)) = x

∴ The functions are inverses; f(g(x)) = x

* Lets find the inverse of h(x)

∵ h(x) = 3x² - 1 where x ≥ 0

- Let h(x) = y

∴ y = 3x² - 1

- Exchange x and y

∴ x = 3y² - 1

- Add 1 to both sides

∴ x + 1 = 3y²

- Divide both sides by 3

∴ $\frac{x + 1}{3}=y^{2}$

- Take √ for both sides

∴ ± $\sqrt{\frac{x+1}{3}}=y$

∵ x ≥ 0

∴ We will chose the positive value of the square root

∴ $\sqrt{\frac{x+1}{3}}=y$

- replace y by $h^{-1}(x)$

∴ $h^{-1}(x)=\sqrt{\frac{x+1}{3}}$

4 0

3 years ago