If rectangle ABCD is translated 5 units right and 3 units down, what are the coordinates of A'.

Mathematics

2 answers:

Kipish [7]2 years ago

8 0

Answer:

Step-by-step explanation:

alexandr402 [8]2 years ago

3 0

The coordinated of A would be 1,1.
Instead of going through the trouble of mentally moving the entire thing, just focus on the one you need to know the answer to, aka A.

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HELPPP

Juli2301 [7.4K]

Answer:

$slope (m) = -\frac{3}{2}$

Step-by-step explanation:

We can find the slope (m) by using coordinate pairs of any 2 points located along the slope of the line that we have on the graph.

This, let's use the coordinate pairs at:

x = -4, y = 2 (-4, 2) => (x2, y2)

x = 0, y = -4 (0, -4) => (x1, y1)

$slope (m) = \frac{y2 - y1}{x2 - x1}$

$slope (m) = \frac{2 -(-4)}{-4 - 0}$

$slope (m) = \frac{2 + 4}{-4 - 0}$

$slope (m) = \frac{6}{-4}$

$slope (m) = \frac{3}{-2}$

$slope (m) = -\frac{3}{2}$

4 0

2 years ago

Can someone help me please?

Olenka [21]

Answer:

sin ∠I = 12/13

Step-by-step explanation:

SOH CAH TOA

↑ sin = opposite side/hypotenuse

sin ∠I = 48/52 => sin ∠I = 12/13

6 0

2 years ago

Read 2 more answers

What is the solution to the equation 3/m+3-m/3-m=m^2+9/m^2-9

Bas_tet [7]

Answer:

no solution

Step-by-step explanation:

If we subtract the left side of the equation, we get ...

$\dfrac{m^2+9}{m^2-9}-\dfrac{3}{m+3}+\dfrac{3}{3-m}=0\\\\\dfrac{m^2+9-3(m-3)-3(m+3)}{(m-3)(m+3)}=0\\\\\dfrac{m^2-6m+9}{(m-3)(m+3)}=0=\dfrac{(m-3)^2}{(m-3)(m+3)}\\\\\dfrac{m-3}{m+3}=0\quad\text{m$\ne$3}$