Ask question

Ask question

All categories

3 years ago

6

At work one day , Ricardo had to go down 3 floor to go to meetings. Then he had to go up 7 floors for another meeting.What will

Ricardo need to go to get back to the floor were he started

1 answer:

mestny [16]3 years ago

4 0

Ricardo would have to go 4 floors down to reach the floor he started in cause
7-3=4 floors

You might be interested in

I HATE word problems can you do this?

KonstantinChe [14]

Answer: $13 dollars

Step-by-step explanation:

5 0

3 years ago

Find the coordinates of the intersection of the diagonals of the parallelogram with the vertices W(−2, 5), X(2, 5), Y(4, 0),

emmasim [6.3K]

Using the midpoint formula, the coordinates of the intersection of the diagonals of the parallelogram is: (1, 2.5).

<h3>What are Diagonals of a Parallelogram?</h3>

The diagonals of a parallelogram bisect each other, therefore, the coordinates of their intersection can be determined using the midpoint formula, which is: $M(\frac{x_2 + x_1}{2}, \frac{y_2 + y_1}{2})$ .

A diagonal is XZ.

X(2, 5) = (x1, y1)

Z(0, 0) = (x2, y2)

Plug in the values

$M(\frac{0 + 2}{2}, \frac{0 + 5}{2})$

= (1, 2.5).

Learn more about the diagonals of a parallelogram on:

brainly.com/question/12167853

#SPJ1

7 0

1 year ago

Solve for y. 3/2x-3y=6

Vlada [557]

X=2(2+y) is your answer

3 0

2 years ago

Read 2 more answers

Simplify the expression.

Nana76 [90]

The answer is the A. I hope this helps!

6 0

3 years ago

Please help solve this system of equations

stepan [7]

Make a substitution:

$\begin{cases}u=2x+y\\v=2x-y\end{cases}$

Then the system becomes

$\begin{cases}\dfrac{2\sqrt[3]{u}}{u-v}+\dfrac{2\sqrt[3]{u}}{u+v}=\dfrac{81}{182}\\\\\dfrac{2\sqrt[3]{v}}{u-v}-\dfrac{2\sqrt[3]{v}}{u+v}=\dfrac1{182}\end{cases}$

Simplifying the equations gives

$\begin{cases}\dfrac{4\sqrt[3]{u^4}}{u^2-v^2}=\dfrac{81}{182}\\\\\dfrac{4\sqrt[3]{v^4}}{u^2-v^2}=\dfrac1{182}\end{cases}$

which is to say,

$\dfrac{4\sqrt[3]{u^4}}{u^2-v^2}=\dfrac{81\times4\sqrt[3]{v^4}}{u^2-v^2}$

$\implies\sqrt[3]{\left(\dfrac uv\right)^4}=81$

$\implies\dfrac uv=\pm27$

$\implies u=\pm27v$

Substituting this into the new system gives

$\dfrac{4\sqrt[3]{v^4}}{(\pm27v)^2-v^2}=\dfrac1{182}\implies\dfrac1{v^2}=1\implies v=\pm1$

$\implies u=\pm27$

Then

$\begin{cases}x=\dfrac{u+v}4\\\\y=\dfrac{u-v}2}\end{cases}\implies x=\pm7,y=\pm13$

(meaning two solutions are (7, 13) and (-7, -13))

8 0

3 years ago