Please do 3 hurry!!!!

(4x-1) /2=x+7<br>4x-1 all over 2 = x +7

Damm [24]

Answer:

x = $\frac{15}{2}$

Step-by-step explanation:

Given

$\frac{4x-1}{2}$ = x + 7 ( multiply both sides by 2 to clear the fraction )

4x - 1 = 2(x + 7) ← distribute

4x - 1 = 2x + 14 ( subtract 2x from both sides )

2x - 1 = 14 ( add 1 to both sides )

2x = 15 ( divide both sides by 2 )

x = $\frac{15}{2}$

7 0

3 years ago

Lizzy and Ella sold 10 boxes of cookies for $55.50. How much does each box of cookies cost? Show all your work and explain how y

densk [106]

55.50/ 10 boxes =5.50 that is cost of 1 box.

5 0

3 years ago

3/7 divide 2/9 explain pls

lubasha [3.4K]

Answer:

27/14

Step-by-step explanation:

to divide by a fraction flip the fraction over and multiply

3/7 divided by 2/9

flip 2/9

3/7 times 9/2

just multiply the top and bottoms by each other

27/14

6 0

3 years ago

Read 2 more answers

Where I had $480 he used 2/3 of it to buy an electric van he also bought a tea set for $60 how

yawa3891 [41]

2/3 of 480 is 480 divided by 3 and that number times 2, which is 320. The man or woman probably used $60 of this to buy the tea set, which leaves them with $260.

8 0

3 years ago

A small plane took 3 hours to fly 960 km from Ottawa to Halifax with a tail wind. On the return trip, flying into the wind, the

Rina8888 [55]

Answer:

Wind speed: $\rm 40\; km \cdot h^{-1}$ .
Speed of the plane in still air: $\rm 320\; km \cdot h^{-1}$ .

Step-by-step explanation:

This problem involves two unknowns:

wind speed, and
speed of the plane in still air.

Let the speed of the wind be $x \rm \; km \cdot h^{-1}$ , and the speed of the plane in still air be $y\rm \; km \cdot h^{-1}$ . It takes at least two equations to find the exact solutions to a system of two variables.

Information in this question gives two equations:

It takes the plane three hours to travel $\rm 960\; km$ from Ottawa to with a tail wind (that is: at a ground speed of $x + y$ .)
It takes the plane four hours to travel $\rm 960\; km$ from Halifax back to Ottawa while flying into the wind (that is: at a ground speed of $-x + y$ .)

Create a two-by-two system out of these two equations:

$\left\{ \begin{aligned}&3(x + y) = 960 && (1) \\ &4(-x + y) = 960 && (2) \end{aligned}\right.$ .

There can be many ways to solve this system. The approach below avoids multiplying large numbers as much as possible.

Note that this system is equivalent to

$\left\{ \begin{aligned}&4 \times 3 (x + y) = 4\times960 && 4 \times (1) \\ &3\times 4(-x + y) = 3\times 960 && 3 \times (2) \end{aligned}\right.$ .

$\left\{ \begin{aligned}&12 x + 12y = 4\times960 && 4 \times (1) \\ &- 12x + 12y = 3\times 960 && 3 \times (2) \end{aligned}\right.$ .

Either adding or subtracting the two equations will eliminate one of the variables. However, subtracting them gives only $1 \times 960$ on the right-hand side. In comparison, adding them will give $7 \times 960$ , which is much more complex to evaluate. Subtracting the second equation ( $3 \times (2)$ ) from the first ( $4 \times (1)$ ) will give the equation