All categories

4 years ago

George has (-2x+5) dollars. Evan has (6x+77) dollars. George and Evan have the same amount of money. How much do they each have?

Mathematics

1 answer:

algol [13]4 years ago

6 0

Answer:

$23

Step-by-step explanation:

Equal both equations next to each other --

-2x+5=6x+77

Solve for x... x=-9

Plug -9 in to each equation, should get the same value

You might be interested in

The manufacturer of a laser printer reports the mean number of pages a cartridge will print before it needs replacing is 12,425.

snow_lady [41]

A suitable probability app shows the 1st percentile to be at about 11040 pages.

_____

The app shown has an error in the first decimal digit. The number is closer to 11040.82301. We expect the manufacturer would round to the nearest 10 or 50 or 100 in advertising.

8 0

3 years ago

Find the equation of the line with Slope = 5 and passing through (-7,-34). Write your equation in the form y=mx+b .

oksano4ka [1.4K]

$(\stackrel{x_1}{-7}~,~\stackrel{y_1}{-34})\hspace{10em} \stackrel{slope}{m} ~=~ 5 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{(-34)}=\stackrel{m}{ 5}(x-\stackrel{x_1}{(-7)}) \implies y +34= 5 (x +7) \\\\\\ y+34=5x+35\implies {\Large \begin{array}{llll} y=5x+1 \end{array}}$

3 0

1 year ago

To transmit information on the internet, large files are broken into packets of smaller sizes. Each packet has 1,500 bytes of in

Lapatulllka [165]

Answer:

750,000 bytes

Step-by-step explanation:

if you place 500 as p, the answer would be 750,000 bytes.

7 0

2 years ago

Help please !!!!!!!!!!

harkovskaia [24]

Answer:

[see below]

Step-by-step explanation:

Using the Pythagorean Theorem:

$6^2+4.5^2=m^2\\\\36+20.25=m^2\\\\5\boxe6.25=m^2\\\\\boxed{m=7.5}$

Hope this helps.

6 0

4 years ago

Read 2 more answers

.....................................

babunello [35]

$\dfrac{q}{q^2+5q+6}+\dfrac{1}{q^2+3q+2}=\dfrac{q}{q^2+3q+2q+6}+\dfrac{1}{q^2+q+2q+2}\\\\=\dfrac{q}{q(q+3)+2(q+3)}+\dfrac{1}{q(q+1)+2(q+1)}\\\\=\dfrac{q}{(q+3)(q+2)}+\dfrac{1}{(q+1)(q+2)}\\\\=\dfrac{q(q+1)}{(q+1)(q+2)(q+3)}+\dfrac{1(q+3)}{(q+1)(q+2)(q+3)}\\\\=\dfrac{q^2+q+q+3}{(q+1)(q+2)(q+3)}=\dfrac{q^2+2q+3}{(q+1)(q+2)(q+3)}\to B.$

8 0

3 years ago