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3 years ago

Solve the following system of equations graphically on the set of axes below. y=3x−3 , x+y=5

Mathematics

1 answer:

topjm [15]3 years ago

6 0

Answer:

x=2, y=3

Step-by-step explanation:

First, graph both equations on a rectangular coordinate system (see attached screenshot). Then, simply figure out where these two lines intersect. Since they intersect at (2,3), x=2 and y=3.

Plugging these values in the equations for x and y, you can see if these are the correct answers:

(3)=3(2)-3, (2)+(3)=5

Since both of these are true, you have the right answer.

Hope this helps!

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if a recipe has 3/8 cups of water and 2 cups of flour what is the ratio and then what is the ratio in a fraction or simplified?

Lady bird [3.3K]

Answer: The ratio of eggs, eggs, to two cups of flour. Let me write two cups of flour. Cups of flour. In either case is, is four eggs for every three cups of flour 5/2

Step-by-step explanation: Im not sure through

7 0

3 years ago

Find the difference 58.84 - 2.78

VladimirAG [237]

Answer:

56.06

Step-by-step explanation:

58.84

-- 2.78

______

56.06

I don't really know how else to explain this, but there ya go! Goodluck on your test or whatever you're doing! Stay safe ♥♥♥

8 0

2 years ago

Read 2 more answers

PLS HELP!<br> solve by completing the square<br> x²+4x=8

masya89 [10]

Step-by-step explanation:

x²+4x=8

rearrange

x² + 4x - 8 = 0

half of 4 is 2

x² + 4x + 2² - 2² - 8 = 0

do x² + 4x + 2²

multiplies to give 4 adds to give 4

2+ 2

(x + 2)²

(x + 2)² - 2² - 8

(x + 2)² - 4 - 8

(x + 2)² - 12

(x + 2 + √12) (x + 2 - √12)

√12 = √4 * √3 = 2 √3

(x + 2 + 2√3) (x + 2 - 2√3)

4 0

2 years ago

Suppose Kaitlin places $6500 in an account that pays 12% interest compounded each year.

Leya [2.2K]

$\bf ~~~~~~ \textit{Compound Interest Earned Amount \underline{for 1 year}} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$6500\\ r=rate\to 12\%\to \frac{12}{100}\dotfill &0.12\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annually, thus once} \end{array}\dotfill &1\\ t=years\dotfill &1 \end{cases}$

$\bf A=6500\left(1+\frac{0.12}{1}\right)^{1\cdot 1}\implies A=6500(1.12)\implies A=7280 \\\\[-0.35em] ~\dotfill$

$\bf ~~~~~~ \textit{Compound Interest Earned Amount \underline{for 2 years}} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$6500\\ r=rate\to 12\%\to \frac{12}{100}\dotfill &0.12\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annually, thus once} \end{array}\dotfill &1\\ t=years\dotfill &2 \end{cases}$

$\bf A=6500\left(1+\frac{0.12}{1}\right)^{1\cdot 2}\implies A=6500(1.12)^2\implies A=8153.6$

6 0

3 years ago

Rollie was succesful in losing weight he had a goal weight in mind he went on a diet for three months each month he would lose o

Llana [10]

Let's say that in the beginning he weighted x and at the end he weighted x-y, y being the number of kg he wanted to loose.

first month he lost
y/3

then he lost:
(y-y/3)/3
this is
(2/3y)/3=2/9y
explanation: ((y-y/3) is what he still needed to loose: y minus what he lost already

and then he lost
(y-2/9y-1/3y)/3+3 (the +3 is his additional 3 pounts)
(y-2/9y-1/3y)/3-3=(7/9y-3/9y)/3+3=4/27y+3

it's not just y/3 because each month he lost one third of what the needed to loose at the current time, not in totatl

and the weight at the end of the 3 months was still x-y+3, 3 pounds over his goal weight!

so: x -y/3-2/9y-4/27y-3=x-y+3

we can subtract x from both sides:
-y/3-2/9y-4/27y-3=-y+3
add everything up:

-19/27y=-y+6

which means
-19/27y=-y+6

y-6=19/27y

8/27y=6
4/27y=3
y=20.25

so... that's how much he wanted to loose, but he lost 3 less than that, so 23.25

ps. i hope I didn't make a mistake in counting, let me know if i did. In any case you know HOW to solve it now, try to do the calculations yourself to see if they're correct!

5 0

3 years ago