2 cm= 150 miles

Write the equation of a line parallel to the line 2x + y = 3 that passes through the point (2,6)

Lady bird [3.3K]

Answer <u>(assuming it can be in slope-intercept form)</u>:

$y = -2x +10$

Step-by-step explanation:

1) First, we need to find the slope of the given equation. To do that easily, convert it to slope-intercept form, represented by the formula $y = mx + b$ . Isolate y:

$2x + y = 3\\y = -2x + 3$

The number in place of $m$ , or the coefficient of the x-term, represents the slope of the equation. So, the slope of the given equation is -2.

Lines that are parallel share the same slope. So, the slope of the new equation will be -2 as well.

2) Now, remember that the slope-intercept form is represented by the formula $y = mx + b$ . In order to write an equation of a line using that formula, we need to substitute values for $m$ and $b$ . We know $m$ is the slope, so we already know what that equals. Now, we just need to find $b$ .

To do that, substitute -2 for $m$ in the slope-intercept formula. Additionally, substitute the x and y values of the point (2,6) for the x and y in the formula as well. This sets up the equation so that we can isolate $b$ and find its values:

$y = mx +b\\6 = (-2)(2)+b\\6 = -4 + b\\10 = b$

So, $b$ = 10.

3) Substitute values for $b$ and $m$ into the slope-intercept formula to write the slope-intercept form of the line:

$y = -2x +10$

6 0

3 years ago

What is the domain for the function below?<br> Х 1 3 5 7 Y 2 4 6 8

erma4kov [3.2K]

X is the the answerrrr

3 0

3 years ago

$20000 is invested in an account that earned 6% p.A. Compounding yearly for 3 years. The interest rate then went up to 8% p.A. F

GuDViN [60]

$\bf ~~~~~~ \textit{Compound Interest Earned Amount \underline{for the first 3 years}} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$20000\\ r=rate\to 6\%\to \frac{6}{100}\dotfill &0.06\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{per annum, thus once} \end{array}\dotfill &1\\ t=years\dotfill &3 \end{cases}$

$\bf A=20000\left(1+\frac{0.06}{1}\right)^{1\cdot 3}\implies A=20000(1.06)^3\implies \boxed{A=2382.032} \\\\[-0.35em] ~\dotfill\\\\ ~~~~~~ \textit{Compound Interest Earned Amount \underline{for the next 4 years}}$

$\bf A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$2382.032\\ r=rate\to 8\%\to \frac{8}{100}\dotfill &0.08\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{per annum, thus once} \end{array}\dotfill &1\\ t=years\dotfill &4 \end{cases}$

$\bf A=2382.032\left(1+\frac{0.08}{1}\right)^{1\cdot 4}\implies A=2382.032(1.08)^4\implies \boxed{A\approx 3240.73} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{amount for this period}}{2382.032+3240.73}\implies 5622.762$

4 0

4 years ago

Solve this step by step and explain your answer

WITCHER [35]

Answer:

(-3,5)

Step-by-step explanation:

6x+7y=17

-6x+5y=7

2y=10

y=5

6x+5(5)=7

6x+25=7

6x=7-25

6x=-18

x=-3

3 0

3 years ago

Read 2 more answers

find the measure of the missing side of a right triangle to the nearest tenth if one leg is 36cm in the hypotenuse is 39 cm

4vir4ik [10]

A^2+36^2=39^2
a^2+1296=1521
a^2=225
a=15

5 0

4 years ago