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skad [1K]
3 years ago
10

What is the slope- intercept equation of the line that passes through(0,2) and (6,6)?

Mathematics
2 answers:
charle [14.2K]3 years ago
8 0

Answer:

y=2/3x+2

Step-by-step explanation:

(0,2) (6,6)=6-2/6-0=4/6 or 2/3,the slope is used to find the rise over run.and the y-intercept is when is 0 and y is another number.

Vsevolod [243]3 years ago
7 0

Answer:

y = 2/3x +2

Step-by-step explanation:

First we need to find the slope

m = (y2-y1)/(x2-x1)

    = (6-2)/(6-0)

    =4/6

   =2/3

The slope intercept form of the equation for a line is

y= mx+b

y =2/3 x +b

We know the y intercept is 2  (the y intercept is when x=0)

y = 2/3x +2

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Solve this inequality for x: -46 -8x >22.
olga2289 [7]

Answer:

Step-by-step explanation:

The key to answering this is to know that the only time you'll change the direction of the sign is when you multiply or divide by a negative number.

-46 - 8x > 22

-8x > 22 + 46

-8x > 68

x < 68/-8        (we divided by -8 so the inequality changed direction)

x < -17/2

3 0
2 years ago
Read 2 more answers
Error Analysis! Answer both parts of the question.
Maurinko [17]

Answer:

Kaitlin's answer is wrong because she is supposed to by dividing. When trying to get rid of a number from a variable you have to divide the number by itself and whatever you do to one side you have to do to the other side

so you would do 5/5 - cancels out

15/5=3

y=3

Step-by-step explanation:

have a great day :D

8 0
2 years ago
Can someone help me? please
Vilka [71]
Its the third one- 1/2
7 0
2 years ago
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Work out the size of angle x 126 degrees exterior and 43 degrees interior
grigory [225]

Answer:

The answer is 83°

Step-by-step explanation:

by the sum of iner and outer angle=180°

180-126= 54°

x=180-43+54

x=180-97

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hope that helped.?<3

8 0
2 years ago
You decide to put $5000 in a savings account to save $6000 down payment on a new car. If the account has an interest rate of 7%
bezimeni [28]

\bf ~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\dotfill&\$6000\\ P=\textit{original amount deposited}\dotfill &\$5000\\ r=rate\to 7\%\to \frac{7}{100}\dotfill &0.07\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{monthly, thus twelve} \end{array}\dotfill &12\\ t=years \end{cases}

\bf 6000=5000\left(1+\frac{0.07}{12}\right)^{12\cdot t}\implies \cfrac{6000}{5000}\approx (1.0058)^{12t}\implies \cfrac{6}{5}\approx(1.0058)^{12t} \\\\\\ \log\left( \cfrac{6}{5} \right)\approx \log[(1.0058)^{12t}]\implies \log\left( \cfrac{6}{5} \right)\approx 12t\log(1.0058) \\\\\\ \cfrac{\log\left( \frac{6}{5} \right)}{12\log(1.0058)}\approx t\implies 2.63\approx t\impliedby \textit{about 2 years, 7 months and 16 days}

6 0
3 years ago
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