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LekaFEV [45]
3 years ago
13

What equation represents a line that when if graphed, passes through the points (0, -1) and

Mathematics
1 answer:
Keith_Richards [23]3 years ago
5 0
Y= 2x-1
Explanation: First, you find slope. If the first point it x-axis point is 0 and the second x-axis point is 1 then you already know the answer is over 1, so you have to figure out how many up you are going on the y-axis from -1 to 1. Which is 2, so your slope is 2/1 or just 2. Next, you find your y-intercept, which is where the line crosses the y-axis, and the easy way to find this is finding what y is equal to when x is equal to 0, in the problem you are told (0, -1) as a point, so they gave you your y-intercept right there, which is -1. Finally you right the equation in slip intercept form (y= mx+b) which is y= 2x-1
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3 years ago
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HELP!!<br>Find the third term of (x^2+3y)^3
lana66690 [7]

Answer:

T_{3}=27{x}^{2}y^2

Step-by-step explanation:

The given binomial expression is:

( {x}^{2} + 3y)^{3}

When we compare to:

{(a +b)}^{n}

We have

a =  {x}^{2}

b = 3y \\ n = 3

The nth term is given by;

T_{r+1}=^nC_ra^{n-r}b^r

To find the 3rd term, we put:

r + 1 = 3 \\ r = 2

We substitute into the formula to get:

T_{3}=^3C_2( {x}^{2} )^{3-2}(3y)^2

We simply:

T_{3}=3( {x}^{2} )^{1} \times 9y^2

T_{3}=27{x}^{2}y^2

3 0
3 years ago
Which ordered pair is a solution to the system of linear equations 1/2x-3/4y=11/60 and 2/5x+1/6y=3/10
natka813 [3]

ANSWER

( \frac{2}{3} , \frac{1}{5} )

EXPLANATION

The first equation is

\frac{1}{2} x -  \frac{3}{4} y =  \frac{11}{60} ...(1)

The second equation is

\frac{2}{5} x  +  \frac{1}{6} y =  \frac{3}{10} ...(2)

We want to eliminate y, so we multiply the first equation by

\frac{4}{5}

\frac{4}{5}  \times \frac{1}{2} x - \frac{4}{5}    \times \frac{3}{4} y =  \frac{11}{60}  \times  \frac{4}{5}

\frac{2}{5} x - \frac{3}{5} y =  \frac{11}{75} ...(3)

We now subtract equation (3) from (2)

(\frac{2}{3} x  -  \frac{2}{3} x )+ ( \frac{1}{6} y -  -  \frac{3}{5}y ) =(  \frac{3}{10}  -  \frac{11}{75} )

\frac{1}{6} y  +  \frac{3}{5}y  =\frac{3}{10}  -  \frac{11}{75}

\frac{23}{30}y =  \frac{23}{150}

Multiply both sides by

\frac{30}{23}

\implies \:  \frac{30}{23} \times  \frac{23}{30}y=  \frac{23}{150}  \times  \frac{30}{23}

\implies \: y =  \frac{1}{5}

Substitute into the first equation to solve for x .

\frac{1}{2} x -  \frac{3}{4}  \times \frac{1}{5} =  \frac{11}{60}

Multiply to obtain

\frac{1}{2} x -  \frac{3}{20} =  \frac{11}{60}

\frac{1}{2} x = \frac{11}{60} + \frac{3}{20}

\frac{1}{2} x = \frac{1}{3}

Multiply both sides by 2.

2 \times \frac{1}{2} x =2 \times  \frac{1}{3}

x = \frac{2}{3}

The solution is

( \frac{2}{3} , \frac{1}{5} )

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How do you set this up?
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