The 2-point form of the equation for a line can be used.
... y = (y₂-y₁)/(x₂-x₁)·(x -x₁) +y₁
Filling in the given information, you have
... y = (4-1)/(4-3)·(x-3) +1 . . . . an equation for the line
... y = 3x -8 . . . . . . . . . . . . . . simplified to slope-intercept form
... 3x -y = 8 . . . . . . . . . . . . . .. rearranged to standard form
Answer:
y²-2y+3
Step-by-step explanation:
We write the dividend, y³-y²+y+3, under the box and the divisor, y+1, to the left of the box.
We first divide y³ by y; this is y². We write this above the box, over -y². We multiply the divisor by y²:
y²(y+1) = y³+y²
This goes under the divisor. We then subtract:
(y³-y²)-(y³+y²) = -2y². We then bring down the next term, y; this gives us -2y²+y.
We then divide -2y² by y; this is -2y. This goes above the box beside the y² in the quotient. We then multiply the divisor by -2y:
-2y(y+1) = -2y²-2y
We now subtract:
(-2y²+y)-(-2y²-2y) = 3y. We bring down the last term, 3; this gives us 3y+3. We divide 3y by y; this is 3. This goes beside the -2y in the quotient. We then multiply this by the divisor:
3(y+1) = 3y+3. We then subtract: (3y+3)-(3y+3) = 0
This makes the quotient y²-2y+3.
f(x) = 2 x^2 + 5 sqrt(x+2)
f(0)= 2 (0)^2 + 5 sqrt(2)
5* sqrt (0)= <span>7.07
hope that helps</span>
Answer:
If y=2x+7 were changed to y=5x+7, then the graph of the new function would be steeper than the graph of the original function. On the other hand, the y-intercept would be unchanged; ( 0, 7)
Step-by-step explanation:
If y=2x+7 were changed to y=5x+7, then the graph of the new function would be steeper than the graph of the original function. On the other hand, the y-intercept would be unchanged; ( 0, 7)
Anwser would be 18/14 simplified to 9/7
