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3 years ago

6) What is the slope of the line graphed by the equation 3y = 2x + 3?

Mathematics

1 answer:

Ghella [55]3 years ago

6 0

c. 2/3

you divide each side by 3 to get y alone y=2/3x+1

so your slope is 2/3 and 1 is your y-intercept.

just a little math work to show u glad to help peace.

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A quadratic function is a function of the form y=ax^2+bx+c where a, b, and c are constants. Given any 3 points in the plane, the

pochemuha

Answer:

The quadratic function whose graph contains these points is $y=-x^{2}-2x-2$

Step-by-step explanation:

We know that a quadratic function is a function of the form $y=ax^{2}+bx+c$ . The first step is use the 3 points given to write 3 equations to find the values of the constants a,b, and c.

Substitute the points (0,-2), (-5,-17), and (3,-17) into the general form of a quadratic function.

$-2=a*0^{2}+b*0+c\\c=-2$

$-17=a*-5^{2}+b*-5+c\\c=-25a+5b-17$

$-17=a*3^{2}+b*3+c\\ c=-9a-3b-17$

We can solve these system of equations by substitution

Substitute $c=-9a-3b-17$

$-9a-3b-17=25a+5b-17\\-9a-3b-17=-2$

Isolate a for the first equation

$-9a-3b-17=-25a+5b-17\\a=\frac{b}{2}$

Substitute $a=\frac{b}{2}$ into the second equation

$-9\left(-\frac{b}{2}\right)-3b-17=-2$

Find the value of b

$-9\left(-\frac{4b}{17}\right)-3b-17=-2\\ b=-2$

Find the value of a

$a=\frac{b}{2}\\ a=-1$

The solutions to the system of equations are:

b=-2,a=-1,c=-2

So the quadratic function whose graph contains these points is

$y=-x^{2}-2x-2$

As you can corroborate with the graph of this function.

8 0

3 years ago

Which set of coefficients of the terms in the expansion of the binomial (x+y)^3 is correct ?

BigorU [14]

We are technically FOILing this out... with a power of 3.

(x+y)(x+y)(x+y)

So we can first factor out the first two "x+y"s.

$( x^{2} +2xy + y^{2} )$ , multiplied by the last "x+y".

$x^{3} +3 x^{2} y +3x y^{2} + y^{3}$

Coefficients are the number that comes in front of a variable.

In this case, 1 comes in front of $x^{3}$ , 3 comes in front of $x^{2} y$ , 3 comes in front of $x y^{2}$ , and 1 comes in front of $y^{3}$ .

Thus: 1, 3, 3, 1.
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