All of the following are equivalent except 1) 4 y 2) (4) (Y) 3) 4•Y 4)4+ y

Mathematics

1 answer:

xz_007 [3.2K]2 years ago

5 0

4 because all of the others involve multiplication and 4 involves addition.

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Write the equation of a line through points (5,5) and (1, 0) in point-slope form?

algol [13]

Answer:

Step-by-step explanation:

I’m assuming that the equation is to be written in slope intercept form (y=mx+b), so for this, you want to find the slope and y-intercept. To find the slope, use the equation in picture 1.

0 - 5 = -5

1 - 5 = -4

-5 / -4 = 1.25

So m, or the slope, is equal to 1.25.

To find the y-intercept, make a table like the second picture. Choose one of the ordered pairs as the given. I used 1,0.

0,?

1,0

Now find the change in x, which in this case, is 1. Then set up something like picture three.

1.25/1 = ?/1

?=1.25

Now backtrack to get:

0, -1.25

1, 0

Now we know what is y when x is 0, we know the y intercept and can make the equation:

y = 1.25x - 1.25

5 0

2 years ago

Use the quadratic formula to solve the following equation -3x^2-x-3=0

tigry1 [53]

<u>Answer:</u>

$x=-\frac{1}{6}-\frac{\sqrt{35}}{6} i \text { and } x=-\frac{1}{6}+\frac{\sqrt{35}}{6} i$ are two roots of equation $-3 x^{2}-x-3=0$

<u>Solution:</u>

Need to solve given equation using quadratic formula.

$-3 x^{2}-x-3=0$

General form of quadratic equation is $a x^{2}+b x+c=0$

And quadratic formula for getting roots of quadratic equation is

$x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

In our case b = -1 , a = -3 and c = -3

Calculating roots of the equation we get

$\begin{array}{l}{x=\frac{-(-1) \pm \sqrt{(-1)^{2}-4(-3)(-3)}}{2 \times-3}} \\\\ {x=-\frac{1}{6} \pm\left(-\frac{\sqrt{-35}}{6}\right)}\end{array}$

Since $b^{2}-4 a c$ is equal to -35, which is less than zero, so given equation will not have real roots and have complex roots.

$\begin{array}{l}{\text { Hence } x=-\frac{1}{6}-\frac{\sqrt{35}}{6} i \text { and } x=-\frac{1}{6}+\frac{\sqrt{35}}{6} i \text { are two roots of equation - }} \\ {3 x^{2}-x-3=0}\end{array}$