All categories

3 years ago

Find the x- and y-intercept of the line with equation -3x+9y=18

1 answer:

8 0

Plug in “0” for y, rearrange and solve.
Remember when you want to find the intercept of a variable, you plug in 0 for the opposite variable. Same goes for finding the x intercept.

When you want to find…
Y intercept: X=0
X intercept: Y=0

Answers:
X int= -6 or (-6,0)
Y int=2 or (0,2)

You might be interested in

If 6 pounds of bananas for $4.20!, what is the unit price

vredina [299]

Answer:

70 cents

Step-by-step explanation:

Try 4.20 Divided by 6

8 0

3 years ago

Solve for a: 3a – 5(a – 2) = 34 a = -12 a = 0 a = -22 a = 1

Evgen [1.6K]

3a - 5(a - 2) = 34

Distribute -5 inside the parentheses

3a - 5a + 10 = 34

Combine like terms

-2a + 10 = 34

Subtract 10 from both sides

-2a = 24

Divide both sides by -2

a = -12

8 0

4 years ago

Read 2 more answers

What is the solution to the equation?

AleksandrR [38]

Solution of the equation: $x=6\frac{1}{2}$

Step-by-step explanation:

The equation that we have to solve in this problem is:

$13\frac{3}{4}+x=7\frac{1}{4}$

The first step to do is to rewrite the mixed fractions as improper fractions. We have:

$13\frac{3}{4}=\frac{13\cdot 4+3}{4}=\frac{52+3}{4}=\frac{55}{4}$

And

$7\frac{1}{4}=\frac{7\cdot 4+1}{4}=\frac{29}{4}$

So the equation becomes

$\frac{29}{4}+x=\frac{55}{4}$

Now we multiply by 4 each term on both sides, and we get

$29+4x=55$

Now we subtract 29 from both sides,

$4x=55-29=26$

And finally, we divide both sides by 4:

$x=\frac{26}{4}=\frac{13}{2}=6\frac{1}{2}$

Learn more about equations:

brainly.com/question/11306893

brainly.com/question/10387593

#LearnwithBrainly

4 0

3 years ago

Read 2 more answers

Quadratic Formula to solve the equation x^2- 4x = - 7

mylen [45]

roots of the equation x^2- 4x = - 7 are 2+3i and 2-3i

What are the natures of root of quadratic equation?

Case I: $4ac > b^2$

The roots of the quadratic equation $ax^2 + bx + c = 0$ are real and unequal when a, b, and c are real numbers, a 0, and the discriminant is positive.

Situation II: $b^2$ $-4ac$ =0

The roots and of the quadratic equation $ax^2 + bx + c = 0$ are real and equal when a, b, and c are real numbers, a 0, and the discriminant is zero.

Case III: $b^2$ $-4ac$ <0

The quadratic equation $ax^2 + bx + c = 0$ has roots when a, b, and c are real numbers, a 0, and the discriminant is negative, but these roots are not equal and are not real. We refer to the roots in this instance as fictitious.

Case IV: Perfect square and $b^2 - 4ac > 0$

The roots of the quadratic equation $ax^2 + bx + c = 0$ are real, rational, and unequal when a, b, and c are real numbers, a 0, and the discriminant is positive and perfect square.

Quadratic Formula $x={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}$