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2 years ago

Consider the following equation. Graph 5x + 6y = 30 PLEASE SHOW STEPS

Mathematics

1 answer:

MissTica2 years ago

6 0

Answer:

Step-by-step explanation:

Convert the equation into slope-intercept form.

5x-6y =30 Subtract 5x from each side

-6y = -5x + 30 Multiply each term by -1

6y = 5x – 30 Divide each side by 6

y = ⅚x - 5

2. Create a table of different x and y values,

x ⅚x – 5 y

-12 ⅚(-12) – 5 -15

-6 ⅚(-6) – 5 -10

0 ⅚(0) – 5 -5

6 ⅚(6) – 5 0

12 ⅚(12) – 5 5

3. Convert the table to ordered pairs

(-12,-15)

(-6, -10)

(0, -5)

(6, 0)

(12, -5)

4. Plot the points and draw a line through them.

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I need help solving this problem.

madam [21]

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2x^2 + 3x + 4x - 9 = 2x^2 +7x -9

2x^2 + 7x - 9 - x^2 -7x = x^2 -9

The answer is : x^2 - 9

6 0

3 years ago

Read 2 more answers

Find the value of x.

zavuch27 [327]

Answer:

Step-by-step explanation:

(6)² = 3x

then x= 36/3

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7 0

2 years ago

What is the quotient when 4x3 + 2x + 7 is divided by x + 3?

Arte-miy333 [17]

Answer:

The quotient of this division is $(4x^2 -12x + 38)$ . The remainder here would be $-26$ .

Step-by-step explanation:

The numerator $4x^3 + 2x + 7$ is a polynomial about $x$ with degree $3$ .

The divisor $x + 3$ is a polynomial, also about $x$ , but with degree $1$ .

By the division algorithm, the quotient should be of degree $3 - 1 = 2$ , while the remainder shall be of degree $1 - 1 = 0$ (i.e., the remainder would be a constant.) Let the quotient be $a\,x^2 + b\, x + c$ with coefficients $a$ , $b$ , and $c$ .

$4x^3 + 2x + 7 = \left(a\,x^2 + b\, x + c\right)(x + 3)$ .

Start by finding the first coefficient of the quotient.

The degree-three term on the left-hand side is $4 x^3$ . On the right-hand side, that would be $a\, x^3$ . Hence $a = 4$ .

Now, given that $a = 4$ , rewrite the right-hand side:

$\begin{aligned}&\left(4\,x^2 + b\, x + c\right)(x + 3) \cr =& \left(4x^2 + (b\, x + c)\right)(x + 3) \cr =& 4x^2(x + 3) + (bx + c)(x + 3) \cr =& 4x^3 + 12x^2 + (bx + c)(x + 3)\end{aligned}$ .

Hence:

$4x^3 + 2x + 7 = 4x^3 + 12x^2 + (b\,x + c)(x + 3)$

Subtract $\left(4x^3 + 12x^2\right$ from both sides of the equation:

$-12x^2 + 2x + 7 = (b\,x + c)(x + 3)$ .

The term with a degree of two on the left-hand side has coefficient $(-12)$ . Since the only term on the right hand side with degree two would have coefficient $b$ , $b = -12$ .

Again, rewrite the right-hand side:

$\begin{aligned}&\left(-12 x + c\right)(x + 3) \cr =& \left(-12 x+ c\right)(x + 3) \cr =& (-12x)(x + 3) + c(x + 3) \cr =& -12x^2 -36x + (bx + c)(x + 3)\end{aligned}$ .