All categories

3 years ago

Convert the following standard form equation into slope-intercept form: 6x + 5y = -15

Mathematics

1 answer:

olasank [31]3 years ago

6 0

Answer:

y= $\frac{-6){5}$ x-3

Step-by-step explanation:

given 6x+5y=-15

5y=-6x-15

y= $\frac{-6}{5}$ x- $\frac{15}{5}$

y= $\frac{-6}{5}$ x-3

slope=[tex]\frac{-6}{5}[tex], intercept=-3

You might be interested in

How do you figure out your ratios

soldi70 [24.7K]

Hello
To find an equal ratio, you can either multiply or divide each term in the ratio by the same number (but not zero). For example, if we divide both terms in the ratio 3:6 by the number three, then we get the equal ratio, 1:2.

7 0

3 years ago

Read 2 more answers

Complete the double number to line show percentages of $50. Fill in the blanks below to match the order of the number line and i

kramer

So the unit rate would just be the numbers all added up

4 0

3 years ago

4/n+9=2/7 adsdwddddfdsfefsef

beks73 [17]

Answer: n= −28 /61

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Please prove this........

Crazy boy [7]

Answer: see proof below

Step-by-step explanation:

Given: A + B + C = π → C = π - (A + B)

→ sin C = sin(π - (A + B)) cos C = sin(π - (A + B))

→ sin C = sin (A + B) cos C = - cos(A + B)

Use the following Sum to Product Identity:

sin A + sin B = 2 cos[(A + B)/2] · sin [(A - B)/2]

cos A + cos B = 2 cos[(A + B)/2] · cos [(A - B)/2]

Use the following Double Angle Identity:

sin 2A = 2 sin A · cos A

Proof LHS → RHS

LHS: (sin 2A + sin 2B) + sin 2C

$\text{Sum to Product:}\qquad 2\sin\bigg(\dfrac{2A+2B}{2}\bigg)\cdot \cos \bigg(\dfrac{2A - 2B}{2}\bigg)-\sin 2C$

$\text{Double Angle:}\qquad 2\sin\bigg(\dfrac{2A+2B}{2}\bigg)\cdot \cos \bigg(\dfrac{2A - 2B}{2}\bigg)-2\sin C\cdot \cos C$

$\text{Simplify:}\qquad \qquad 2\sin (A + B)\cdot \cos (A - B)-2\sin C\cdot \cos C$

$\text{Given:}\qquad \qquad \quad 2\sin C\cdot \cos (A - B)+2\sin C\cdot \cos (A+B)$

$\text{Factor:}\qquad \qquad \qquad 2\sin C\cdot [\cos (A-B)+\cos (A+B)]$

$\text{Sum to Product:}\qquad 2\sin C\cdot 2\cos A\cdot \cos B$

$\text{Simplify:}\qquad \qquad 4\cos A\cdot \cos B \cdot \sin C$

LHS = RHS: 4 cos A · cos B · sin C = 4 cos A · cos B · sin C $\checkmark$

7 0

3 years ago

What is a simpler form of the expression? (2n^2 + 5n +3)(4n - 5)

coldgirl [10]

Answer:

8n³ + 10n² - 13n - 15

Step-by-step explanation:

Distribute the factors by multiplying each term in the first factor by each term in the second factor, that is

4n(2n² + 5n + 3) - 5(2n² + 5n + 3) ← distribute both parenthesis

= 8n³ + 20n² + 12n - 10n² - 25n - 15 ← collect like terms

= 8n³ + 10n² - 13n - 15

5 0

3 years ago

Read 2 more answers