All categories

3 years ago

How do change slope-intercept form to standard form

2 answers:

7 0

Suppose you are given the slope-intercept form y = 5x - 1/16 and are to change it to standard form. Here's what you'd do:

1) multiply all 3 terms by 16 to eliminate the fraction 1/16:

16y = 80x - 1

Subtr. 16y from both sides: 0 = 80x - 16y - 1

This equation is in std. form.

olganol [36]3 years ago

6 0

Example. Convert y=54x+5 y = 5 4 x + 5 , graphed on the right, to standard form. The only fraction is $$ \frac { 5}{ \red 4} $$ so you can multiply everything by 4. Solve for the y-intercept (or "b" in the slope interceptequation) which is 20 in this example.

You might be interested in

0.034 is 1/10 of 340

sergeinik [125]

Nope

1/10 of 240 is 3.4

7 0

3 years ago

Read 2 more answers

A bag contains 2 blue marbles, 3 green marbles, 1 white marble and 4 red marbles.

Dima020 [189]

Answer:

2/5

Step-by-step explanation:

add all marbles together: 10 marbles

since only 4 are red, the fraction would be 4/10

4/10 can be simplified to 2/5

3 0

2 years ago

Read 2 more answers

Determine whether each equation is True or False. In case you find a "False" equation, explain why is False.

elixir [45]

Answer:

(1) TRUE.

(2) FALSE.

(3) FALSE.

(4) TRUE.

(5) FALSE.

Step-by-step explanation:

(1) $\sqrt{32} = 2^{\frac{5}{2} }$

$2^{\frac{5}{2} } = (\sqrt{2} )^5 = (\sqrt{2} \ \times \ \sqrt{2} \ \times \ \sqrt{2} \ \times \ \sqrt{2} \ \times \ \sqrt{2}) = 4\sqrt{2}\\\\\sqrt{32} = \sqrt{16 \ \times \ 2}\ = \ \sqrt{16} \ \times \ \sqrt{2} \ = \ 4\sqrt{2}$

Thus, the equation is TRUE.

(2) $16^{\frac{3}{8} } = 8^2$

$16^{\frac{3}{8} } =(2^4)^{\frac{3}{8} } = 2^\frac{3}{2} }= (\sqrt{2} )^3 = (\sqrt{2} \ \times \ \sqrt{2} \ \times \ \sqrt{2}) = 2\sqrt{2} \\\\8^2 = 64$

Thus, the equation is FALSE.

(3) $4^{\frac{1}{2} } = \sqrt[4]{64}$

$4^{\frac{1}{2} }= \sqrt{4} = 2\\\\\sqrt[4]{64} = (64)^{\frac{1}{4} } = (2^6)^{\frac{1}{4} }= 2^{\frac{6}{4} } = 2^{\frac{3}{2} }=(\sqrt{2} )^3 = (\sqrt{2} \times \sqrt{2} \times \sqrt{2} ) = 2\sqrt{2}$

Thus, the equation is FALSE.

(4) $2^8 = (\sqrt[3]{16} )^6$

$2^8 = 256\\\\ (\sqrt[3]{16} )^6 = (16)^{\frac{6}{3} } = (2^4)^{\frac{6}{3} } = (2)^{\frac{24}{3} } = 2^8 = 256$

Thus, the equation is TRUE.

(5) $(\sqrt{64} )^{\frac{1}{3} } = 8^{\frac{1}{6} }\\\\$

$8^{\frac{1}{6} } = (2^3)^{\frac{1}{6} } = 2^{\frac{3}{6} } = 2^{\frac{1}{2} } = \sqrt{2} \\\\(\sqrt{64} )^{\frac{1}{3} } = (2^6)^{\frac{1}{3} } = 2^{\frac{6}{3} } = 2^2 = 4$