2 years ago

What is the equation of the line in standard form?

Mathematics

1 answer:

madam [21]2 years ago

8 0

Answer:

x+6y=9 - > y= (-1/6)x+(3/2)

x - 6y=-9 - > y= (1/6)x+(3/2)

-9x - 6y = -9 - > y = (-3/2)x +(3/2)

3x - y = - 6 - > y= 3x + 6

-63x - y = - 6 - > y = - 63x + 6

3x + y = 6 - > y = -3x +6

Step-by-step explanation:

The standard form of the equation of a line is: y=ax+b

You seperate y from x and the constant value(the number) and then devide tge equation with the coefficient of y. Obviously it mustnt be division by zero.

You might be interested in

Find m <2 and M<4 kekekskskkskskskskskksksksksks

adoni [48]

Answer:

ok this is a good question

Step-by-step explanation:

i don't even know it

5 0

2 years ago

Lab tests of a new drug indicate a 70% success rate in completely curing the targeted disease. The doctors at the lab created th

Katyanochek1 [597]

Answer: 30%?

Step-by-step explanation:

7 0

2 years ago

HELP GUYS IM REALLY NOT SURE,,

Solnce55 [7]

What u not sure bout?

5 0

3 years ago

Easy! please help! <br><br> 17% is equivalent to? A. 17/100, B. 17 to 100 C. 17:100

nika2105 [10]

The answer is a obviously

3 0

2 years ago

Read 2 more answers

Suppose that prices of recently sold homes in one neighborhood have a mean of $265,000 with a standard deviation of $9300. Using

Hatshy [7]

Answer:

Range = (237100, 292900)

Step-by-step explanation:

Using Chebyshevs Inequality:

$P(|X - \mu | \le k \sigma )\ge 1 -\dfrac{1}{k^2}= 0.889$

$1 -\dfrac{1}{k^2}= 0.889$

$\dfrac{1}{k^2}= 1- 0.889$

$\dfrac{1}{k^2}=0.111$

$k = \sqrt{\dfrac{1}{0.111}}$

$k \simeq 3$

Thus, 88.9% of the population is within 3 standard deviation of the mean with the Range = μ ± kσ

where;

μ = 265000

σ = 9300

Range = 265000 ± 3(9300)

Range = 265000 ± 27900

Range = (265000 - 27900, 265000 + 27900)

Range = (237100, 292900)

3 0

2 years ago