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ryzh [129]
3 years ago
13

line v has an equation of y=10/9x +1. Line w includes the point (-3,3) and is perpendicular to line v. What is the equation of l

ine w?
Mathematics
1 answer:
NARA [144]3 years ago
3 0

Answer:

y = -9/10x + 3/10 or y = 3/10(1 - 3x)

Step-by-step explanation:

Perpendicular lines have slopes that are negative reciprocals of one another

so slope of line w is -9/10

Using (-3,3) >> y = -9/10x + b

3 = -9/10(-3) + b

3 = 27/10 + b

b = 3 - 27/10 = 30/10 - 27/10 = 3/10

y = -9/10x + 3/10 or y = 3/10(1 - 3x)

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aniked [119]
Just for future reference always put ^ for exponents. The answer is y-6 though. 
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7 0
3 years ago
Read 2 more answers
4x^2+2xy-10y^2+9=0 what is the value of a,b,c and the discriminant?
zavuch27 [327]

basically you treat y like a number and not a variable

Answer:

a is 4

b is 2y

c is -10y²+9

discriminant is 4(41y²-36)

Step-by-step explanation:

4x²+2xy-10y²+9=0

rewrite in standard form of a quadratic equation like ax² + bx + c = 0

4x²+2yx-10y²+9=0

basically you treat y like a number and not a variable

a is the number with the x²

right away we know a is 4 because of 4x²

b is the one with x so in this formula b is 2y

c is the number without the x which in this case is -10y²+9

discriminant is

b² - 4ac

(2y)²- (4)(4)(-10y²+9)

4y²-(16)(-10y²+9)

4y²-(16)(-10y²+9)

4y²+160y²-144

164y²-144 =

4(41y²-36)

6 0
2 years ago
Please help will give brainliest
Gelneren [198K]

Answer:

B

Step-by-step explanation:

It is B because to find the volume you multiply all three values together, which gets you 68.448. I multiplied 9.2x3.1, which got me 28.52, and then I multiplied 28.52 by 2.4 which got me 68.448.

6 0
3 years ago
How do u solve this?? I'm confused
rjkz [21]
You will take 100 = 15.5x - 7 and solve.

the end product comes out to be 6 = x or (OCD is kicking) x = 6.

hope this helps (; hmu if u need more info
7 0
4 years ago
Write the following numbers in order starting with the smallest 1.75kg 1975g 0.9kg 1800g
Jobisdone [24]

0.9kg

1.75kg

1800g

1975g

6 0
3 years ago
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