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Sav [38]
3 years ago
5

PLEASE HELP ASAP 15 BRAINLY POINTS

Mathematics
2 answers:
lana66690 [7]3 years ago
5 0

Answers

1) 1

2) 11

3) 4

4) D) Yes, the constant of variation is 3

Explanation

Q1

The slope of a line is given by

Slope = (change in y)/(change in x)

= (0 - -2)/(-4 - -6)

= 2/2

= 1

Q2

Slope = Δy/Δx

= (8 - -3)/(2 - 1)

= 11/1

= 11

Q3

Y varies directly as X. This is written as,

Y ∞ X

Put and equal sign then introduce the constant of proportionality, K.

Y = KX

K = Y/X

K = 6/18 = 1/3

∴ Y = (1/3)X

Y = (1/3) × 12

Y = 4

Q4

32Y = 96X

Dividing both sides by 32,

Y = 96/32 X

Y = 3X

3 been the constant, Y varies directly as X.

So, the equation 32y = 96x is a direct variation.

Constant of variation = 3

joja [24]3 years ago
3 0

the answers are 1) 1 2) 11 3) 4 4) D) 3

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Given Angle 1 = Angle 2,<br> find x.
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Answer:

x = 6

Step-by-step explanation:

Given ∠ 1 = ∠ 2 then the segment is an angle bisector and the ratios of sides to base are equal, that is

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x(x - 4) = 12 ← distribute left side

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8 0
4 years ago
Please help me ASAP!
Over [174]
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4 0
3 years ago
Find two consecutive integers whose product is 50
GarryVolchara [31]

n, n+1 - two consecutive integers

n(n + 1) = 50     <em>use distributive property</em>

n² + n = 50     <em>subtract 50 from both sides</em>

n² + n - 50 = 0

-----------------------------------------------------

ax² + bx + c =0

if b² - 4ac > 0 then we have two solutions:

[-b - √(b² - 4ac)]/2a and [-b - √(b² + 4ac)]/2a

if b² - 4ac = 0 then we have one solution -b/2a

if  b² - 4ac < 0 then no real solution

----------------------------------------------------------

n² + n - 50 = 0

a = 1, b = 1, c = -50

b² - 4ac = 1² - 4(1)(-50) = 1 + 200 = 201 > 0 → two solutions

√(b² - 4ac) = √(201) - it's the irrational number

Answer: There are no two consecutive integers whose product is 50.

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4 years ago
Simplify: –3(y + 2)2 – 5 + 6yWhat is the simplified product in standard form?
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Answer by Mimiwhatsup:

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6 0
4 years ago
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Whitepunk [10]
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2(10-24•1/4)+8^2
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8+64=72





4 0
4 years ago
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