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ale4655 [162]
3 years ago
14

Which function is most likely graphed on the coordinate plane below? What is the equation of the line perpendicular to 2x – 3y =

13 that passes through the point (–6, 5)?
Y=2/3 x +9
Y=-3/2x-4
Y=-3/2 x -13
Y=2/3x-1
Mathematics
1 answer:
masha68 [24]3 years ago
4 0

Answer:

y=\frac{-3}{2}x-4

Step-by-step explanation:

the equation of the line perpendicular to 2x – 3y = 13

It passes through the point (–6, 5)

2x -3y = 13

Subtract 2x from both sides

-3y =-2x+13

Divide both sides by -3

y=\frac{2}{3}x-\frac{13}{3}

Slope = 2/3

Slope of perpendicular lines are negative reciprocal of one another

slope of perpendicular line is \frac{-3}{2}

Point (-6,5)

y-y1=m(x-x1)

y-5=\frac{-3}{2}(x+6)

y-5=\frac{-3}{2}x-9

Add 5 on both sides

y=\frac{-3}{2}x-4

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spayn [35]

Answer:

-5

Step-by-step explanation:

-5 is the slope because it is the multiplier of x

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2 years ago
Factor 16x+40y I don't know how to do this
Marizza181 [45]

Answer:

8(2x + 5y)

Step-by-step explanation:

You cant fact (take out) 8 from both 16 and 40 since 8 is a factor for both.

4 0
3 years ago
Read 2 more answers
<img src="https://tex.z-dn.net/?f=%285y%20%2B%209%29%286y%20-%201%29" id="TexFormula1" title="(5y + 9)(6y - 1)" alt="(5y + 9)(6y
mestny [16]
Hey there :)

( 5y + 9 )( 6y - 1 )

We need to use FOIL to expand, that is
First Terms
Outer Terms
Inner Terms
Last Terms

     First          Outer         Inner      Last
( 5y )( 6y ) + ( 5y )( - 1 ) + 9 ( 6y ) + 9 ( - 1 )
    30y²     -        5y      +    54y   -       9

Combine, if any, the like-terms
30y² + 49y - 9
7 0
3 years ago
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Given the pdf of f(x) =2x for 0&lt;x&lt;2 and zero elsewhere, find the cdf​
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Hope it helps.........

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3 years ago
Water whose temperature is at 100∘C is left to cool in a room where the temperature is 60∘C. After 3 minutes, the water temperat
Tju [1.3M]

Answer:

21.68 minutes ≈ 21.7 minutes

Step-by-step explanation:

Given:

T=60+40e^{kt}

Initial temperature

T = 100°C

Final temperature = 60°C

Temperature after (t = 3 minutes) = 90°C

Now,

using the given equation

T=60+40e^{kt}

at T = 90°C and  t = 3 minutes

90=60+40e^{k(3)}

30=40e^{3k}

or

e^{3k}=\frac{3}{4}

taking the natural log both sides, we get

3k = \ln(\frac{3}{4})

or

3k = -0.2876

or

k = -0.09589

Therefore,

substituting k in 1 for time at temperature, T = 65°C

65=60+40e^{( -0.09589)t}

or

5=40e^{( -0.09589)t}

or

e^{( -0.09589)t}=\frac{5}{40}

or

e^{( -0.09589)t}=0.125

taking the natural log both the sides, we get

( -0.09589)t = ln(0.125)

or

( -0.09589)t = -2.0794

or

t = 21.68 minutes ≈ 21.7 minutes

6 0
3 years ago
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