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Greeley [361]
3 years ago
13

A line is perpendicular to y=x/3-2 and passes through point( 6,2) write its equation

Mathematics
1 answer:
Katarina [22]3 years ago
6 0

First find slope of perpendicular:

M=1/3, so new slope m= -3

y=mx + b, and subsitute (6,2) and m

2=-3(6) + b, move -18 to other side

b=20

y = -3x+20

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Reduce reduce 36% to its lowest term ​
AVprozaik [17]

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9/25

Step-by-step explanation:

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2 years ago
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The intensity of a light source at a distance is directly proportional to the strength of the source and inversely proportional
jolli1 [7]

Answer:

x=\frac{16}{\sqrt[3]{2}+1 }

Step-by-step explanation:

Q= illumination

I = intensity

Q= I/d^2

Q_total = \frac{I_1}{d_1^2}+\frac{I_2}{d_2^2}

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⇒I{-\frac{2}{x^3}}+\frac{4}{(16-x)^3}

x=\frac{16}{\sqrt[3]{2}+1 }

[/tex]\frac{1}{x^3} = \frac{2}{(16-x)^3}

x=\frac{16}{\sqrt[3]{2}+1 }

is the required point

5 0
3 years ago
Help please!! How do I solve this
egoroff_w [7]
This is an exponential equation. We will solve in the following way. I do not have special symbols, functions and factors, so I work in this way
 2 on (2x) - 5 2 on x + 4=0 =>. (2 on x)2 - 5 2 on x + 4=0  We will replace expression ( 2 on x) with variable t => 2 on x=t  =. t2-5t+4=0 => This is quadratic equation and I solve this in the folowing way => t2-4t-t+4=0 =>     t(t-4) - (t-4)=0 => (t-4) (t-1)=0 => we conclude t-4=0 or t-1=0 => t'=4 and t"=1 now we will return t' => 2 on x' = 4 => 2 on x' = 2 on 2 => x'=2 we do the same with t" => 2 on x" = 1 => 2 on x' = 2 on 0 => x" = 0 ( we know that every number on 0 gives 1). Check 1: 2 on (2*2)-5*2 on 2 +4=0 =>                   2 on 4 - 5 * 4+4=0 => 16-20+4=0 =. 0=0 Identity proving solution. 
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1-5+4=0 => 0=0  Identity provin solution.
5 0
3 years ago
John took 3 exams so far this semester. He scored 45, 65 and 80 on those 3 exams . What score does he need on the fourth exam to
Olegator [25]

Answer:

He \: needs \: to \: score \: 90 \: or \: more

Step-by-step explanation:

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3 0
3 years ago
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