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3 years ago

BEST ANSWER GETS BRAINLIESTT!! PLEASE HELP

1 answer:

7 0

The independent variable is "The number of his friends" 
Equation:
5x
which is same as 5×x

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6/9 = y/6 solve for y

uranmaximum [27]

6/9 = y/6

multiply each side by 6

6*6/9 = y

36/9 =y

4 =y

3 0

3 years ago

Read 2 more answers

What is 10 x 10 to the power 8

barxatty [35]

Your answer is

1,000,000,000 (one billion)

Hope this helps.

5 0

2 years ago

Which prime number do the prime factorization of 130 and 165 have in common?

Marizza181 [45]

Answer:

They both have the #5 and they both have 3 #'s that are prime?

Step-by-step explanation:

5 0

3 years ago

Marine fuel is a combination of gasoline and motor oil. The standard gasoline and oil mixture is about 98% gasoline and about 2%

Georgia [21]

Answer:

0.98g+ 0.02m= 7.1224 gallons standard mixture

0.96g+ 0.04m= 6.25 gallons break oil

3.92 gallons of gasoline is required for 4 gallons the standard mixture

Step-by-step explanation:

If we have to use 6 gallons of gasoline this means that 98%= 6 gallons and 2% would be 6/98*2

Gasoline: Motor oil

6 : x

98 : 2

Using the product rule

x= 6*2/98= 0.1224 gallons of motor oil is required.

Let the gasoline be represented by g and motor oil by m then for the standard oil the equation would be

0.98g+ 0.02m= 7.1224 gallons standard mixture

For break in oil

Gasoline: Motor oil

6 : x

96 : 4

Using the product rule

x= 6*4/96= 0.25 gallons of motor oil is required.

Let the gasoline be represented by g and motor oil by m then for the break oil the equation would be

0.96g+ 0.04m= 6.25 gallons break oil

Part b. For 4 gallons of the "break in" mixture

96% of the 4 gallons of the "break in" mixture is gasoline = 3.84 gallons of gasoline

4% of 4 = 0.16 gallons of motor oil

Now the standard mixture 4 gallons would contain 98 % of gasoline and 2 % of motor oil

98% of 4 gallons= 3.92 gallons of gasoline

2% of 4 gallons= 0.08 gallons of motor oil.

6 0

3 years ago

A triangle is formed from the points L(-3, 6), N(3, 2) and P(1, -8). Find the equation of the following lines:

Dima020 [189]

Answer:

Part A) $y=\frac{3}{4}x-\frac{1}{4}$

Part B) $y=\frac{2}{7}x-\frac{5}{7}$

Part C) $y=\frac{2}{7}x+\frac{8}{7}$

see the attached figure to better understand the problem

Step-by-step explanation:

we have

points L(-3, 6), N(3, 2) and P(1, -8)

Part A) Find the equation of the median from N

we Know that

The median passes through point N to midpoint segment LP

step 1

Find the midpoint segment LP

The formula to calculate the midpoint between two points is equal to

$M(\frac{x1+x2}{2},\frac{y1+y2}{2})$

we have

L(-3, 6) and P(1, -8)

substitute the values

$M(\frac{-3+1}{2},\frac{6-8}{2})$

$M(-1,-1)$

step 2

Find the slope of the segment NM

The formula to calculate the slope between two points is equal to

$m=\frac{y2-y1}{x2-x1}$

we have

N(3, 2) and M(-1,-1)

substitute the values

$m=\frac{-1-2}{-1-3}$

$m=\frac{-3}{-4}$

$m=\frac{3}{4}$

step 3

Find the equation of the line in point slope form

$y-y1=m(x-x1)$

we have

$m=\frac{3}{4}$

$point\ N(3, 2)$

substitute

$y-2=\frac{3}{4}(x-3)$

step 4

Convert to slope intercept form

Isolate the variable y

$y-2=\frac{3}{4}x-\frac{9}{4}$

$y=\frac{3}{4}x-\frac{9}{4}+2$

$y=\frac{3}{4}x-\frac{1}{4}$

Part B) Find the equation of the right bisector of LP

we Know that

The right bisector is perpendicular to LP and passes through midpoint segment LP

step 1

Find the midpoint segment LP

The formula to calculate the midpoint between two points is equal to

$M(\frac{x1+x2}{2},\frac{y1+y2}{2})$

we have

L(-3, 6) and P(1, -8)

substitute the values

$M(\frac{-3+1}{2},\frac{6-8}{2})$

$M(-1,-1)$

step 2

Find the slope of the segment LP

The formula to calculate the slope between two points is equal to

$m=\frac{y2-y1}{x2-x1}$

we have

L(-3, 6) and P(1, -8)

substitute the values

$m=\frac{-8-6}{1+3}$

$m=\frac{-14}{4}$

$m=-\frac{14}{4}$

$m=-\frac{7}{2}$

step 3

Find the slope of the perpendicular line to segment LP

Remember that

If two lines are perpendicular, then their slopes are opposite reciprocal (the product of their slopes is equal to -1)

$m_1*m_2=-1$

we have

$m_1=-\frac{7}{2}$

$m_2=\frac{2}{7}$

step 4

Find the equation of the line in point slope form

$y-y1=m(x-x1)$

we have

$m=\frac{2}{7}$

$point\ M(-1,-1)$ ----> midpoint LP

substitute

$y+1=\frac{2}{7}(x+1)$

step 5

Convert to slope intercept form

Isolate the variable y

$y+1=\frac{2}{7}x+\frac{2}{7}$

$y=\frac{2}{7}x+\frac{2}{7}-1$

$y=\frac{2}{7}x-\frac{5}{7}$

Part C) Find the equation of the altitude from N

we Know that

The altitude is perpendicular to LP and passes through point N

step 1

Find the slope of the segment LP

The formula to calculate the slope between two points is equal to

$m=\frac{y2-y1}{x2-x1}$

we have

L(-3, 6) and P(1, -8)

substitute the values

$m=\frac{-8-6}{1+3}$

$m=\frac{-14}{4}$

$m=-\frac{14}{4}$

$m=-\frac{7}{2}$

step 2

Find the slope of the perpendicular line to segment LP

Remember that

If two lines are perpendicular, then their slopes are opposite reciprocal (the product of their slopes is equal to -1)

$m_1*m_2=-1$

we have

$m_1=-\frac{7}{2}$

$m_2=\frac{2}{7}$

step 3

Find the equation of the line in point slope form

$y-y1=m(x-x1)$

we have

$m=\frac{2}{7}$

$point\ N(3,2)$

substitute

$y-2=\frac{2}{7}(x-3)$

step 4

Convert to slope intercept form

Isolate the variable y

$y-2=\frac{2}{7}x-\frac{6}{7}$

$y=\frac{2}{7}x-\frac{6}{7}+2$

$y=\frac{2}{7}x+\frac{8}{7}$

7 0

3 years ago