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3 years ago

Is the graph of y=3/2x-4 paralel to the graph of y=-3/2x-4

Mathematics

1 answer:

Sholpan [36]3 years ago

3 0

The slope of the first function is 3/2, and the slope of the second is -3/2.

For two graphs to be parallel, the slope must be the same.

3/2 does not equal -3/2, and therefore the two graphs are not parallel.

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What’s 16+5x+x=1+3x ??

kati45 [8]

Answer:

-5

Step-by-step explanation:

1: simplify both sides of the equation. (combine like terms)

2: subtract 3x from both sides

3: subtract 16 from both sides

4: divide both sides by 3

i hope that helped!

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Step-by-step explanation:

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3 years ago

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A plane flies at x miles per hour in still air. Flying with a tailwind, its speed is 485 miles per hour. Against the wind, its a

Svetradugi [14.3K]

Answer:20mph

Step-by-step explanation:

speed of plane = x mph

Speed of plane with wind = 485 mph

So therefore, speed of wind =

485 - x

Speed of plane against wind = 445 mph

So therefore, speed of wind =

x - 445

So therefore, equating the 2 equations of speed of wind

485 - x = x - 445

Collect like terms

485 + 445 = x + x

930 = 2x

2x = 930

x = 930/2 = 465 mph

Speed of wind = x = 465mph

So we put this value on any of the equations for speed of wind

Speed of wind = x - 445 = 465 - 445

= 20 mph

So therefore, speed of wind = 20mph

7 0

2 years ago

If n is a positive integer, how many 5-tuples of integers from 1 through n can be formed in which the elements of the 5-tuple ar

erma4kov [3.2K]

Answer:

$\frac{(n+4)*(n+3)*(n+2)*(n+1)*n}{120}$

Step-by-step explanation:

Given

5 tuples implies that:

$n = 5$

$(h,i,j,k,m)$ implies that:

$r = 5$

Required

How many 5-tuples of integers $(h, i, j, k,m)$ are there such that $n\ge h\ge i\ge j\ge k\ge m\ge 1$

From the question, the order of the integers h, i, j, k and m does not matter. This implies that, we make use of combination to solve this problem.

Also considering that repetition is allowed: This implies that, a number can be repeated in more than 1 location

So, there are n + 4 items to make selection from

The selection becomes:

$^{n}C_r => ^{n + 4}C_5$

$^{n + 4}C_5 = \frac{(n+4)!}{(n+4-5)!5!}$

$^{n + 4}C_5 = \frac{(n+4)!}{(n-1)!5!}$

Expand the numerator

$^{n + 4}C_5 = \frac{(n+4)!(n+3)*(n+2)*(n+1)*n*(n-1)!}{(n-1)!5!}$

$^{n + 4}C_5 = \frac{(n+4)*(n+3)*(n+2)*(n+1)*n}{5!}$

$^{n + 4}C_5 = \frac{(n+4)*(n+3)*(n+2)*(n+1)*n}{5*4*3*2*1}$

$^{n + 4}C_5 = \frac{(n+4)*(n+3)*(n+2)*(n+1)*n}{120}$

Solved

6 0

2 years ago

Solve the equation for y Y minus 5 equals 1/3 (x plus 12)

Vinvika [58]

Answer:

ok so this problem you are basically just rearranging it. you're going to want y by itself.

Step-by-step explanation:

let's start with the original equation: 5-y=1/3x +12

so first you have to move the five over, so you subtract 5 from both sides. -y=1/3x+12-5

then you subtract the 5 from the twelve since they're like terms: -y=1/3x+7

and finally you have to move the negative over. when you move the negative think of it like dividing negative one from both sides, everything on the other side has to get the negative.

so we're left with y=-1/3x-7

I hope this helps!

5 0

2 years ago