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3 years ago

What is the constant of variation k of the line y=kx through (5,8)

Mathematics

2 answers:

mina [271]3 years ago

8 0

Answer:

k= 8/5

Step-by-step explanation:

here x= 5 and y= 8

we have y=kx

k= y/x

k= 8/5 this is constant of variation

patriot [66]3 years ago

4 0

Answer:

$k=\frac{8}{5}$

Step-by-step explanation:

find the constant of variation k of the line y=kx through (5,8)

Use equation of line $y=kx$

k is the constant of variation

LEts find out k using the point (5,8)

x=5 and y=8, plug in the values

$8=k(5)$

To get k alone, divide by 5 on both sides

$k=\frac{8}{5}$

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What is 2 15/16 Ian’s a decimal?

Arada [10]

Answer:

2.9375

Step-by-step explanation:

15 divided by 16 equals .9375 so 2 15/16 is 2.9375

5 0

2 years ago

Darnell ran 550 yards in 3 minutes. At this rate, how many yards can he run in 6 minutes?

Elan Coil [88]

Answer:

Darnell can run 1100 yards in 6 minutes

Step-by-step explanation:

There are 2 ways you can solve this

Method 1: You would have to divide 550 by 3 which you would get about 183, (the exact answer is very long so I recommend using a calculator). Then you multiply that by 6 to get 1100.

Method 2: This one would probably be easier. Since 3 x 2 is 6 then you can multiply 550 x 2 to get 1100.

I hope this helps!

4 0

2 years ago

Read 2 more answers

Help with Pre Calculus

marshall27 [118]

Answer:

The first option

Step-by-step explanation:

The domain of a rational function should be all real numbers except for when the denominator is equal to 0. To find when the denominator is equal to 0 you simply need to find the zeroes of the denominator... but in this case you can do that through factoring and using the quadratic equation.

So first step is going to be to factor out the GCF, which in this case is x. This gives you the equation. $x(2x^2-x-15)$ . So one of the zeroes is when x=0. Now to find the other two zeroes you can use the quadratic equation which is $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\$ . So to find the other zeroes you simply plug the values in. a=2, b=-1, c=-15

$x=\frac{-(-1)\pm\sqrt{(-1)^2-4(2)(-15)}}{2(2)}\\\\x=\frac{1\pm\sqrt{1-4(2)(-15)}}{4}\\\\x=\frac{1\pm\sqrt{121)}}{4}\\\\x=\frac{1\pm11}{4}\\\\x=\frac{12}{4}\\x=3\\\\x=\frac{-10}{4}\\x=-\frac{5}{2}$