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3 years ago

What is a markup that is more than 200%?

Mathematics

1 answer:

RideAnS [48]3 years ago

6 0

If I understand the concept of markups correctly I would say 500%

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Solve the Equation p+5/8=15/16?

Sphinxa [80]

So the first thing you want to do when faced with two fractions with different denominators (when subtracting or adding) is to make the denominators the same. So for this equation they would turn out to be p+10/16=15/16 (because 16 is the lowest common denominator, 8 times two) so then you want to subtract 10/16 from 15/16 to isolate the variable (p) which would get:
p=5/16
This is the final answer because it cannot be simplified.
Hope this helps!

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3 years ago

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Answer: A

Step by step explanation:

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3 years ago

In simplest radical form, what are the solutions to the quadratic equation 6 = x2 – 10x?

vichka [17]

Answer:

$x = 5+\sqrt{31}\,\, and\,\, x=5-\sqrt{31}$

Step-by-step explanation:

We need to solve the quadratic equation

6 = x^2 -10x

Rearranging we get,

x^2-10x-6=0

Using quadratic formula to solve the quadratic equation

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

a= 1, b =-10 and c=6

Putting values in the quadratic formula

$x=\frac{-(-10)\pm\sqrt{(-10)^2-4(1)(-6)}}{2(1)}\\x=\frac{10\pm\sqrt{100+24}}{2}\\x=\frac{10\pm\sqrt{124}}{2}\\x=\frac{10\pm\sqrt{2*2*31}}{2}\\x=\frac{10\pm\sqrt{2^2*31}}{2}\\x=\frac{10\pm2\sqrt{31}}{2}\\x = 5\pm\sqrt{31}$

So, $x=5+\sqrt{31}\,\, and\,\, x=5-\sqrt{31}$

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3 years ago

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