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2 years ago

Find the trade discount amount for an order of merchandise with a list price of $25,000 less trade discounts of 30/25/10.

Mathematics

1 answer:

nasty-shy [4]2 years ago

5 0

Answer:

i wish i could help you but...

Step-by-step explanation:

I don't know the answer to this question. If you have any other questions on your work i could help you..

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3[10-(27÷9) HELP PLEASE

steposvetlana [31]

Answer:

Step-by-step explanation:

You do the parentheses first, so 27/9 = 3

10-3 = 7

3 * 7 = 21

8 0

3 years ago

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Determine the median of the numbers: 6 5 2 2 5 00

zavuch27 [327]

Answer:

Step-by-step explanation:

6 0

3 years ago

The angle between the minute hand and the hour hand of a clock is 90 degrees. What time is it? Justify your answer.

tangare [24]

If it is exactly 90 degrees you are looking for, your answer would be, 3:0:0. Which is 3:00.

3 0

3 years ago

What is the value for x

Step2247 [10]

this is an isosceles triangle... meaning that the two base angles will be equal.

180 - 38 = 142

142 / 2 = 71

x = 71

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3 years ago

Y=-x^2+2x+10 y=x+2 Substitution Please show your work Need ASAP

erik [133]

Answer:

The solutions of the system of equations are the points

$(\frac{1-\sqrt{33}} {2},\frac{5-\sqrt{33}} {2})$

$(\frac{1+\sqrt{33}} {2},\frac{5+\sqrt{33}} {2})$

Step-by-step explanation:

we have

$y=-x^{2} +2x+10$ ----> equation A

$y=x+2$ ----> equation B

Solve the system by substitution

substitute equation B in equation A

$x+2=-x^{2} +2x+10$

solve for x

$-x^{2} +2x+10-x-2=0$

$-x^{2} +x+8=0$

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

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

in this problem we have

$-x^{2} +x+8=0$

$a=-1\\b=1\\c=8$

substitute in the formula

$x=\frac{-1\pm\sqrt{1^{2}-4(-1)(8)}} {2(-1)}$

$x=\frac{-1\pm\sqrt{33}} {-2}$

$x=\frac{-1+\sqrt{33}} {-2}$ -----> $x=\frac{1-\sqrt{33}} {2}$

$x=\frac{-1-\sqrt{33}} {-2}$ -----> $x=\frac{1+\sqrt{33}} {2}$

Find the values of y

For $x=\frac{1-\sqrt{33}} {2}$

$y=x+2$

$y=\frac{1-\sqrt{33}} {2}+2$ ----> $y=\frac{5-\sqrt{33}} {2}$

For $x=\frac{1+\sqrt{33}} {2}$

$y=x+2$

$y=\frac{1+\sqrt{33}} {2}+2$ ----> $y=\frac{5+\sqrt{33}} {2}$

therefore

The solutions of the system of equations are the points

$(\frac{1-\sqrt{33}} {2},\frac{5-\sqrt{33}} {2})$

$(\frac{1+\sqrt{33}} {2},\frac{5+\sqrt{33}} {2})$

5 0

3 years ago