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2 years ago

15

Find the selling price. Cost to store 130$. Markup 85%

1 answer:

AlexFokin [52]2 years ago

4 0

Answer:

$240.5

Step-by-step explanation:

Multiply 0.85 and 130 to get 110.5. Add this to the original price, 130, to get 240.5.

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Y – 1 = 2(x – 2), solve for y

solmaris [256]

Answer:

y = 2x-3

Step-by-step explanation:

y – 1 = 2(x – 2)

Distribute

y-1 =2x-4

Add 1 to each side

y-1+1 = 2x-4+1

y = 2x-3

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2 years ago

Which inequality has no solution?<br> 6(x+2) > x-3<br> 3+4x2(1+ 2x)<br> -2(x+6) X-9 <3(x-3)

Juli2301 [7.4K]

Answer:

$6(x + 2) > x - 3$

$6x + 12 > x - 3$

$6x - x > - 12 - 3$

$5x > - 9$

$3 + 4 \times 2(1 + 2x)$

$3 + 8(1 + 2x)$

$3 + 8 + 16x$

$11 + 16x$

$- 2(x + 6)x - 9 < 3(x - 3)$

$- 2( {x}^{2} + 6x) - 9 < 3x - 9$

$- 2 {x}^{2} - 12x - 9 < 3x - 9$

5 0

2 years ago

Please answer this asap.

vodomira [7]

The answer would be 39.78

4 0

2 years ago

Read 2 more answers

Find the solution to each equation.

Alex

Answer:

k = 16

p = 21

m = 24

c = 23

g = 18

Step-by-step explanation:

All you had to do was divide the product by the number with the variable

7 0

1 year ago

Which equation has the solutions x=1+or-square root of 5?

stiv31 [10]

We will proceed to solve each case to determine the solution of the problem.

<u>case a)</u> $x^{2}+2x+4=0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$x^{2}+2x=-4$

Complete the square. Remember to balance the equation by adding the same constants to each side.

$x^{2}+2x+1=-4+1$

$x^{2}+2x+1=-3$

Rewrite as perfect squares

$(x+1)^{2}=-3$

$(x+1)=(+/-)\sqrt{-3}\\(x+1)=(+/-)\sqrt{3}i\\x=-1(+/-)\sqrt{3}i$

therefore

case a) is not the solution of the problem

<u>case b)</u> $x^{2}-2x+4=0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$x^{2}-2x=-4$

Complete the square. Remember to balance the equation by adding the same constants to each side.

$x^{2}-2x+1=-4+1$

$x^{2}-2x+1=-3$

Rewrite as perfect squares

$(x-1)^{2}=-3$

$(x-1)=(+/-)\sqrt{-3}\\(x-1)=(+/-)\sqrt{3}i\\x=1(+/-)\sqrt{3}i$

therefore

case b) is not the solution of the problem

<u>case c)</u> $x^{2}+2x-4=0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$x^{2}+2x=4$

Complete the square. Remember to balance the equation by adding the same constants to each side.

$x^{2}+2x+1=4+1$

$x^{2}+2x+1=5$

Rewrite as perfect squares

$(x+1)^{2}=5$

$(x+1)=(+/-)\sqrt{5}\\x=-1(+/-)\sqrt{5}$

therefore

case c) is not the solution of the problem

<u>case d)</u> $x^{2}-2x-4=0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$x^{2}-2x=4$

Complete the square. Remember to balance the equation by adding the same constants to each side.

$x^{2}-2x+1=4+1$

$x^{2}-2x+1=5$

Rewrite as perfect squares

$(x-1)^{2}=5$

$(x-1)=(+/-)\sqrt{5}\\x=1(+/-)\sqrt{5}$

therefore

case d) is the solution of the problem

therefore

<u>the answer is</u>

$x^{2}-2x-4=0$

5 0

2 years ago

Read 2 more answers