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2 years ago

Nine less twice a number is the same as three more than the number in standard form

Mathematics

1 answer:

ad-work [718]2 years ago

6 0

Answer:

Step-by-step explanation:

Writing the equation, we get

2x - 9 = x + 3

2x -x = 9+3

x = 12.

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Anna11 [10]

Answer: -104

Explanation: x=-9 y=2
2(y-3xy)
=2(2-3(-9)(2))
=4-2(3(-18))
=4-108
=-104

Hope it helped

4 0

1 year ago

The selling price of an item is 720 it is marked down 30% but the sale price is still marked up from the cost of 420 find the ma

aliya0001 [1]

Answer:

$84

Step-by-step explanation:

720x30=21600

21600/100=216

216=markdown (as a number instead of a percentage)

Now subtract that like this

720-216=504

504= sale price

Then do this

504-420=84

Therefore it is $84

Hope you understand!

3 0

2 years ago

F(x)=x^2+3x-2 find f(-2)

Rasek [7]

Answer:

-12

Step-by-step explanation:

6 0

2 years ago

Please help with math

Sergio [31]

Answer: x = 2

Step-by-step explanation:

$x+y=8\\y=3x$

Replace y in the first equation by the value of y in the second equation.

$x+y=8\\x+3x=8\\4x=8\\x=\frac{8}{4}\\ x=2$

4 0

2 years ago

Read 2 more answers

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

2 years ago