Ask question

Ask question

All categories

Zigmanuir [339]

3 years ago

6

A bounce house is rented at $400 per day plus a charge per hour of use. The bounce house was used for 6 hours one day, and the t

otal charge was $950. Which equation can be used to find the charge x, in dollars, per hour of use?
400x + 6x = 950
400 + 6 = 950x
400 + 6x = 950
400x + 6 = 950

2 answers:

Nata [24]3 years ago

7 0

Answer:400x+6=950...

andriy [413]3 years ago

5 0

Answer:

400x + 6 = 950

Step-by-step explanation:

x represents the charge in dollars.

It is given the bounce house was used for 6 hours one day, and the total charge was $950.

The equation that will best represent to find x per hour of use is :

400x+6=950.

You might be interested in

PLEASE MATH HELP! Find the measures of the complement and the supplement of an angle with a measure of 60 degrees.

Assoli18 [71]

Your answer will be choice A recall complementary angles total 90 degrees and supplementary angles total 180 degrees

6 0

4 years ago

Read 2 more answers

PLEASE HELP OR I WON'T GET MY CREDITS!!!!!!!!!!!!!!!!!!!!!!!!!!

sergeinik [125]

Answer:

$y = \frac14 x - 4$

Step-by-step explanation:

The slope is 1/4 since y increases 1 when x increases by 4 (you divide deltaY by deltaX).

Then, to make it go through (0,-4) you subtract -4.

7 0

3 years ago

Solve for g 2-4g=14

Mama L [17]

Answer: g=-3

Step-by-step explanation: 2-4g=14

-2 -2

-4g=12

-4g/-4=g 12/-4=-3

5 0

4 years ago

Read 2 more answers

H(x)=-x^2+6x. So what is the value of h(2)

crimeas [40]

✩ Answer:

✧･ﾟ: *✧･ﾟ:*✧･ﾟ: *✧･ﾟ:*✧･ﾟ: *✧･ﾟ:*

$\bold{Hello!}\\\bold{Your~answer~is~below!}$

✩ Step-by-step explanation:

✧･ﾟ: *✧･ﾟ:*✧･ﾟ: *✧･ﾟ:*✧･ﾟ: *✧･ﾟ:*

✺ Quadratic polynomials can be factored using the transformation $ax^2+bx+c=a(x-x_{1})(x-x_{2} )$ , where $x_{1}$ and $x_{2}$ are the solutions of the quadratic equation $ax^2+bx+c=0$ :

$-x^2+6x=0$

✺ All equations of the form $ax^2+bx+c=0$ can be solved using the quadratic formula:

$-b=\frac{+}\\\sqrt{b^2-4ac}\\~~~~~~~~~~~~~2a$

✺ The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction:

$x=\frac{\sqrt{-6\frac{+}\\\sqrt{6^2}}}{2(-1)}$

✺ Take the square root of $6^2$ :

$x=\frac{\sqrt{-6\frac{+}\\{6}}}{2(-1)}$

✺ Multiply $2$ times $-1$ :

$x=\frac{\sqrt{-6\frac{+}\\{6}}}{-2}$

✺ Now solve the equation $x=\frac{\sqrt{-6\frac{+}\\{6}}}{-2}$ when ± is plus. Add $-6$ to $6$ :

$x=\frac{0}{-2}$

✺ Divide $0$ by $-2$ :

$x=0$

-OR-

✺ Now solve the equation $x=\frac{\sqrt{-6\frac{+}\\{6}}}{-2}$ when ± is minus. Subtract $6$ from $-6$ :

$x=\frac{-12}{-2}$

✺ Divide $-12$ by $-2$ :

$x=6$

✺ Optional : Factor the original expression using $ax^2+bx+c=a(x-x_{1})(x-x_{2} )$ . Substitute $0$ for $x_{1}$ and $6$ for $x_{2}$ :

$-x^2+6x=-x(x-6)$

✩ Answer:

✺ <u>Factored Form</u>: $x(x-6)$

✺ <u>Exact Form</u>: $x=6$

✺ <u>Graph Point Form</u>: $x=(6,0)$

$Hope~this~helps~and,\\Best~of~luck!\\\\~~~~~-TotallyNotTrillex$

"ටᆼට"

8 0

4 years ago

Lines AB and CD (if shown) are straight lines. Find x.

lakkis [162]

Answer:

Can you do plsssssssss help meeeeeeeeeeeeee

7 0

3 years ago