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2 years ago

6

A downtown parking lot charges $5 per hour

2 answers:

Tom [10]2 years ago

7 0

Parking charges for first 2hours =$5 per hour after 2 hours parking charges = $2.50per hour

lilavasa [31]2 years ago

5 0

Parking charges for first 2hours =$5 per hour after 2 hours parking charges = $2.50per hour

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DerKrebs [107]

Answer:

m=-3

Step-by-step explanation:

m - 7 = -10

Ok so,

in order to get m by itself you just need to do the opposite of -7 which is +7 so you have to add 7 to both sides. On the left, both of the 7's cancel out and on the right -10 + 7 = -3. So, now it should look like this.

m=-3

To check you just need to plug in -3 into the question

Hope this helps! Have a good day

4 0

3 years ago

Find the perimeter of the image bellow: A:32.1 units B: 35.8 units C: 37.6 units D: 39.2 units

masha68 [24]

A is the closest I got. My answer was 32.665 using thousandth place. I found the perimeter by adding up the lengths. I found all of the lengths by using pathat oran theorem.

7 0

3 years ago

You are the owner of a small bakery. This week the bakery has orders for 48 birthday cakes. Each cake sells for $52. Suppose

ser-zykov [4K]

Answer:

Answer:$624

Total sales: $52 x 48 = $2,496

Cost of ingredients:

$2,496

4

= $624

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

51: Y = 3: 5 value of Y

ivann1987 [24]

Answer:Y=25:3

Step-by-step explanation:

7 0

2 years ago

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The graph of which function has a minimum located at (4, –3)?

valentinak56 [21]

Answer:

None of the options is the answer to the question

Step-by-step explanation:

we know that

The equation of a vertical parabola into vertex form is equal to

$f(x)=a(x-h)^{2}+k$

where

(h,k) is the vertex of the parabola

case A) we have

$f(x)=x^{2}+4x-11$

In this case the x-coordinate of the vertex will be negative

therefore

case A is not the solution

case B) we have

$f(x)=-2x^{2}+16x-35$

This case is a vertical parabola open downward (the vertex is a maximum)

The vertex is the point $(4,-3)$ <u>but is not a minimum</u>

see the attached figure

therefore

case B is not the solution

case C) we have

$f(x)=x^{2}-4x+5$

Convert into vertex form

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

$f(x)-5=x^{2}-4x$

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

$f(x)-5+4=(x^{2}-4x+4)$

$f(x)-1=(x^{2}-4x+4)$

Rewrite as perfect squares

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

$f(x)=(x-2)^{2}+1$ --------> vertex form

The vertex is the point $(2,1)$

therefore

case C is not the solution

case D) we have

$f(x)=2x^{2}-16x+35$

Convert into vertex form

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

$f(x)-35=2x^{2}-16x$

Factor the leading coefficient

$f(x)-35=2(x^{2}-8x)$

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

$f(x)-35+32=2(x^{2}-8x+16)$

$f(x)-3=2(x^{2}-8x+16)$

Rewrite as perfect squares

$f(x)-3=2(x-4)^{2}$

$f(x)=2(x-4)^{2}+3$ --------> vertex form

The vertex is the point $(4,3)$

therefore

case D is not the solution

The answer to the question will be the function

$f(x)=2x^{2}-16x+29$

7 0

3 years ago

Read 2 more answers