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2 years ago

A valet has 34 cars to park in a rectangular array, what are tge different ways the valet could park the cars

Mathematics

1 answer:

MrMuchimi2 years ago

8 0

17 by 2 because 34 divided by 2 is 17

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Dana has $25, she went to the store and spent $10. She then received $50 for a birthday present. Dana was excited until she real

MissTica

Dana:

Starts with 25 = 25

Spends 10 = 25 - 10 = 15

Received 50 = 15 + 50 = 65

New tires cost 75 = 65 - 75 = -10

Tracey:

Starts with 50 = 50

Gets gas for 25 = 50 - 25 = 25

Received 100 = 25 + 125 = 125

Purchased boots for 150 = 125 - 150 = -25

Tracey ends up with the least amount of money after their purchases.

Hope this helps! :)

8 0

3 years ago

Read 2 more answers

Someone please finish this for me

sergij07 [2.7K]

Think these are the answers to the first few hope it helps

7 0

2 years ago

Freddie made the number line below to model the laps he walks around a track at his

Elena-2011 [213]

Answer:

On a standard running 400-meter track a mile is 4 laps… plus 9 more meters if you want to get all technical about it. A mile is specifically 1,609.3 meters.

Step-by-step explanation:

6 0

3 years ago

-X² + 5x - 6 = 2<br> Using the quadratic formula

Ivenika [448]

$-x^2+5x-6=2\\-x^2+5x-6-2=0\\-x^2+5x-8=0\\$

↝ Form into $ax^2+bx+c = 0$ where a > 0 (Optional)

$x^2-5x+8=0$ ↝ Multiply -1 for both sides.

↝ Then proceed with the Quadratic Formula.

x = [-b ± (√b²-4ac)]/2a ↝ Quadratic Formula

$x=\frac{-(-5)+-\sqrt{(-5)^2-4(1)(8)}}{2(1)}\\x=\frac{5+-\sqrt{25-32}}{2}\\$

↝ As you can see that inside the square root, the numbers cannot be negative. Therefore, the solutions are imaginary.

↝ Solution with Real Numbers System ↝ There are no real solutions.

↝ Solution with Complex Number System ↝

$x=\frac{5+-\sqrt{25-32}}{2}\\x=\frac{5+-\sqrt{-7}}{2}\\x=\frac{5+-\sqrt{7}i }{2}$

Therefore, the answer for Complex Number System is (5±√7i)/2

3 0

2 years ago

an astronomer wishes to measure on a photograph the distance between the image of a certain star and three star other star image

Greeley [361]

The astronomer has to choose 3 star images out of 8.This is a problem of combinations. This can be expressed as 8C3 i.e. combination of 8 objects taken 3 at a time.

$8C3= \frac{8!}{3!*5!}=56$

This means, if 8 images are nearby the astronomer will have 56 choices <span>of the three stars.</span>

6 0

3 years ago