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3 years ago

A ball begins rolling down a hill. after 5 seconds, its velocity is 10 meters per second. what is the acceleration of the ball?

Mathematics

2 answers:

egoroff_w [7]3 years ago

8 0

Answer:

2m/s/s

Step-by-step explanation:

It was 0 m/s/s before, and now it is 10 / 5 = 2 m/s/s

Hatshy [7]3 years ago

8 0

Answer:

nce

Step-by-step explanation:

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Please help me quick! <br><br> Find the sums and differences 8 -- 0.21

Rama09 [41]

Answer:

8 + .21 = 8.21

8 - .21 = 7.79

Step-by-step explanation:

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3 years ago

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Write a system of equations to describe the situation below, solve using substitution, and fill

Lynna [10]

Answer:

The cost must be $32 since that is the only cost possible for the 1st deal

Step-by-step explanation:

Let C represent the total cost of each deal

Let T be the number of trips down the water slide

C = 32 for the 1st deal since the cost is the same regardless of how much she rides

C = 18 + T for the 2nd deal

To find when the two costs are equal set them equal to each other and solve for T

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The cost must be $32 since that is the only cost possible for the 1st deal

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3 years ago

a rectangular dog kennel is to be constructed alongside a house with 60m of fencing. If the house serves as one side of the kenn

Basile [38]

Answer:

$450\text{ m}^2$ .

Step-by-step explanation:

Let x represent side of kennel opposite to house and y represent other sides.

We have been given that a rectangular dog kennel is to be constructed alongside a house with 60 m of fencing.

Since fencing will cover 3 sides of kennel, so perimeter of kennel would be:

$x+y+y=60$

$x+2y=60$

Let us solve for x.

$x+2y-2y=60-2y$

$x=60-2y$

The area of the kennel would be product of its sides that is:

$\text{Area}=x\cdot y$

Now, we will substitute $x=60-2y$ in area equation as:

$A(y)=(60-2y)\cdot y$

$A(y)=60y-2y^2$

Let us find the derivative as shown below:

$A'(y)=60-4y$

Now, we will set derivative equal to 0 and solve for y.

$60-4y=0$

$60=4y$

$\frac{60}{4}=\frac{4y}{4}$

$y=15$

Upon substituting $y=15$ in area function, we will greatest possible area.

$A(y)=60y-2y^2$

$A(15)=60(15)-2(15)^2$

$A(15)=900-2(225)$

$A(15)=900-450$

$A(y)=450$

Therefore, the greatest possible area that can be enclosed by 60 m of fencing is 450 square meters.

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