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2 years ago

5

The diagram shows a garden plot. The area of the garden is 72 square feet. The length of fencing required to completely enclose

the garden is how many feet.

1 answer:

QveST [7]2 years ago

8 0

Answer:

The area of the garden is 72 square feet. The length of fencing required to completely enclose the garden is 36 feet.

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What is m? 49° 77° 98° 161°

Mariana [72]

we know that

The measurement of <u>the external angle</u> is the semi-difference of the arcs it includes.

In this problem

$21\°=\frac{1}{2}[arc\ RU-arc\ SU]$

Solve for the measure of arc SU

$42\°=[arc\ RU-arc\ SU]$

$arc\ SU=arc\ RU-42\°$

$arc\ SU=119\°-42\°=77\°$

therefore

the answer is

The measure of the arc SU is $77\°$

4 0

2 years ago

Read 2 more answers

A fox sees a rabbit 35 feet away and starts chasing it. As soon as the fox starts moving the rabbit sees it and starts running a

Artemon [7]

The speed of the fox should be greater than the rabbit's speed in other to catch up, Hence, the speed of the fox must be 35/t + 40

The distance between the rabbit and fox, d = 35 feets

The rabbit's speed = 40 feets per second

Fox's speed = F

In other to obtain a specific speed for the fox, the time at which the fox is required to catch up with the rabbit should be given.

Let the time = t

Recall :

Speed = distance / time

Distance = speed × time

Rabbit's distance from fox after time t will be :

35 + (40 × t) = 35+40t

Fox's distance after time, t = fox speed × t = F × t

In other to catch up ; both fox and rabbit must be at the same distance ;

Rabbit's distance at time t = Fox's distance at time t

35+40t = Ft

To solve for F which is the fox's speed:

Divide both sides by t

(35+40t) / t = Ft/t

35/t + 40 = F

Hence, the fox' s speed must be : 35/t + 40

To obtain the exact speed of the fox, we must be given a specific time value.

Learn more : brainly.com/question/18112348

4 0

1 year ago

The distance d (in miles) that you can see to the horizon with your eye level h feet above the water is given by d=3h2−−−√. You

liubo4ka [24]

Answer: He cannot see a ship that is 4.5 miles away.

Step-by-step explanation:

Since we have given that

$d=3\sqrt{h}$

Here, h is the eye level above the water

d is the distance away.

If h = 12 feet

We will get :

$d=3\sqrt{12}\\\\d=3\times 2\sqrt{3}\\\\d=6\sqrt{3}\\\\d=10.39$

Hence, it is 10.39 feet away when eye level is 12 feet above the water.

Therefore , he cannot see a ship that is 4.5 miles away.

7 0

1 year ago

If a cow has a mass of 9×102 kilograms, and a blue whale has a mass of 1.8×105 kilograms, which of these statements is true?

lana66690 [7]

Answer:

The mass of the Blue whale is 200 times the mass of the cow

Step-by-step explanation:

Given

Mass of Cow = 9 * 10² kg

Mass of Blue Whale = 1.8 * 10⁵ kg

Required

Determine the relationship between both weights

Represent the mass of the cow with C and the mass of the whale with B

$C = 9 * 10^2kg$

$C = 9 *100kg$

$C = 900kg$

$B = 1.8 * 10^5kg$

$B = 1.8 * 100000kg$

$B = 180000kg$

Divide the bigger weight by the smaller weight

$\frac{B}{C} = \frac{180000kg}{900kg}$

$\frac{B}{C} = \frac{180000}{900}$

$\frac{B}{C} = {200}{}$

Multiply both sides by C

$C * \frac{B}{C} = {200}{} * C$

$B = {200}{} * C$

$B = 200}C$

<em>From the expression above, it can be concluded that the mass of the Blue whale is 200 times the mass of the cow</em>

7 0

1 year ago

A small cube with side length 6y is placed inside a larger cube with side length 4x^2. What is the difference in the volume of t

fomenos

the different volume is the numbers. 4*2*6=?

3 0

2 years ago

Read 2 more answers