Determine whether each statement about the design from the rug is true.

Help me you guys for brainlest

ziro4ka [17]

Answer:

20

Step-by-step explanation:

$\frac{10^{2} }{5^{1}}$

$\frac{100}{5}$

$\frac{20}{1}$

$20$

6 0

2 years ago

A garden wall is 4 feet in height and has a 6-foot shadow. A tree in the garden casts a 24-foot shadow at the same time of day.

4vir4ik [10]

Answer:

7height

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Hi uwu pls help me?? i really need it :'( dont make me cry

densk [106]

Answer:

4z-(3z) z

Step-by-step explanation:

I hope that helps

8 0

3 years ago

Consider a diving board that is 10 feet above a pool. If the ladder is 5 feet away from the base of the diving board, approximat

OleMash [197]

Answer: 15 feet

Step-by-step explanation:

From the question, we are informed that a diving board is 10 feet above a pool. Since the ladder is 5 feet away from the base of the diving board, to calculate how tall is the ladder is, we subtract -5 from 10. This will be:

= 10 - (-5)

= 10 + 5

= 15 feet

The answer is option C.

5 0

2 years ago

A rancher wishes to build a fence to enclose a 2250 square yard rectangular field. Along one side the fence is to be made of hea

Bess [88]

Answer:

The least cost of fencing for the rancher is $1200

Step-by-step explanation:

Let x be the width and y the length of the rectangular field.

Let C the total cost of the rectangular field.

The side made of heavy duty material of length of x costs 16 dollars a yard. The three sides not made of heavy duty material cost $4 per yard, their side lengths are x, y, y. Thus

$C=4x+4y+4y+16x\\C=20x+8y$

We know that the total area of rectangular field should be 2250 square yards,

$x\cdot y=2250$

We can say that $y=\frac{2250}{x}$

Substituting into the total cost of the rectangular field, we get

$C=20x+8(\frac{2250}{x})\\\\C=20x+\frac{18000}{x}$

We have to figure out where the function is increasing and decreasing. Differentiating,

$\frac{d}{dx}C=\frac{d}{dx}\left(20x+\frac{18000}{x}\right)\\\\C'=20-\frac{18000}{x^2}$

Next, we find the critical points of the derivative

$20-\frac{18000}{x^2}=0\\\\20x^2-\frac{18000}{x^2}x^2=0\cdot \:x^2\\\\20x^2-18000=0\\\\20x^2-18000+18000=0+18000\\\\20x^2=18000\\\\\frac{20x^2}{20}=\frac{18000}{20}\\\\x^2=900\\\\\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}\\\\x=\sqrt{900},\:x=-\sqrt{900}\\\\x=30,\:x=-30$

Because the length is always positive the only point we take is $x=30$ . We thus test the intervals $(0, 30)$ and $(30, \infty)$

$C'(20)=20-\frac{18000}{20^2} = -25 < 0\\\\C'(40)= 20-\frac{18000}{20^2} = 8.75 >0$

we see that total cost function is decreasing on $(0, 30)$ and increasing on $(30, \infty)$ . Therefore, the minimum is attained at $x=30$ , so the minimal cost is

$C(30)=20(30)+\frac{18000}{30}\\C(30)=1200$

The least cost of fencing for the rancher is $1200

Here’s the diagram:

3 0

2 years ago