Marcs is growing three bean plants as part of a science experiment. Plant A 1/3 inch tall. Plant B is 3/4 inch tall. Plant C is 5/6 inch tall. Order the plants from shortest to tallest.

Answers

Answer 1

Answer: 1/3 , 3/4, 5/6

1÷3= 0.333333

3÷4= 0.75

5÷6= 0.8333333333333333

Hope I This Helps :D

Related Questions

Whitch simplified fraction is equal to 0.53
Please someone answer this question as quick as possible please
Cone A has a height of 3 meters and a radius of 2 meters. Cone B has the same radius, but the height is 6 meters. Calculate the volume of each cone. Which conclusion is correct?A) The volume of cone A is twice the volume of cone B. Eliminate B) The volume of cone B is twice the volume of cone A. C) The volume of cone B is 4 times the volume of cone A. D) The volume of cone B is three times the volume of cone A.
GEOMETRY. PLEASE CHECK MY ANSWER! THANKS.Mario’s company makes unusually shaped imitation gemstones. One gemstone had 12 faces and 10 vertices. How many edges did the gemstone have? A. 23 edges B. 22 edges C. 25 edges D. 20 edges****
Expanded form of 3(z +5)?

Factor completely: 3x^2 + 12x + 7

Answers

Answer:

6x+12

Step-by-step explanation:

A rectangular lot that measures 125 feet by 185 feet is completely fenced. What is the length, in feet, of the fence?Can you please explain with some details?

Answers

620, because 125+125+185+185=620, and that would be the perimeter of the rectangle.

Calculati aria unui patrat care are perimetrul de 7,2 cm

Answers

$A=l^2=\left((P)/(4)\right)^2=\left((7,2)/(4)\right)^2=1,8^2=3.24\;cm^2.$

Green eyes.

A carpenter is building a fence around a swimming pool. One side of the pool is next to the house and does not need fencing. The carpenter has 120 feet of fencing and plans to use it all. Solve the equation w + 76 + w = 120 to find the unknown dimension of the enclosed rectangular area.a. 12 ft

c. 20 ft

b. 22 ft

d. 18 ft

Answers

The answer is B. 22ft
The equation goes w + 76 + w = 120 so I'm guessing you're solving for w since that's the only remaining side you haven't discovered so first of all collect like terms:
2w + 76 = 120
Secondly minus 76 from the equation to leave 2w on its own
2w - 76 = 120
BUT whatever you do to one side of the equation you have to do it to the side
2w - 76 = 120 - 76
Which will get you to
2w = 44
To get w on it's own you divide the equation by 2 and like before you must do the same to the other side of the equation to balance it out.
w = 44/2
Therefore w= 22ft

Use the functions f(x) = 5x − 12 and g(x) = 11x + 2 to complete the function operations listed below. Part A: Find (f +g. (x). Show your work. (3 points) Part B: Find (f ⋅
g. (x). Show your work. (3 points) Part C: Find f[g(x)]. Show your work. (4 points)

Answers

Answer: The completed function operations are :

$(f+g)(x)=16x-10,\n\n(f.g)(x)=55x^2-122x-24,\n\nf[g(x)]=55x-2.$

Step-by-step explanation: We are given two functions as follows :

$f(x)=5x-12,\n\ng(x)=11x+2.$

We are to complete the following function operations.

Part A : (f + g)(x)

We have

$(f+g)(x)\n\n=f(x)+g(x)\n\n=(5x-12)+(11x+2)\n\n=16x-10.$

Part B : (f . g)(x)

We have

$(f.g)(x)\n\n=f(x).g(x)\n\n=(5x-12).(11x+2)\n\n=55x^2+10x-132x-24\n\n=55x^2-122x-24.$

Part C : f[g(x)]

We have

$f[g(x)]\n\n=f(11x+2)\n\n=5(11x+2)-12\n\n=55x+10-12\n\n=55x-2.$

Thus, the completed function operations are :

$(f+g)(x)=16x-10,\n\n(f.g)(x)=55x^2-122x-24,\n\nf[g(x)]=55x-2.$

$(f+g)(x)=f(x)+g(x)\n(f\cdot g)(x)=f(x)\cdot g(x)\n\nA.\n(f+g)(x)=5x-12+11x+2=16x-10\n\nB.\n(f\cdot g)(x)=(5x-12)(11x+2)\n(f\cdot g)(x)=55x^2+10x-132x-24\n(f\cdot g)(x)=55x^2-122x-24\n\nC.\nf(g(x))=5(11x+2)-12=55x+10-12=55x-2$

If you sleep 7 hours every night how many hours are you awake in a year?

Answers

24h - 7h = 17h. 17 * 365= 6 205 h a year. In a leap year, it is 6 222 h a year.

Other Questions

Does anyone know how to do this HELP !!!

What is the surface area of this rectangular pyramid?3 cmD4 cm4 cmsquare centimetersSubmit??

Convert 53 grams to kilograms. 530 kg 53,000 kg 0.053 kg 5.3 kg

How can you use 2, 3, 5 and 8 to get 104?