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Darina [25.2K]
1 year ago
10

Kevin planted ome corn in the back yard and meaure it height regularly. Currently, the talk are 75 inche tall. That i 25% taller

than the lat time he checked. How tall were the talk then?
Mathematics
1 answer:
Pavlova-9 [17]1 year ago
8 0

The height of the tree was 56 inches.

We can calculate the percentage of a certain amount by :

Subtracting  New - Original / New multiplied by 100

this implies it will be

75 - x / 75 × 100 = 25 %

7500 - 100 x = 1875

this implies 7500 - 1875 = 100 x

5625 = 100 x

Thus , On dividing both sides by 100 we get

56.25 = x

Now on rounding off this number we will get

56 = x

Thus the original height of the corn plant was 56 inches

To know more about percentage you may visit the link which is mentioned below :

brainly.com/question/10168902

#SPJ1

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The area of a quadrilateral is 64 in^2. The length of the quadrilateral is 32 in. How wide is the quadrilateral?
Mars2501 [29]

Answer: 2 inches

Explanation: The area formula is length times width (L*w=a). We know that the area is 64in^2 (a=64) and the length is 32in (L=32), so we put those values into the equation and get 32*w=64. We then get w by itself by dividing both sides by 32, and get w=2. so the width is 2 inches.

7 0
3 years ago
Please help me with this
juin [17]

Answer:

x=30, <A=60, <C=30

Step-by-step explanation:

A triangle always has its total angles added up to 180 degrees, so we can set up the following equation:

90+2x+x=180

90+3x=180

3x=90

x=30

Since x=30, this means <C is 30 degrees. For <A, it's twice as much as <C, so 2*30=60.

So x=30, <A=60, <C=30

Hope this helped!

8 0
3 years ago
Read 2 more answers
The area of a rectangular city park is 25/54 square miles. The length of the park is 5/9 miles. What is the width of the park?
alexdok [17]
Hi

We know that Area= length * width right?

So this is what we do.

Total area/length

25/54 / 5/9

You get the width to be 5/6

CHECK:
5/6*5/9=25/54

Hope this helps :)


4 0
3 years ago
A cylindrical tank has a base of diameter 12 ft and height 5 ft. The tank is full of water (of density 62.4 lb/ft3).(a) Write do
saw5 [17]

Answer:

a.  71884.8 π lb/ft-s²∫₀⁵(9 - y)dy

b.  23961.6 π lb/ft-s²∫₀⁵(5 - y)dy

c. 99840π lb/ft-s²∫₀⁶rdr

Step-by-step explanation:

.(a) Write down an integral for the work needed to pump all of the water to a point 4 feet above the tank.

The work done, W = ∫mgdy where m = mass of cylindrical tank = ρA([5 + 4] - y) where ρ = density of water = 62.4 lb/ft³, A = area of base of tank = πd²/4 where d = diameter of tank = 12 ft.( we add height of the tank + the height of point above the tank and subtract it from the vertical point above the base of the tank, y to get 5 + 4 - y) and g = acceleration due to gravity = 32 ft/s²

So,

W = ∫mgdy

W = ∫ρA([5 + 4] - y)gdy

W = ∫ρA(9 - y)gdy

W = ρgA∫(9 - y)dy

W = ρgπd²/4∫(9 - y)dy

we integrate W from  y from 0 to 5 which is the height of the tank

W = ρgπd²/4∫₀⁵(9 - y)dy

substituting the values of the other variables into the equation, we have

W = 62.4 lb/ft³π(12 ft)² (32 ft/s²)/4∫₀⁵(9 - y)dy

W = 71884.8 π lb/ft-s²∫₀⁵(9 - y)dy

.(b) Write down an integral for the fluid force on the side of the tank

Since force, F = ∫PdA where P = pressure = ρgh where h = (5 - y) since we are moving from h = 0 to h = 5. So, P = ρg(5 - y)

The differential area on the side of the tank is given by

dA = 2πrdy

So.  F = ∫PdA

F = ∫ρg(5 - y)2πrdy

Since we are integrating from y = 0 to y = 5, we have our integral as

F = ∫ρg2πr(5 - y)dy

F = ∫ρgπd(5 - y)dy    since d = 2r

substituting the values of the other variables into the equation, we have

F = ∫₀⁵62.4 lb/ft³π(12 ft) × 32 ft/s²(5 - y)dy

F = 23961.6 π lb/ft-s²∫₀⁵(5 - y)dy

.(c) How would your answer to part (a) change if the tank was on its side

The work done, W = ∫mgdr where m = mass of cylindrical tank = ρAh where ρ = density of water = 62.4 lb/ft³, A = curved surface area of cylindrical tank = 2πrh  where r = radius of tank, d = diameter of tank = 12 ft. and h =  height of the tank = 5 ft and g = acceleration due to gravity = 32 ft/s²

So,

W = ∫mgdr

W = ∫ρAhgdr

W = ∫ρ(2πrh)hgdr

W = ∫2ρπrh²gdr

W = 2ρπh²g∫rdr

we integrate from r = 0 to r = d/2 where d = diameter of cylindrical tank = 12 ft/2 = 6 ft

So,

W = 2ρπh²g∫₀⁶rdr

substituting the values of the other variables into the equation, we have

W = 2 × 62.4 lb/ft³π(5 ft)² × 32 ft/s²∫₀⁶rdr

W = 99840π lb/ft-s²∫₀⁶rdr

7 0
3 years ago
making 10 pounds of Apple Jam requires five pounds of Apple how many pounds of Apple are needed to make eight pounds of apple ja
allochka39001 [22]
10lb. of jam needs 5lb. apple, or 2lb. of jam needs 1lb. apple. 
Now we want 8lb. of jam.
Set up a proportion:
2/1=8/x 
Solve for x by cross-multiplying.
2x=8 
x=4 lb. of apple

5 0
3 years ago
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