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2 years ago

If the area of square 2 is 225 units?, and

Mathematics

1 answer:

murzikaleks [220]2 years ago

3 0

Answer:

25 square units

Step-by-step explanation:

square 1 = 100 units (perimeter)

therefore each side = 25 units

square 2 = 225 square units (area)

therefore each side = 15 units

The sequence gradually decreases by 10.

Therefore square 3 each side would be 5 units.

To find the Area = 5×5

= 25 square units.

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Identify all of the problems that would result in a product of 2 5 ?

mars1129 [50]

1 multiplied by 25 and 5 multiplied by 5

8 0

3 years ago

Which expression best estimates 11 1/5 divided by 2 3/4? 4 6 11 22

Maksim231197 [3]

11 1/5 = (11*5+1)/5=56/5
2 3/4= (2*4+3)/4=11/4
11 1/5 : 2 3/4=56/5:11/4=56/5*4/11=(56*4)/(5*11)=224/55≈4.07≈4
Answer is 4

5 0

3 years ago

PLEASE HELP ME The length of the one of the legs of a right triangle is 8m and the hypotenuse is 12m. Which of the following cho

Bas_tet [7]

Answer:

The length of the other leg is 8.94 m.

Step-by-step explanation:

Given that,

Hypotenuse of a right triangle is 12 m

The length of the one of the legs of a right triangle is 8 m

We need to find the length of the other leg. It is based on the concept of Pythagoras theorem. Let the other leg is h. So, according to this theorem,

$H^2=b^2+h^2$

H = hypotenuse = 12 m

b = base = 8 m

$h^2=H^2-b^2\\\\h^2=12^2-8^2\\\\h=\sqrt{80}\\\\h=8.94\ m$

So, the length of the other leg is 8.94 m.

8 0

3 years ago

X+52+120+105 What's my answer

blagie [28]

Your answer is =x+277

3 0

3 years ago

Read 2 more answers

A pyramid-shaped vat has square cross-section and stands on its tip. The dimensions at the top are 2 m × 2 m, and the depth is 5

Strike441 [17]

Answer:

the water level increases at a rate of 1.718 m/min when the depth is 4 m

Step-by-step explanation:

the volume of the pyramid is

V = (1/3)(height)(area of base) = 1/3 H*L²

For the diagonal in the pyramid

tg Ф = Side Length/ Height = L / H = x / h

where h= depth of water , x= side of the corresponding cross section

therefore x= L *h/H

the volume of the water is

v= 1/3 h*x² = 1/3 (L/H)² h³

in terms of time

v = Q*t

then

Q*t = 1/3 (L/H)² h³

h³ = 3*(H/L)² *Q *t

h = ∛(3*(H/L)² *Q *t) = ∛(3*(H/L)² *Q) *∛t = k* ∛t , where k=∛(3*(H/L)² *Q)

h = k* ∛t

then the rate of increase in depth is dh/dt

dh/dt = 1/3*k* t^(-2/3)

since

t = (h/k)³

dh/dt = 1/3*k* t^(-2/3) = 1/3*k* (h/k)³ ^(-2/3) = 1/3*k* (h/k)^(-2) = 1/3 k³ / h²

= 1/3 (3*(H/L)²*Q) / h² = (H/L)²*Q /h²

dh/dt= [H/(h*L)]²*Q

replacing values, when h=4m

dh/dt= [H/(h*L)]²*Q = [5m/(4m*2m)]² * (3m³/min)= 1.718 m/min

8 0

3 years ago