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exis [7]
3 years ago
14

7/30+9/20 what is the answer pls hurry im so confused

Mathematics
1 answer:
Svetradugi [14.3K]3 years ago
8 0

Here is your answer:

  • \frac{7}{30} + \frac{9}{20}
  • Find common denominator:
  • 30 \times 20 = 60 /20 \times 30 = 60
  • Common denominator is 60:
  • \frac{7}{30} + \frac{9}{20} = \frac{140}{60} + \frac{270}{60}
  • 140 + 270 = 410
  • = \frac{410}{60}
  • \frac{410}{60} \div 2 = \frac{205}{30} \div 5 = \frac{41}{6}
  • = \frac{41}{6} or 6\frac{5}{6}

Hope this helps!

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The rectangle below has an area of x^2+8x+15x
Darina [25.2K]

Answer:

The length of the rectangle is x+5.

Step-by-step explanation:

Given : The rectangle below has an area of x^2+8x+15 square meter and a width of  x+3 meter.

To find : What expression represents the length of the rectangle?

Solution :

The area of the rectangle is given by,

\text{Area}=\text{Length}\times \text{Breadth}

\text{Area}=x^2+8x+15

\text{Breadth}=x+3

x^2+8x+15=\text{Length}\times(x+3)

\text{Length}=\frac{x^2+8x+15}{x+3}

\text{Length}=\frac{(x+5)(x+3)}{x+3}

\text{Length}=x+5

Therefore, the length of the rectangle is x+5.

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3 years ago
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3 years ago
A flower garden has 12 sunflowers for every 45 irises. What is the unit rate for the number of irises per sunflower? Explain how
m_a_m_a [10]

It is given in the question that,

A flower garden has 12 sunflowers for every 45 irises.

And we have to find the unit rate  for the number of irises per sunflower.

And for that we have to divide number of Irises by number of Sunflower. That is

= \frac{45 Irises }{12 Sunflower} = 3.75 Irises \ per \ sunflower .

And that's the required unit rate .

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3 years ago
A pond forms as water collects in a conical depression of radius a and depth h. Suppose that water flows in at a constant rate k
Scrat [10]

Answer:

a. dV/dt = K - ∝π(3a/πh)^⅔V^⅔

b. V = (hk^3/2)/[(∝^3/2.π^½.(3a))]

The small deviations from the equilibrium gives approximately the same solution, so the equilibrium is stable.

c. πa² ≥ k/∝

Step-by-step explanation:

a.

The rate of volume of water in the pond is calculated by

The rate of water entering - The rate of water leaving the pond.

Given

k = Rate of Water flows in

The surface of the pond and that's where evaporation occurs.

The area of a circle is πr² with ∝ as the coefficient of evaporation.

Rate of volume of water in pond with time = k - ∝πr²

dV/dt = k - ∝πr² ----- equation 1

The volume of the conical pond is calculated by πr²L/3

Where L = height of the cone

L = hr/a where h is the height of water in the pond

So, V = πr²(hr/a)/3

V = πr³h/3a ------ Make r the subject of formula

3aV = πr³h

r³ = 3aV/πh

r = ∛(3aV/πh)

Substitute ∛(3aV/πh) for r in equation 1

dV/dt = k - ∝π(∛(3aV/πh))²

dV/dt = k - ∝π((3aV/πh)^⅓)²

dV/dt = K - ∝π(3aV/πh)^⅔

dV/dt = K - ∝π(3a/πh)^⅔V^⅔

b. Equilibrium depth of water

The equilibrium depth of water is when the differential equation is 0

i.e. dV/dt = K - ∝π(3a/πh)^⅔V^⅔ = 0

k - ∝π(3a/πh)^⅔V^⅔ = 0

∝π(3a/πh)^⅔V^⅔ = k ------ make V the subject of formula

V^⅔ = k/∝π(3a/πh)^⅔ -------- find the 3/2th root of both sides

V^(⅔ * 3/2) = k^3/2 / [∝π(3a/πh)^⅔]^3/2

V = (k^3/2)/[(∝π.π^-⅔(3a/h)^⅔)]^3/2

V = (k^3/2)/[(∝π^⅓(3a/h)^⅔)]^3/2

V = (k^3/2)/[(∝^3/2.π^½.(3a/h))]

V = (hk^3/2)/[(∝^3/2.π^½.(3a))]

The small deviations from the equilibrium gives approximately the same solution, so the equilibrium is stable.

c. Condition that must be satisfied

If we continue adding water to the pond after the rate of water flow becomes 0, the pond will overflow.

i.e. dV/dt = k - ∝πr² but r = a and the rate is now ≤ 0.

So, we have

k - ∝πa² ≤ 0 ---- subtract k from both w

- ∝πa² ≤ -k divide both sides by - ∝

πa² ≥ k/∝

5 0
3 years ago
Divide X^2+77x+20 by x+9 what is the quotient and remainder ?
adelina 88 [10]
X + 68 with a remainder of (-592)

5 0
3 years ago
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