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Sati [7]
3 years ago
14

Help me plzzzzzzzzzzz

Mathematics
1 answer:
dimulka [17.4K]3 years ago
7 0

Answer:

Top row, third one from left to right

Step-by-step explanation:

Use test points:

when x = 0 , y = 2(0) - 4 = 0 - 4 = -4

(0, -4)

when y = 0, 0 = 2x - 4, 4 = 2x, 2 = x

(2, 0)

Top row, third one from left to right

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Perimeter of rectangle is 48 inches area is 40 inches what are the length of the side
Aloiza [94]
\bf \textit{perimeter of a rectangle}=p=side+side+side+side
\\\\
or\implies p=w+w+l+l\implies p=2w+2l
\\\\
p=2(w+l)\qquad 
\begin{cases}
w=width\\
l=length\\
------\\
p=48
\end{cases}\implies 40=2(w+l)\\\\
-----------------------------\\\\
\textit{area of it}=A=w\cdot l\qquad \begin{cases}
w=width\\
l=length\\
------\\
A=40
\end{cases}\implies 40=wl\\\\
-----------------------------\\\\

\bf thus\qquad 
\begin{cases}
40=2(w+l)\to \frac{40}{2}=w+l\to\frac{40}{2}-l=\boxed{w}
\\\\
40=wl\\
--------------\\
40=\left( \boxed{\frac{40}{2}-l} \right)\cdot l
\end{cases}

solve for "l" to find its length
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3 years ago
How do I write 1,234,678 in word form
ANTONII [103]

Answer:

one million two hundred thirty four thousand six hundred seventy eight

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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/∝

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Step-by-step explanation:

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You would need 6 quarts of milk. You would do 24(2) then you take 48 and divide it by 288 and you get 6.
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3 years ago
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