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shtirl [24]
3 years ago
9

What is the equation of the line whose y-intercept is 3 and slope is 1? y = x - 3 y = x 3 y = 3x 1

Mathematics
2 answers:
katrin [286]3 years ago
4 0
The equation is y=1x+3. The line equation is y=mx+b. m being the slope and the b being the y-intercept.
Lynna [10]3 years ago
4 0

Answer:

The answer is y=x+3

Step-by-step explanation:

An equation for a line which is not vertical can be written in the form

y=mx+b

where m is its  slope and b is its y-intercept.

So, if you replace m and b in the last equation by 1 and 3 respectively, you have the equation of the line:

y=1x +3

but, this is the same as

y=x+3

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2 years ago
Lyn puts dvds on shelves. She has 236 dvds.each full shelf has 28 dvds.how many full shelves of dvds does Lyn have?
cluponka [151]

Answer:

8 shelves

Step-by-step explanation:

As 28 multiple by 2 is 224 and 28*9 is 252 so she will have 8 full shelves

3 0
3 years ago
Read 2 more answers
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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ZanzabumX [31]

Answer:

-0.8333

Step-by-step explanation:

used omni calculator

https://www.omnicalculator.com/math/slope

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The diagonals of rhombus abcd intersect at point
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