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4 years ago

7

How to equally 4/3 feet of rope

1 answer:

Gemiola [76]4 years ago

8 0

If I you mean to get a mixed number, the answer is 1 1/3.

You might be interested in

You get reported if u don't know but answer.

Maksim231197 [3]

Answer:

A

Step-by-step explanation:

A point estimate is the value of a sample characteristic (obtained from the sample)

Sample mean is a point estimate of the population parameter

8 0

3 years ago

Please helppppp me Idk how to answer this

BARSIC [14]

C. Plug in 0 into the equation
y = (0)^2 + 10
y = 10
It would be (0,10)

3 0

3 years ago

For (triangle)DEF, you are required to construct the median DG to EF. Which of the following must be performed first?

iren2701 [21]

Answer: C. Bisect DE at G.

Step-by-step explanation:

In a triangle, a median is a line segment drawn from a vertex to the midpoint of the opposite side of the vertex.

So, for triangle DEF , if we draw the median DG to the opposite side EF.

Here EF is the opposite side to vertex D.

So, G must be the midpoint of line segment EF.

So , we need to first mark G on EF.

i.e. we need to bisect EF at G.

So first step would be : Bisect EF at G.

8 0

3 years ago

Let c be a positive number. A differential equation of the form dy/dt=ky^1+c where k is a positive constant, is called a doomsda

stich3 [128]

Answer:

The doomsday is 146 days

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Step-by-step explanation:

Given

$\frac{dy}{dt} = ky^{1 +c}$

First, we calculate the solution that satisfies the initial solution

Multiply both sides by

$\frac{dt}{y^{1+c}}$

$\frac{dt}{y^{1+c}} * \frac{dy}{dt} = ky^{1 +c} * \frac{dt}{y^{1+c}}$

$\frac{dy}{y^{1+c}} = k\ dt$

Take integral of both sides

$\int \frac{dy}{y^{1+c}} = \int k\ dt$

$\int y^{-1-c}\ dy = \int k\ dt$

$\int y^{-1-c}\ dy = k\int\ dt$

Integrate

$\frac{y^{-1-c+1}}{-1-c+1} = kt+C$

$-\frac{y^{-c}}{c} = kt+C$

To find c; let t= 0

$-\frac{y_0^{-c}}{c} = k*0+C$

$-\frac{y_0^{-c}}{c} = C$

$C =-\frac{y_0^{-c}}{c}$

Substitute $C =-\frac{y_0^{-c}}{c}$ in $-\frac{y^{-c}}{c} = kt+C$

$-\frac{y^{-c}}{c} = kt-\frac{y_0^{-c}}{c}$

Multiply through by -c

$y^{-c} = -ckt+y_0^{-c}$

Take exponents of $-c^{-1$

$y^{-c*-c^{-1}} = [-ckt+y_0^{-c}]^{-c^{-1}$

$y = [-ckt+y_0^{-c}]^{-c^{-1}$

$y = [-ckt+y_0^{-c}]^{-\frac{1}{c}}$

i.e.

$y(t) = [-ckt+y_0^{-c}]^{-\frac{1}{c}}$

Next:

$t= 3$ i.e. 3 months

$y_0 = 2$ --- initial number of breeds

So, we have:

$y(3) = [-ck * 3+2^{-c}]^{-\frac{1}{c}}$

-----------------------------------------------------------------------------

We have the growth term to be: $ky^{1.01}$

This implies that:

$ky^{1.01} = ky^{1+c}$

By comparison:

$1.01 = 1 + c$

$c = 1.01 - 1 = 0.01$

$y(3) = 16$ --- 16 rabbits after 3 months:

-----------------------------------------------------------------------------

$y(3) = [-ck * 3+2^{-c}]^{-\frac{1}{c}}$

$16 = [-0.01 * 3 * k + 2^{-0.01}]^{\frac{-1}{0.01}}$

$16 = [-0.03 * k + 2^{-0.01}]^{-100}$

$16 = [-0.03 k + 0.9931]^{-100}$

Take -1/100th root of both sides

$16^{-1/100} = -0.03k + 0.9931$

$0.9727 = -0.03k + 0.9931$

$0.03k= - 0.9727 + 0.9931$

$0.03k= 0.0204$

$k= \frac{0.0204}{0.03}$

$k= 0.68$

Recall that:

$-\frac{y^{-c}}{c} = kt+C$

This implies that:

$\frac{y_0^{-c}}{c} = kT$

Make T the subject

$T = \frac{y_0^{-c}}{kc}$

Substitute: $k= 0.68$ , $c = 0.01$ and $y_0 = 2$

$T = \frac{2^{-0.01}}{0.68 * 0.01}$

$T = \frac{2^{-0.01}}{0.0068}$

$T = \frac{0.9931}{0.0068}$

$T = 146.04$

<em>The doomsday is 146 days</em>

4 0

3 years ago

The value of a car after 5 years is a quantities data

Levart [38]

Answer:

12

Step-by-step explanation:

3 0

4 years ago

Read 2 more answers