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kolbaska11 [484]

3 years ago

8

Round decimal numbers to the nearest tenth, hundredth, or whole number,

1 answer:

finlep [7]3 years ago

4 0

Answer:

C

Step-by-step explanation:

because the nearest is to ones tenth hundred thousand

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Which expression would you use to find the number of outcomes for flipping 4 coins?

Papessa [141]

The answer would be 2(4)

3 0

3 years ago

Read 2 more answers

Alyssa is looking at a car. If she buys the car, she will pay $750 when she picks up the car (a down payment) and pay $314 each

soldier1979 [14.2K]

Answer:

a very expensive waste of money.

Step-by-step explanation:

3 0

3 years ago

In circle E, AC=4, CG=6, and AH=3, Find HB and EG. If necessary, round to the tenths place.

bezimeni [28]

Using theorem about secant segments we can write,
AB*AH=AG*AC
AC=4,
CG=6
AG=AC+CG=4+6=10
AH=3
AB= AH+HB=AH+x=3+x
(3+x)*3=10*4
9+3x=40
3x=40-9
3x=31
x=31/3≈10.3
HB≈10.3
EG=HB/2 (as radius and diameter)
EG=10.3/2≈5.2

3 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

<em></em>

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

A local movie theater charges $12 for an adult ticket and $10 for a child's ticket. a group of eight people spent a total of $86

oksian1 [2.3K]

3 adults and 5 kids.

You can get this by setting up the following system of equations.

x + y = 8 (number of people)
10x + 12y = 86 (cost)

7 0

3 years ago