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kolbaska11 [484]
3 years ago
8

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

Mathematics
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
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