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3 years ago

14

Is 36 T-shirts in 3 boxes equivalent to 60 T-shirts in 6 boxes?

2 answers:

kolbaska11 [484]3 years ago

7 0

Answer:no it is not equivalent

Step-by-step explanation:

36x3=108 so there is 108 shirts in that box

60x6=360 so 360 is in that box

so not equivalent because they have different amounts in the boxes one is more and the other one less

your welcome

Nadusha1986 [10]3 years ago

3 0

Answer:

No

Step-by-step explanation:

Since the unit rates 12 t-shirts/1 box

and 10 t-shirts/1 box are not the same, the rates are not equivalent.

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KATRIN_1 [288]

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5/2x, 5/2 y

Step-by-step explanation:

An enlargement means the scale factor must be greater than one

The only choice with a scale factor greater than one is 5/2x, 5/2 y

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3 years ago

Which is greater 1 millennium or 10 centuries?

Bingel [31]

They are both the same
1 millenium - 1000 yrs
1 century -100 yrs
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4 years ago

Read 2 more answers

There are 48⋅5^{6} cows spread across 32⋅5^{4} square miles of land. Find the average number of cows per square mile. Round your

stich3 [128]

Answer:

69.1

Step-by-step explanation:

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

<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

Meg invested $16,000 in a savings account. If the annual interest rate is 6%, how much will be in the account in 5 years for qua

nexus9112 [7]

Answer:

In 5 years the account will be $ 21,549.68.

Step-by-step explanation:

Given that Meg invested $ 16,000 in a savings account, if the annual interest rate is 6%, to determine how much will be in the account in 5 years for quarterly compounding, the following calculation must be performed:

16,000 x (1 + 0.06 / 4) ^ 5x4 = X

16,000 x 1,015 ^ 20 = X

16,000 x 1.34685500 = X

21,549.68 = X

Therefore, in 5 years the account will be $ 21,549.68.

8 0

3 years ago