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tankabanditka [31]

3 years ago

9

a full container of juice holds 64 ounces. How many 7 ounce servings of juice are in a full container?

2 answers:

strojnjashka [21]3 years ago

4 0

9 servings with 1 ounce left over as64 divided by 7 is 9 r 1 hope this helps!

musickatia [10]3 years ago

3 0

There is 9 servings and 1 ounce left over.

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Can somebody help me with this problem

kherson [118]

The 2nd answer

looking at the graph
when she was born she was 0 years old. and at the 0 year, the height is at 30 in

7 0

3 years ago

I need helppp help meeeeeeee

Rama09 [41]

1) $\frac{11-13}{4-7}= \frac{2}{3}$

2) $\frac{-2-7}{-10-(-7)}=\frac{-9}{-3} =3$

3) $\frac{9-3}{-11-(-14)} =\frac{6}{3}=2$

4) $\frac{8-4}{11-(-9)} =\frac{4}{20}= \frac{1}{5}$

ok done. Thank to me :>

8 0

2 years ago

7n +9 = -1+6(n+4)<br>can someone walk me through this please?

Ira Lisetskai [31]

Answer:

n = 14

Step-by-step explanation:

7n +9 = -1+6(n+4)

Distribute

7n+9 = -1 +6n+24

Combine like terms

7n +9 = 6n +23

Subtract 6n from each side

7n -6n+9 = 6n-6n +23

n+9 = 23

Subtract 9 from each side

n+9-9=23-9

n = 14

6 0

3 years ago

1. Consider the population model dP dt = 0.2P 1 − P 135 , where P(t) is the population at time t. (a) For what values of P is th

Kryger [21]

Answer:

a

$P = 0$ OR $P = 135$

b

$P > 0$ and $P < 135$

OR

$P > 0$ and $P < 135$

c

Generally the carrying capacity is can be defined as the highest amount of population and environment can support for an unlimited duration or time period

d

$P = 67.5$

Step-by-step explanation:

From the question we are told that

The population model is $\frac{dP}{dt} = 0.2P(1 - \frac{P}{135} )$

Generally at equilibrium

$\frac{dP}{dt} = 0$

So

$0.2P = 0$

=> $P = 0$

Or

$(1 - \frac{P}{135} ) = 0$

=> $P = 135$

Thus at equilibrium P = 0 or P = 135

Generally when the population is increasing we have that

$\frac{dP}{dt} > 0$

So

$0.2P > 0$

=> $P > 0$

and

$(1 - \frac{P}{135} ) > 0$

$P < 135$

Now when the first value of P i.e $P< 0$ for $\frac{dP}{dt} > 0$

$P_2 > 135$

So when population increasing the values of P are

$P > 0$ and $P < 135$

OR

$P > 0$ and $P < 135$

So to obtain initial values of P where the population converge to the carrying capacity as $t \to [\infty]$

The rate equation can be represented as

$\frac{dP}{dt} = \frac{1}{5}P (1 - \frac{P}{135} )$

So we will differentiate the equation again we have that

$\frac{d^2 P}{dt^2} = \frac{(1 - \frac{P}{135} )}{5} - \frac{P}{675}$

Now as $t \to [\infty]$

$\frac{d^2 P}{dt^2} \to 0$

So

$\frac{(1 - \frac{P}{135} )}{5} - \frac{P}{675} = 0$

=> $\frac{(1 - \frac{P}{135} )}{5} = \frac{P}{675}$

=> $P = 67.5$

5 0

3 years ago

Find the product of 10a³

svlad2 [7]

Answer:

$1000a^3$

Step-by-step explanation:

$10a^3$ is already fully simplified.

If you meant $(10a)^3$ , then:

$(10a)^3\\10^3a^3\\1000a^3$

7 0

2 years ago