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3 years ago

8

Rewrite 24 + 48 using GCF and the distributive property.

1 answer:

Nutka1998 [239]3 years ago

6 0

The correct answer is B

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Can someone please help meee

pshichka [43]

Answer:

160

Step-by-step explanation:

Angle ABE is a 180 as ABE forms a line

so angle ABC + ANGLE CBE = 180

x + 20 = 180

x = 180-20

= 160

I hope im right!!

6 0

3 years ago

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I'm not sure what they mean by "relations that represents a function". I don't understand this.

sergeinik [125]

Means, for every 1 input, there is 1 coresponding output
normally x is input and y is output in form (x,y)
so just look for the option(s) that has/have every first number repeat with only 1 2nd number

A. -7 repeats with 5 and -3, not a function

B. no repeats of first number (6,-8) and (-5,-8) are fine because first numbers don't repeat, das is funciton

C. no repeats, function

D. no repeats, function

answer is B,C,D

4 0

3 years ago

Initially a tank contains 10 liters of pure water. Brine of unknown (but constant) concentration of salt is flowing in at 1 lite

zhenek [66]

Answer:

Therefore the concentration of salt in the incoming brine is 1.73 g/L.

Step-by-step explanation:

Here the amount of incoming and outgoing of water are equal. Then the amount of water in the tank remain same = 10 liters.

Let the concentration of salt be a gram/L

Let the amount salt in the tank at any time t be Q(t).

$\frac{dQ}{dt} =\textrm {incoming rate - outgoing rate}$

Incoming rate = (a g/L)×(1 L/min)

=a g/min

The concentration of salt in the tank at any time t is = $\frac{Q(t)}{10}$ g/L

Outgoing rate = $(\frac{Q(t)}{10} g/L)(1 L/ min)$ $\frac{Q(t)}{10} g/min$

$\frac{dQ}{dt} = a- \frac{Q(t)}{10}$

$\Rightarrow \frac{dQ}{10a-Q(t)} =\frac{1}{10} dt$

Integrating both sides

$\Rightarrow \int \frac{dQ}{10a-Q(t)} =\int\frac{1}{10} dt$

$\Rightarrow -log|10a-Q(t)|=\frac{1}{10} t +c$ [ where c arbitrary constant]

Initial condition when t= 20 , Q(t)= 15 gram

$\Rightarrow -log|10a-15|=\frac{1}{10}\times 20 +c$

$\Rightarrow -log|10a-15|-2=c$

Therefore ,

$-log|10a-Q(t)|=\frac{1}{10} t -log|10a-15|-2$ .......(1)

In the starting time t=0 and Q(t)=0

Putting t=0 and Q(t)=0 in equation (1) we get

$- log|10a|= -log|10a-15| -2$

$\Rightarrow- log|10a|+log|10a-15|= -2$

$\Rightarrow log|\frac{10a-15}{10a}|= -2$

$\Rightarrow |\frac{10a-15}{10a}|=e ^{-2}$

$\Rightarrow 1-\frac{15}{10a} =e^{-2}$

$\Rightarrow \frac{15}{10a} =1-e^{-2}$

$\Rightarrow \frac{3}{2a} =1-e^{-2}$

$\Rightarrow2a= \frac{3}{1-e^{-2}}$

$\Rightarrow a = 1.73$

Therefore the concentration of salt in the incoming brine is 1.73 g/L

8 0

3 years ago

You run out of gas and measure the amount of gas it takes to fill the tank. Is this data a type discrete or continuous

ziro4ka [17]

I think it is continuous

7 0

3 years ago

There are 12 plants in 3 rows. How many plants are in each row? multiplication<br> show your work

anygoal [31]

There are 4 per row.
12 divided by 3 = 4.

7 0

3 years ago

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