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2 years ago

6

I need help on this question please

2 answers:

Nadya [2.5K]2 years ago

4 0

The answer is 54 ants. You would divide the perimeter of the sandwich by the measure of each ant. :)

Juliette [100K]2 years ago

4 0

9 x 6 = 54

54 ants.

Hope this helped!

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A sign language club made $627.50 from selling popcorn at a festival. The popcorn cost the club $95.00. Which expression represe

Charra [1.4K]

Answer:

(627.50-95)/m

Step-by-step explanation:

The total value of money earned was $627.50. Then subtract the amount of money the popcorn cost, $95.00. This subtraction would make the difference of $532. This equation would be in parentheses because you have to get the difference before you can divide the profits among the club members. After you put that into parentheses, divide the difference by M. This will give you the equation (627.50-95.00)/m hope this helps!

4 0

2 years ago

Read 2 more answers

What is the rule for the reflection?

Dvinal [7]

Answer:

Incident ray, the reflection ray and the normal to thereflection surface at the point of the incidence lie in the same plane

5 0

2 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

HELP PLEASE 30 POINTS!!

Naya [18.7K]

Answer:

p - w = $25 OR w + $25 = p

Step-by-step explanation:

pencil = p

pen = w

p - w = $25 OR w + $25 = p

4 0

2 years ago

What is 0.2857142857 rounded to the nearest whole percent? pls help me

Grace [21]

29% would be the answer if you are rounding up

6 0

2 years ago