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2 years ago

An envelope measures 8 inches by 15 inches. A pencil is placed in

Mathematics

2 answers:

UNO [17]2 years ago

5 0

Answer:

The length of the pencil is 17 inches.

Step-by-step explanation:

You need to do the Pythagorean theorem. For example, this is what I did, 8^2+15^2=x^2

64+225=x^2

289=x^2

the square root of 289 is 17.

hope this helps :)

Oliga [24]2 years ago

3 0

The length of the pencil would be 17. This could be easily found by using a^2 + b^2 = c^2. Please mark me brainliest it would be pretty cash money of you

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Describe a proportional relationship involving some product or service and its cost that IS proportional. Explain why it is a pr

kvasek [131]

Natalie is an expert archer. The following table shows her scores (points) based on the number of targets she hit.

Targets hit 111 222 555

Points 333 666 151515

Plot the ordered pairs from the table.

Are Natalie's points proportional to the number of targets she hit?

Choose 1 answer:

(Choice A)

Yes

(Choice B)

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2 years ago

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The sum of a positive integer and a negative integer is...

murzikaleks [220]

Depends is the positive integer is larger than the negative interger for example -10+5=-5.
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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

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

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

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

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

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

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

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

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

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3 years ago

Three equivalent ratios for 6: 54

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3 years ago

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Katen [24]

Can you help me with my posted questions

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