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2 years ago

Can somebody tell me what the answer to y-13=8 is? Please and thank you.

Mathematics

1 answer:

makvit [3.9K]2 years ago

4 0

Answer:

Y=21

Step-by-step explanation:

add 13 on both sides you get Y=21

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A puppy and kitten are 180 meters apart when they see each other. The puppy can run at a speed of 25 meters per second, while th

Gnom [1K]

It seems it should be the other way... . How soon will the puppy catch up with the kitten.

The way you have it the kitten will never catch the pup. There is already 180m between them so, the distance will only increase because the puppy runs faster than the cat.

If the puppy runs after the cat, will eventually catch up with it and can be calculated. Not the other way around.

If the pup runs after the cat, the distance between them will decrease and the pup will catch the cat.

RT = D kitten

RT = D+180 puppy

20T = D

25T = D + 180

Substitution

25T = 20T +180

5T = 180

T = 36s

7 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

2/3x8= answer as a mixed fraction?

dimulka [17.4K]

Answer:

5 1/3

Step-by-step explanation:

2/3 x 8 = 2 x 8 / 3 = 16 / 3 = 5 1/3

6 0

3 years ago

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Write two fractions that are equivalent to 5

Dima020 [189]

Answer:

5/1, 25/5

Step-by-step explanation:

8 0

3 years ago

Simplify 6(x + y) + (x - y). 6x 7x 7x + 6y 7x + 5y

poizon [28]

6(x + y) + (x - y)

<em>use distributive property</em>

= 6x + 6y + x - y

<em>combine like terms</em>

= (6x + x) + (6y - y)

8 0

3 years ago

Read 2 more answers