Because y varies inversely as x, therefore
where k = constant.
When x = 6, y = 2.5. Ther efore
Cross multiply.
k = 6 x 2.5 = 15
The equation is
When x = 4, obtain
Answer: a. 3.75
To make it easier, you calculate the volume of the first aquarium.
1st aquarium:
V = L x W x H
V = 8 x 9 x 13
V = 72 x 13
V = 936 in.
Rate: 936 in./2 min.
Now that you've got the volume and rate of the first aquarium, you can find how many inches of the aquarium is filled within a minute, which is also known as the unit rate. To do that, you have to divide both the numerator and denominator by their least common multiple, which is 2. 936 divided by 2 is 468 and 2 divided by 2 is 1.
So the unit rate is 468 in./1 min. Now that you've got the unit rate, you can find out how long it'll take to fill the second aquarium up by finding its volume first.
2nd aquarium:
V = L x W x H
V = 21 x 29 x 30
V = 609 x 30
V 18,270 inches
Calculations:
Now, you divide 18,270 by 468 to find how many minutes it will take to fill up the second aquarium. 18,270 divided by 468 is about 39 (the answer wasn't exact, so I said "about").
2nd aquarium's rate:
18,270 in./39 min.
As a result, it'll take about 39 minutes to fill up an aquarium measuring 21 inches by 29 inches by 30 inches using the same hose. I really hope I helped and that you understood my explanation! :) If I didn't, I'm sorry. I tried. :(
Answer:
Divide by 2
q^2+4q=3/2
q^2+4q(4/2)^2=3/2+(4/2)^2
(q+4/2)^2=3/2+16/4
taking the square root of both side
√(q+4/2)^2=√(3/2+16/4)
Note that the square will cancel the square root then you will take LCM on the right hand side
q+4/2=√6+16/4
q+4/2=√22/4
q= -4/2+-√22/4
q=(-4+_√22/4)
Answer:
There is approximately 5.475 liters of blood in the body.
Step-by-step explanation:
At a resting pulse rate of 75 beats per minute, the human heart typically pumps about 73 ml of blood per beat.
73 ml in liters = 
liters
Now, 75 beats/min x 0.073 liters/beat = 5.475 liters/min
And 5.475 liters/min x 1 min/body = 5.475 liters/body
Hence, there is approximately 5.475 liters of blood in the body.
