PLS HELP ASAPPPP!!!

Ede4ka [16]

Answer:

Sure! Your answer is about 315.33.

Step-by-step explanation:

If you want to figure this out by yourself, just open a calculator and do 946 * 0.33. You could also find this by dividing 946 by 3.

Please like and mark brainliest!

Hope it helps!

3 0

3 years ago

Read 2 more answers

What is the perpendicular line in slope form of the equation (-5,8);y=1/5x-1

denpristay [2]

Answer: Find the negative reciprocal of the slope of the original line and use the slope-intercept form y=mx+b to find the line perpendicular to y=1/5x−1. y=−5x−17

5 0

4 years ago

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In a school, the ratio of soccer players to volleyball players is 4 : 1.

VLD [36.1K]

4:1 means for every four soccer player there is 1 volleyball player

that means there are four times the number of soccer players than volleyball players

7 0

3 years ago

Suppose that a large mixing tank initially holds 500 gallons of water in which 50 pounds of salt have been dissolved. Another br

Lina20 [59]

Answer:

The differential equation for the amount of salt A(t) in the tank at a time t > 0 is $\frac{dA}{dt}=12 - \frac{2A(t)}{500+t}$ .

Step-by-step explanation:

We are given that a large mixing tank initially holds 500 gallons of water in which 50 pounds of salt have been dissolved. Another brine solution is pumped into the tank at a rate of 3 gal/min, and when the solution is well stirred, it is then pumped out at a slower rate of 2 gal/min.

The concentration of the solution entering is 4 lb/gal.

Firstly, as we know that the rate of change in the amount of salt with respect to time is given by;

$\frac{dA}{dt}= \text{R}_i_n - \text{R}_o_u_t$

where, $\text{R}_i_n$ = concentration of salt in the inflow $\times$ input rate of brine solution

and $\text{R}_o_u_t$ = concentration of salt in the outflow $\times$ outflow rate of brine solution

So, $\text{R}_i_n$ = 4 lb/gal $\times$ 3 gal/min = 12 lb/gal

Now, the rate of accumulation = Rate of input of solution - Rate of output of solution

= 3 gal/min - 2 gal/min

= 1 gal/min.

It is stated that a large mixing tank initially holds 500 gallons of water, so after t minutes it will hold (500 + t) gallons in the tank.

So, $\text{R}_o_u_t$ = concentration of salt in the outflow $\times$ outflow rate of brine solution

= $\frac{A(t)}{500+t} \text{ lb/gal } \times 2 \text{ gal/min}$ = $\frac{2A(t)}{500+t} \text{ lb/min }$

Now, the differential equation for the amount of salt A(t) in the tank at a time t > 0 is given by;

= $\frac{dA}{dt}=12\text{ lb/min } - \frac{2A(t)}{500+t} \text{ lb/min }$

or $\frac{dA}{dt}=12 - \frac{2A(t)}{500+t}$ .

4 0

3 years ago

A=1/2bh. A=200ft. H=10. What is b

Svetllana [295]

Answer:

B= $\frac{2A}{H}$

Step-by-step explanation:

If you noticed, A= $\frac{1}{2}$ BH is the formula for finding the area for a triangle. Your goal is to get B by it's self. Your first step will be to clear of the fraction first, so you will multiply both sides by 2. 2(A)=2( $\frac{1}{2}$ BH). On the left, you have 2×A= On the right side, you have 2( $\frac{1}{2}$ BH), but since you have a number in the equation, you will only use 2× $\frac{1}{2}$ . To solve 2× $\frac{1}{2}$ , you will cnacel out both 2's and you have 1. 1×BH will still equal BH, so you are now left will B×H.

(Your new equation looks like this by the way). 2A=BH

Since you need to get B by its self, the way to clear the H away from the B is by dividing. You will now divide the B and H aswell as 2A and H. (It will look like this) $\frac{2A}{H} = \frac{BxH}{H}$ . (Again when you have the same number or letter, you cross it out. When you divide, you won't change anything on the left side, and all you have to do on the right id to cross out the H next to the B and cross out the H on the bottom of the equation). You should be left with $\frac{2A}{H}$ = B. Now you can turn it around for your final answer. B= $\frac{2A}{H}$ .

Please let me know if i helped, how I did, and if you have any questions.

3 0

3 years ago