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IrinaVladis [17]
3 years ago
11

A) Work out the value of 213 + (29)2

Mathematics
1 answer:
Alina [70]3 years ago
6 0

a) 2^13 / 2^18

b) 216

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The linear equation y = −5x − 18 is written in which form?
Savatey [412]
It’s written in slope intercept form .
8 0
2 years ago
NEED ANSWER ASAP!! TYYY SO MUCH IN ADVANCE
Kipish [7]

Answer:

24.84

Step-by-step explanation:

use Pythagoras theorem

h=√p²+b² = √(19)²+(16)² = √617 =24.84

4 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
Find the equation of the line<br> Slope =4/3<br> y-intercept =-5
AlladinOne [14]

Answer:

y=4/3x - 5

Step-by-step explanation:

5 0
3 years ago
pls help! :) it’s my first day knowing about this things, i don’t even think we had a lesson on it— anywayysss
podryga [215]

Answer:

acute and equilateral

Step-by-step explanation:

because all the angles are given equal

3 0
3 years ago
Read 2 more answers
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