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Katarina [22]
2 years ago
6

What is the solution set for

Mathematics
1 answer:
lana [24]2 years ago
5 0
It’s s=-2 and s=2 because the absolute value makes both of the numbers be positive so either way it’s going to be 2+4=6
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Find the volume of each figure round to the near tenth only number 2 pls it’s 20 points
Nana76 [90]

Answer:

2094.4yds cubed

Step-by-step explanation:

V=pi*r^2*h/3

V=pi*100*20/3

V≈2094.4

6 0
3 years ago
Which best describes a transformation that preserves the size, shape, and angles of an object?
vovikov84 [41]
The transformation is ISOMETRY.

In mathematics, an isometry<span> (or congruence, or congruent transformation) is a distance-preserving transformation between metric spaces, usually assumed to be bijective. A composition of two opposite </span>isometries<span> is a direct </span>isometry<span>. A reflection in a line is an opposite </span>isometry<span>, like R </span>1<span> or R </span>2<span> on the image.</span>
3 0
3 years ago
Can you answer question 29, please?
timurjin [86]
From the table, for every 6 containers of water, you need 8 containers of red dye. So, for every one container of red dye you need 0.75 containers of water (6/8). Then, multiply that by 100 (since you wanted to know water necessary for 100 containers of red dye). For 100 containers of red dye, you would need 75 containers of water. 
5 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
HELP
Shtirlitz [24]

x(x + 9-x) = 6^2

9x = 36

x = 4


Answer

c. 4

4 0
2 years ago
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