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2 years ago

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 (or congruence, or congruent transformation) is a distance-preserving transformation between metric spaces, usually assumed to be bijective. A composition of two opposite isometries is a direct isometry. A reflection in a line is an opposite isometry, like R 1 or R 2 on the image.

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