ANSWER QUICKLY PLEASE!!!

This design shows several circles with the same center. The total radius of the design is 8 inches. The angle shown has a measur

Burka [1]

Answer:

11.33 in. to the nearest hundredth.

Step-by-step explanation:

The perimeter of the shaded area = length of the 2 straight lines + the length of the 2 arcs = 4 + length of the 2 arcs.

Calculate the length of the outer arc:

This equals (30 / 360) * perimeter of the largest circle

= 1/12 * 2 π * 8

= 4/3 π in.

The inner circle has a radius of 8 - 2 = 6 ins

so the length of the inner arc

= 1/12 * π * 2 * 6

= π in.

So the perimeter of the shaded region = 4 + 4/3 π + π

= 4 + 7π/3

= 11.33 in.

7 0

3 years ago

Read 2 more answers

Rina is making a nut mixture to sell at the local farmers' market. She mixes 6 pounds of pecans with a nut mixture that is 70% p

Marat540 [252]

So the best way to do these is concentration1 (%) × volume1 = concentration2 × volume2
Or C1V1 + C2V2 = C3V3, where C1 = 100% (bc ALL pecans), V1 = 6 lbs, C2 = 70%, C3 = 82%:
100%×6 + 70%×v2 = 82%×(6+v2)
100%=1.00, 70%=.7, 82%=.82
note: if none is poured out then v3 = v1+v2
6 + .7v2 = .82 (6+v2)
6 + .7v2 = 4.92 + .82v2
6 + .7v2 -.7v2 = 4.92 + .82v2 -.7v2
6 = 4.92 + .12v2
6-4.92 = 4.92-4.92 + .12v2
1.08 = .12v2
.12v2/.12 = 1.08/.12
v2 = 9 lbs
that's only v2!!!

For the final poundage, we need v3:

v3 = 6 + v2 = 6 + 9 = 15 lbs

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

Two pools are being drained. To start, the first pool had 3850 liters of water and the second pool had 4370 liters of water. Wat

SashulF [63]

Answer:

(a)

Amount of water in the first pool (in liters) = 3850 liters - 33 liters per minute * x minutes

Amount of water in the second pool (in liters) = 4370 liters - 43 liters per minute * x minutes

(b)

3850 liters - 33 liters per minute * x minutes= 4370 liters - 43 liters per minute * x minutes

Step-by-step explanation:

(a) With x being the minutes after which you want to calculate the amount of water in the pool, to calculate this amount of water you must subtract the water drained from the initial amount of water that the pool contains.

Knowing that, for example, the water from the first pool drains at a rate of 33 liters per minute, then after x minutes, the total water drained will be 33 liters per minute * x minutes. Then:

Amount of water in the first pool (in liters) = 3850 liters - 33 liters per minute * x minutes

Reasoning in the same way:

Amount of water in the second pool (in liters) = 4370 liters - 43 liters per minute * x minutes

(b) Now want to know when the two pools would have the same amount of water. If they have the same amount of water then you can express:

Amount of water in the first pool = Amount of water in the second pool

Replacing the expressions found in (a):

3850 liters - 33 liters per minute * x minutes= 4370 liters - 43 liters per minute * x minutes

5 0

4 years ago

Quality control is an important issue at ACME Company which manufacturers light bulbs. In order to conduct testing of the life h

mrs_skeptik [129]

Answer: 3424.28

Step-by-step explanation:

Given data : 378, 361, 350, 375, 200, 391, 375, 368, 321

Number of data values = 9

Mean : $\overline{x}=\dfrac{\sum^n_{i=1} x_i}{n}$

$\\\\=\dfrac{378+361+350+375+200+391+375+368+321}{9}\\\\=\dfrac{3119}{9}\approx346.56$

Variance = $\dfrac{\sum^n_{i=1} (x_i-\overline{x})^2}{n-1}$

$\sum^n_{i=1} (x_i-\overline{x})^2 = (31.44)^2+(14.44)^2+(3.44)^2+(28.44)^2+(-146.56)^2+(44.44)^2+(28.44)^2+(21.44)^2+(-25.56)^2\\\\=27394.2224$

Variance = $\dfrac{27394.2224}{8}=3424.2775\approx3424.28$

4 0

3 years ago