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jek_recluse [69]
3 years ago
7

Use the provided ruler. Suppose you recreate the drawing so that it has a scale of 1 cm : 4 m. What will the length and width of

the new scale drawing be?
Mathematics
2 answers:
ki77a [65]3 years ago
5 0

Answer:

1cm 2m

Step-by-step explanation:

Zinaida [17]3 years ago
4 0

Answer:

Include more details

Step-by-step explanation:

i will answer again after.

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How do you round 2.8497 × 10^3 and 349.69 to 3 significant figures?
Kobotan [32]

Your answer for rounding 2.8497 x 10^3 is correct: 2.85 x 10^3.

350.0 is not correct because it has 4 sig figs. The proper rounding would be simply 350. with not additional zeros.

6 0
4 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
Richard filled 3-gallon jug with fruit punch.
Alecsey [184]

Answer:

About 11.4

Step-by-step explanation:

There are 3.78541178 liters in a gallon.

Round that up to the nearest 100 and you get 3.8

Multiply 3.8 by 3.

You get 11.4

7 0
3 years ago
Convert 2 inches into feet. Round your answer to the nearest hundredth.
DiKsa [7]

Answer: 0.17

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
An author is having a book signing at a store. He agrees to give 5% of his profit to literacy programs and 15% of his profit to
quester [9]

Answer:

160

Step-by-step explanation:

15% + 5% = 20%

80%=0.8

200x0.8=160

6 0
3 years ago
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