Round 4.265 to the nearest hundredth

Bob has some 10 lb weights and some 3 lb weights. Together, all his weights add up to 50 lbs. If x represents the number of 3 lb

Romashka [77]

Total weight = 50 lb
x = number of 3-lb weights
y = number of 10-lb weights

weight of 3-lb weights = 3x
weight of 10-lb weights = 10y
total weight = 3x + 10y

equation
3x + 10y = 50

8 0

1 year ago

1: This year Cornell's class took one more than three times as many field trips as last year's class. This Year, Cornell's class

S_A_V [24]

Cornell- 2 trips Jamyra- 23 mins studying James- 182 bucks Kaya- 5.5 hours Runner-up- 4.5 Anthony-160 bucks Jasmine-39 years old Ronnie- 7.5 hours working.

6 0

2 years ago

Enrique earns 10 points for each question that he answers correctly on a geography test.

saul85 [17]

<h3>✽ - - - - - - - - - - - - - - - ~<u>Hello There</u>!~ - - - - - - - - - - - - - - - ✽</h3>

➷ The equation would be y = 10x

➶ Hope This Helps You

➶ Good Luck

➶ Have A Great Day ^-^

↬ ʜᴀɴɴᴀʜ ♡

4 0

2 years ago

Read 2 more answers

A professional writer decides to judge the value of her laptop by the number of times she presses

shusha [124]

Answer:

D.)

15040000 <= k <=19840000

Step-by-step explanation:

V = 1200 -0.00002k

Value V after one year falls in the range of $803.20 and $899.20.

Let's substract these values from 1200.

1200-803.20=396.80

1200-899.20=300.80

So it means that

0.00002k= 396.80

And

0.00002k= 300.80

So

0.00002k= 396.80

K= 396.80/0.00002

K= 19840000

And

0.00002k= 300.80

K = 300.80/0.00002

K= 15040000

So range is between

15040000 <= k <=19840000

7 0

2 years ago

A large tank is partially filled with 100 gallons of fluid in which 20 pounds of salt is dissolved. Brine containing 1 2 pound o

Valentin [98]

Answer:

47.25 pounds

Step-by-step explanation:

$\dfrac{dA}{dt}=R_{in}-R_{out}$

<u>First, we determine the Rate In</u>

Rate In=(concentration of salt in inflow)(input rate of brine)

$=(0.5\frac{lbs}{gal})( 6\frac{gal}{min})\\R_{in}=3\frac{lbs}{min}$

Change In Volume of the tank, $\frac{dV}{dt}=6\frac{gal}{min}-4\frac{gal}{min}=2\frac{gal}{min}$

Therefore, after t minutes, the volume of fluid in the tank will be: 100+2t

Rate Out=(concentration of salt in outflow)(output rate of brine)

$R_{out}=(\frac{A(t)}{100+2t})( 4\frac{gal}{min})\\\\R_{out}=\frac{4A(t)}{100+2t}$

Therefore:

$\dfrac{dA}{dt}=3-\dfrac{4A(t)}{100+2t}\\\\\dfrac{dA}{dt}=3-\dfrac{4A(t)}{2(50+t)}\\\\\dfrac{dA}{dt}=3-\dfrac{2A(t)}{50+t}\\\\\dfrac{dA}{dt}+\dfrac{2A(t)}{50+t}=3$

This is a linear differential equation in standard form, therefore the integrating factor:

$e^{\int \frac{2}{50+t}dt}=e^{2\ln|50+t|}=e^{\ln(50+t)^2}=(50+t)^2$

Multiplying the DE by the integrating factor, we have:

$(50+t)^2\dfrac{dA}{dt}+(50+t)^2\dfrac{2A(t)}{50+t}=3(50+t)^2\\\{(50+t)^2A(t)\}'=3(50+t)^2\\$Taking the integral of both sides\\\int \{(50+t)^2A(t)\}'= \int 3(50+t)^2\\(50+t)^2A(t)=(50+t)^3+C $ (C a constant of integration)\\Therefore:\\A(t)=(50+t)+C(50+t)^{-2}$

Initially, 20 pounds of salt was dissolved in the tank, therefore: A(0)=20

$20=(50+0)+C(50+0)^{-2}\\20-50=C(50)^{-2}\\C=-\dfrac{30}{(50)^{-2}} =-30X50^2=-75000$

Therefore, the amount of salt in the tank at any time t is:

$A(t)=(50+t)-75000(50+t)^{-2}$

After 15 minutes, the amount of salt in the tank is:

$A(15)=(50+15)-75000(50+15)^{-2}\\=47.25$ pounds$

8 0

1 year ago