Please helpppppp pleaseeeeeeeeeee.....PLEASE

Express the following

krok68 [10]

Answer:

1. 32

2. 24

3. $\frac{1}{12}$

4. $\frac{1}{8}$

5. 15

6. $\frac{1}{15}$

Step-by-step explanation:

1. 4÷ $\frac{1}{8} = 4*8=32$

2. 6÷ $\frac{1}{4} = 6*4=24$

3. $\frac{1}{3}$ ÷4= $\frac{1}{3}*\frac{1}{4} =\frac{1}{12}$

4. $\frac{1}{2}$ ÷4= $\frac{1}{2} *\frac{1}{4} =\frac{1}{8}$

5. 5÷ $\frac{1}{3} =5*3=15$

6. $\frac{1}3}$ ÷5= $\frac{1}{3} *\frac{1}{5} = \frac{1}{15}$

3 0

4 years ago

Identify the value of p. Give your answers in simplest radical form. HELP PLEASE!!

VikaD [51]

Answer:

C

Step-by-step explanation:

using the cosine ratio to find p

cos 45° = $\frac{adjacent}{hypotenuse}$ = $\frac{6}{p}$

cross- multiplying gives

p × cos45° = 6 → [ cos 45° = $\frac{1}{\sqrt{2} }$ ]

p × $\frac{1}{\sqrt{2} }$ = 6

multiply both sides by $\sqrt{2}$

⇒ p = 6 $\sqrt{2}$ ( third option on list )

7 0

3 years ago

2. Given square CANE with diagonals

svlad2 [7]

What’s something goes up but never comes down?

6 0

3 years ago

In order to keep a boat from drifting away, the boat is tied to a pier with a 60-foot rope. When the boat drifts as far from the

Rus_ich [418]

The answer is C) hope this helped!

6 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