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3 years ago

What is the coefficient of 1/3d+15

Mathematics

2 answers:

drek231 [11]3 years ago

4 0

1/3d is the coefficient of this problem.

sineoko [7]3 years ago

3 0

1/3d+15=1/3(d+45)
Coefficient=1/3

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A container has peppermint gum and spearmint gum. There are 40 pieces of peppermint gum in the container and the ratio of pepper

Anit [1.1K]

Answer:

16 pieces of spearmint gum

Step-by-step explanation:

For every five pieces of peppermint gum, there are 2 pieces of spearmint gum.

First, divide the total number of peppermint gum by 5:

40/5 = 8

Multiply this number by 2 (pieces of spearmint) to get the number of spearmint gum pieces:

8 x 2 = 16

Proof:

Ratio is 40 : 16, dividing this by 8 will reduce the ratio to 5:2

7 0

3 years ago

A student was asked to use the formula for the perimeter of a rectangle, p = 2l 2w, to solve for l. the student came up with an

timofeeve [1]

<span>A student was asked to use the formula for the perimeter of a rectangle, p = 2l 2w, to solve for l. the student came up with an answer, p -2w=2l. what</span>

6 0

3 years ago

Solve 5^2x+1-5^2x=150 with steps

tiny-mole [99]

Answer:

x = 2.98

Step-by-step explanation:

5^2x + 1 - 5^2x = 150

25x + 1 + 25x = 150

50x + 1 = 150

50x = 149

x = 2.98

5^2x + 1 - 5^2x = ?

5^2(2.98) + 1 - 5^2(2.98) = ?

25(2.98) + 1 + 25(2.98) = ?

74.5 + 1 + 74.5 = ?

149 + 1 = ?

150 = ?

4 0

3 years ago

Read 2 more answers

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

Select the correct answer.

nydimaria [60]

Answer:

It is CD

Step-by-step explanation:

because 300+200=500, juts times it by 3 and divide

5 0

2 years ago