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Tamiku [17]
3 years ago
13

Complete the multiplication sentence.

Mathematics
1 answer:
lana66690 [7]3 years ago
7 0

Answer:

3

Step-by-step explanation:

multiply both sides of the equation by 10

10(1/10x) = 10 (3/10)

simplify both sides of the equation

left side: 10 (1/10x)

x=10(3/10)

Right side:

x=10 (3/10)- cancel the common factor, then rewrite the expression

x=3

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There are 110 calories per 177.4 grams of cereal X. Find how many calories are in 284.5 grams of this cereal.
Firlakuza [10]

Answer:

176 calories in 284.5 grams of cereal X

Step-by-step explanation:

110/177.4=.62

.62(284.5)=176.4

7 0
3 years ago
a fountain in a pond has a pump that recirculates 75 gallons of water in 1/4 of an hour.express this rate as a unit rate in gall
olga55 [171]
So if in 1/4 of an hour 75 gallons are pumped you would multiply 75 by four which will give you 300. This means 300 gallons are pumped in an hour
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
I'm confused- how do I form the equation with 2 pts?
klio [65]

Answer:

the equation of the line is   y = -3x - 6

Step-by-step explanation:

Note that since (0, -6) is the y-intercept, we can write the slope-intercept equation of the line as y = mx - 6.  The other given point is (-2, 0) (which happens to be the x-intercept also).  Starting with y = mx - 6, replace y with 0 and x with -2:  

0 = m(-2) - 6.  We now solve this for the slope, m:  0 = -2m - 6 becomes

2m = -6, or m = -3.

With m = -3 and b = -6, the equation of the line is   y = -3x - 6

4 0
3 years ago
Connor gathered data about the height of customers entering a store and the time that it took them to gather their items and ent
sattari [20]

0.88. you can simply see this by using the right use of the data

7 0
3 years ago
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