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suter [353]
2 years ago
15

A square piece of cardboard has sides that are 7/12 meter long. What is the carb words perimeter

Mathematics
1 answer:
gavmur [86]2 years ago
6 0

Since it’s a square all sides are even. Meaning you multiply 7/12 by 4. Giving you 7/3 or 2 ⅓. Hope this helps!

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Ummm what this ?? Helping my sister
Radda [10]

Answer:

ok this the answer   By definition, a rectangle is a parallelogram because its pairs of opposite sides are parallel. A rectangle also has the special characteristic that all of its angles are right angles; all four of its angles are congruent. The other special case of a parallelogram is a special type of rectangle, a square.

Step-by-step explanation:

6 0
2 years ago
Solve the system by substitution 5x+3y=8 3x+y=8
Yakvenalex [24]
3x+y=8
y=8-3x

5x+3(8-3x)=8
5x+24-9x=8
-4x+24=8
-4x=-16
x=4

5x+3y=8
5(4)+3y=8
20+3y=8
3y=-12
Y=-4

5x+3y=8
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4 0
3 years ago
What is a whole number between the square root of 27 and the square root of 39
Vinvika [58]
That number would be 6.
Since 5^2 = 25 and 6^2 = 36, we can see that the only possible whole number between sqrt(27) and sqrt(39) has to be 6.


8 0
3 years ago
If P is the circumcenter of triangle ABC, find each measure.
seropon [69]

Answer:

Step-by-step explanation:

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5 0
2 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
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