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Jet001 [13]
2 years ago
6

Help need it done for tomorrow ​

Mathematics
1 answer:
Gelneren [198K]2 years ago
4 0

Step-by-step explanation:

the law of sines :

a/sin(A) = b/sin(B) = c/sin(C)

with the sides being opposite to the angles.

so,

14/sin(80) = x/sin(60)

x = 14×sin(60)/sin(80) ≈ 12.3cm

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What combination of transformations is shown below?
Anastasy [175]

Answer:

Rotation then reflection

Step-by-step explanation:

The triangle starts where the light green one is. Then it is rotated 90 degrees clockwise around the point where the hypotenuse and longer leg meet. From the sark green triangle it is reflectd to where the purple triangle is.

8 0
2 years ago
Write as a mixed number 73/8
lyudmila [28]
Your answer would be 9 1/8
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3 years ago
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5. While working out, Susie doos 25 sit-ups for every 12 push-ups. If she does 00 push-ups, how
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2500 is the answer ...
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Draw a diagram that shows 1/5 times 30 equals 6
MrMuchimi

Total number = 30

1/5 of 30 = 1/5 x 30 = 6

The attached figures show the total number of filled circles in five lines. Each line have “6” filled circles.

So 5 lines having 6 filled circles = 5 x 6 = 30

<span>Every line represents “1/5” of total amount “30”. Which proved that 1/5 of 30 is equal to 6. </span>

3 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
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