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vlada-n [284]
1 year ago
13

A circle has center

Mathematics
1 answer:
Fed [463]1 year ago
5 0
Note that the equation of the circle is

(x-h)² +(y-k)² =r²

where centre is (h,k)

the equation of the circle based on the information given

(x-3)² +(y-4)² =r²
and the point on the circle (3,-2)

substitute into the equation

(3-3)² +(-2-4)² =r²

r=6 or r=-6

since r is radius, we reject r=-6 since radius must be nonnegative.

the radius is 6
You might be interested in
How to solve for -3x squared + 2y squared + 5xy - 2y + 5x squared - 3y squared when x = 0.5 and y = -1/10??
Pani-rosa [81]

Answer:

0.44

Step-by-step explanation:

-3x^2 + 2y^2 + 5xy - 2y +5x^2 - 3y^2

Combine like terms

-3x^2 + 5x^2 = 2x^2 2y^2 - 3y^2 = -1y^2

2x^2 - 1y^2 + 5xy - 2y

Now plug in the solutions Note: it is easier if you have all decimals or all fractions (-1/10=-.1

2(0.5)^2 - 1(-0.1)^2 + 5(0.5)(-0.1) - 2(-0.1)

Simplify:

0.5 - 0.01 - 0.25 + 0.2

0.5 + 0.2 - 0.01 - 0.25

0.7 - 0.26

0.44

5 0
3 years ago
Answer?I need it pleaseeee
Elan Coil [88]

Answer: 20 dallors

Step-by-step explanation:

8 0
2 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
Which answer choice is equivalent to the expression below?
anyanavicka [17]
In order to know the equivalent expression of the given above, all we need to do is to simplify it. So, for the exponents, what we need to do is just subtract.
The final answer would be 3/2xy^3. So the answer is option C. Hope this is the answer that you are looking for. 
8 0
3 years ago
Read 2 more answers
Please help: <br> Simplify: -1/2x + 3 - 4x + 5 + 3/2x
KATRIN_1 [288]
Wait, didn't I answer this yesterday?
eh

group like terms
-1/2x-4x+3/2x+3+5
add
-4x-1/2x+3/2x+8
-4x+2/2x+8
-4x+1x+8
-3x+8
8 0
3 years ago
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