Answer:
Step-by-step explanation:
hello :
2x²+12x+2y²−16y−150=0
divide by 2 : (x²+6x)+(y²-8y)-75 =0
(x²+6x+9)-9+(y²-8y+16)-16-75 =0
(x+3)²+(y-4)² =10²
The coordinates of the center are (−3,4), and the length of the radius is 10 units.
Answer:
16^25
Step-by-step explanation:
The applicable relation is ...
(a^b)^c = a^(bc)
2^100 = 16^x = (2^4)^x = 2^(4x)
Equating exponents, we have ...
100 = 4x
25 = x
Then 2^100 = 16^25.
That is a quadratic trinomial.
Its largest degree is 2 (which makes it quadratic), and it has 3 terms separated by addition operators (which makes it trinomial).
The area of the rectangular pen is 10,000 m² and the area of the circular pen is 12,738.95 m².
For the rectangular pen, we want the sides to be congruent. To minimize perimeter and maximize area, the figure needs to be equilateral.
400/4 = 100 m for each side
This means that the area is 100(100) = 10,000 m².
If the pen is circular, this means the circumference is 400 m. Circumference is given by the formula C=πd:
400 = 3.14d
Divide both sides by 3.14:
400/3.14 = 3.14d/3.14
127.389 = d
The radius is half of the diameter:
r = 127.389/2 = 63.6945
The area of the circle is given by A=πr²:
A=3.14(63.6945)² = 12738.9465 ≈ 12738.95 m²
Answer:
15
Step-by-step explanation:
Let n, d, q represent the numbers of nickels, dimes, and quarters. The problem statement tells us ...
n +d +q = 37
n = d +4
q = n +2
___
Rearranging the second equation gives ...
d = n -4
Substituting that into the first, we get ...
n + (n -4) +q = 37
2n +q = 41 . . . . . . . add 4 and simplify
Rearranging the third original equation gives ...
n = q -2
Substituting into the equation we just made, we get ...
2(q -2) +q = 41
3q = 45 . . . . . . . . add 4 and simplify
q = 15 . . . . . . . . . divide by 3
Joe has 15 quarters.
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<em>Check</em>
The number of nickels is 2 fewer, so is 13. The number of dimes is 4 fewer than that, so is 9. The total number of coins is 15 + 13 + 9 = 37, as required.
