The base of a solid s is the region bounded by the circle x^2 y^2=16
2 answers:
The center is (0, 0) and the radius is 4 units.
<h3>What is the circle?</h3>A circle is a round-shaped figure that has no corners or edges. In geometry, a circle can be defined as a closed, two-dimensional curved shape.
The base of a solid is s is the region bounded by the circle is;
The standard equation of a circle is:
(x - h)² + (y - k)² = r²
Where (h, k) is the circle center and r is the radius.
On comparing with the standard equation of a circle;
h = 0 , k = 0 and r =4
Hence. the center is (0, 0) and the radius is 4 units.
Learn more on the circle at:
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Answer:
Step-by-step explanation:
<u>Key indices rule: When dividing powers, subtract the indices and keep the base numbers constant.</u>
<u /> Based on the rule above, the indices are determined by 3--8, which is 11.
The base number is kept constant.
Therefore, the answer is .
Short Answer:Choice C [Third one down]
Remark
Start with the number so I can talk about the basic principle at work. The way to work it is to factor 162 into prime factors and hope there are at least 3 that are the same.
162 = 2 * 81
162 = 2 * 3 * 3 * 3 * 3
Now When you take the cube root of that, you get![\sqrt[3]{2*2*2*2*3}](https://tex.z-dn.net/?f=%20%5Csqrt%5B3%5D%7B2%2A2%2A2%2A2%2A3%7D%20)
Here's the rule for a cube root. For every 3 prime factors under the cuberoot sign, you take out one and throw the other two away. So for cube root of 81,
you would factor it as ∛(81) = ∛(3 * 3 * 3 * 3) = 3∛3.
So out of four 3s under the cube root sign, you have 1 outside the root sign and one inside the root sign. 2 of the four 3s have been thrown away.
Continuation.
X first
You want 2 xs outside the root sign
∛(x * x * x * x * x *x ) = (x * x) You have thrown away 2 xs for every x outside the cube root sign
C = 6 There are no left overs.
Y second
For the ys, you need 1 outside and 2 inside the root sign. that's because you need 5 altogether.
∛(y * y * y * y * y) = y ∛y^2
Answer
c = 6; d = 2
Choice C <<<<<< answer
Remark
Start with the number so I can talk about the basic principle at work. The way to work it is to factor 162 into prime factors and hope there are at least 3 that are the same.
162 = 2 * 81
162 = 2 * 3 * 3 * 3 * 3
Now When you take the cube root of that, you get
Here's the rule for a cube root. For every 3 prime factors under the cuberoot sign, you take out one and throw the other two away. So for cube root of 81,
you would factor it as ∛(81) = ∛(3 * 3 * 3 * 3) = 3∛3.
So out of four 3s under the cube root sign, you have 1 outside the root sign and one inside the root sign. 2 of the four 3s have been thrown away.
Continuation.
X first
You want 2 xs outside the root sign
∛(x * x * x * x * x *x ) = (x * x) You have thrown away 2 xs for every x outside the cube root sign
C = 6 There are no left overs.
Y second
For the ys, you need 1 outside and 2 inside the root sign. that's because you need 5 altogether.
∛(y * y * y * y * y) = y ∛y^2
Answer
c = 6; d = 2
Choice C <<<<<< answer