A 16 karat gold means that there are 16 parts of gold in
24 parts of the gold chain. Therefore the weight of the pure gold is:
pure gold = (16 / 24) * 2.6 ounces
<span>pure gold = 1.73 ounces</span>
Answer:
CD = two square root of 10 end square root
Step-by-step explanation:
To find the length of a segment, use the distance formula. Substitute the order pairs for the endpoints of the segment. CD has the end points (-7, -4) and (-1, -2).

The maximum amount of profit the carnival makes based on the number of tickets sold is 6.5 thousand of dollars,
<h3>How to determine the difference?</h3>
The function is given as:
f(x) = -0.5x^2 + 5x - 6
Differentiate the function
f'(x) = -x + 5
Set to 0
-x + 5 = 0
Make x the subject
x = 5
Substitute x = 5 in f(x)
f(5) = -0.5 * 5^2 + 5 * 5 - 6
f(5) = 6.5
The table is not given.
Hence, the maximum amount of profit the carnival makes based on the number of tickets sold is 6.5 thousand of dollars,
Read more about quadratic functions at:
brainly.com/question/18797214
Answer:
<h3>A Angle 1 and Angle 7 </h3>
Step-by-step explanation:
<h3> I think the Answer is A...</h3>
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Answer:
No, mn is not even if m and n are odd.
If m and n are odd, then mn is odd as well.
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Proof:
If m is odd, then it is in the form m = 2p+1, where p is some integer.
So if p = 0, then m = 1. If p = 1, then m = 3, and so on.
Similarly, if n is odd then n = 2q+1 for some integer q.
Multiply out m and n using the distribution rule
m*n = (2p+1)*(2q+1)
m*n = 2p(2q+1) + 1(2q+1)
m*n = 4pq+2p+2q+1
m*n = 2( 2pq+p+q) + 1
m*n = 2r + 1
note how I replaced the "2pq+p+q" portion with r. So I let r = 2pq+p+q, which is an integer.
The result 2r+1 is some other odd number as it fits the form 2*(integer)+1
Therefore, multiplying any two odd numbers will result in some other odd number.
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Examples:
- 3*5 = 15
- 7*9 = 63
- 11*15 = 165
- 9*3 = 27
So there is no way to have m*n be even if both m and n are odd.
The general rules are as follows
- odd * odd = odd
- even * odd = even
- even * even = even
The proof of the other two cases would follow a similar line of reasoning as shown above.