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liq [111]
3 years ago
9

Calculate the minimum value of the quadratic f(X)=x^2-6x+10​

Mathematics
1 answer:
tangare [24]3 years ago
8 0
The answer is (-3,-1)
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The least common multiple of two numbers is 22. One number is even, and one number is odd.
Nata [24]
22 = 2 · 11

2 is even
11 is odd

LCM(2; 11) = 22

but we have other answer:

22 = 1 · 22

22 is even
1 is odd

LCM(1; 22) = 22

Answer: 2 and 11 or 22 and 1.
4 0
3 years ago
choose the equation below that represents the line passing through the point (1,-4) with a slope of 1/2
Archy [21]
Well, we can use the form y = mx + b, where m equals the slope and x and y are the coordinate values:
-4 = 1/2(1) + b
b = 9/2
Therefore, the equation of the line is:
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6 0
3 years ago
What is the answer <br> Need quick <br> Help
FrozenT [24]
Your answer would be A b=a+5
5 0
3 years ago
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If N has a remainder of 2 when divided by 3,4,5, or 6 and N is a 3 digit number, what is the largest possible value for N?
coldgirl [10]

Short Answer 962

Remark

What an interesting problem. I didn't see it until a moment ago. Take the lowest common multiple of 3,4,5,6 and choose a value close to 999 that is divisible by the Lowest Common Multiple.


Step One

Find the lowest common multiple of 3,4,5 and 6. That's done by listing all the prime factors.


3 is prime so there's at least 1 three in the LCM.

4 = 2 * 2 so there must be two 2s in the LCM

5 is prime so the LCM has 1 prime in the LCM

6 = 2*3 and we have everything we need from the other numbers. The LCM is


3*2*2*5 = 60.


Step Two

The first step is to find the largest number closest to 999 that is evenly divided by 60. You'll see why in step 3.


Start backwards from 999 to see which number(s) can be divided by 60 evenly. There must be at least 1.


But which numbers can we choose?

60 = 10*6

The number we want must end in zero if it is to be evenly divided by 60


Our choices are

900 910 920 930 940 950 960 970 980 990


Suppose you think it is 980 If you divide this by 60 you 16.33 and no matter how you juggle that, it will never lead you to a remainder of 2. A sudden inspiration hits you and you try 960.


960 divide by 60 = 16. It's the number you are looking for.


Is there one bigger? You can try 870 980 and 990. They all leave remainders and are not evenly divisible by 60. But 960 is not the answer you want. But it will give it to you.


Step Three

Add 2 onto 960 to give you 962. That number will leave a remainder of 2 when you divide by 3,4,5,6


Step Four

Confirm that 962 leaves a remainder of 2


962 divide by 3 = 320 with a remainder of 2

962 divide by 4 = 240 with a remainder of 2

962 divide by 5 = 192 with a remainder of 2

962 divide by 6 = 160 with a remainder of 2


Comment

How do you know that the next 38 numbers don't contain one that will do the same thing? One of the ways you know is that if you divide by 60, the remainder can be quite large but under 60. For example


990 / 60 = 16 with a remainder of 30.

963 / 60 = 16.05 which translates into 16 with a remainder of 3 [0.05 * 60 = 3 which is what you might expect]. So going from 962 to 999 will give remainders in terms of under 60 but increasing by 1


963 remainder 3

964 remainder 4

965 remainder 5


and so on.


966 gives a remainder of 6. when divided by 60.

6 0
3 years ago
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