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o-na [289]
3 years ago
6

Determine the quadrant in which the given angle terminates with negative numbers

Mathematics
1 answer:
Pavel [41]3 years ago
6 0
If both x and y are negative, your answer would be the 3rd Quadrant. Hope I could answer your question correctly!
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Is 9 a factor of 54x + 63​
Zolol [24]

Answer:

yes

Step-by-step explanation:

It is because 9 divides into 54 and 63

7 0
3 years ago
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How do u write a expression
zhuklara [117]

Answer:

<h2><u>it depends on what expression u want </u></h2>

Step-by-step explanation:

An example of a mathematical expression with a variable is 2x + 3. All variables must have a coefficient, a number that is multiplied by the variable. In the expression 2x + 3, the coefficient of x is the number 2, and it means 2 times x plus 3.

3 0
3 years ago
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Find the zeros of polynomial function and solve polynomials equations.<br> f(x)=x^2-x-90
victus00 [196]

The zeros are:

10 and -9

(ten and negative-nine)

6 0
3 years ago
Prove for any positive integer n, n^3 +11n is a multiple of 6
suter [353]

There are probably other ways to approach this, but I'll focus on a proof by induction.

The base case is that n = 1. Plugging this into the expression gets us

n^3+11n = 1^3+11(1) = 1+11 = 12

which is a multiple of 6. So that takes care of the base case.

----------------------------------

Now for the inductive step, which is often a tricky thing to grasp if you're not used to it. I recommend keeping at practice to get better familiar with these types of proofs.

The idea is this: assume that k^3+11k is a multiple of 6 for some integer k > 1

Based on that assumption, we need to prove that (k+1)^3+11(k+1) is also a multiple of 6. Note how I've replaced every k with k+1. This is the next value up after k.

If we can show that the (k+1)th case works, based on the assumption, then we've effectively wrapped up the inductive proof. Think of it like a chain of dominoes. One knocks over the other to take care of every case (aka every positive integer n)

-----------------------------------

Let's do a bit of algebra to say

(k+1)^3+11(k+1)

(k^3+3k^2+3k+1) + 11(k+1)

k^3+3k^2+3k+1+11k+11

(k^3+11k) + (3k^2+3k+12)

(k^3+11k) + 3(k^2+k+4)

At this point, we have the k^3+11k as the first group while we have 3(k^2+k+4) as the second group. We already know that k^3+11k is a multiple of 6, so we don't need to worry about it. We just need to show that 3(k^2+k+4) is also a multiple of 6. This means we need to show k^2+k+4 is a multiple of 2, i.e. it's even.

------------------------------------

If k is even, then k = 2m for some integer m

That means k^2+k+4 = (2m)^2+(2m)+4 = 4m^2+2m+4 = 2(m^2+m+2)

We can see that if k is even, then k^2+k+4 is also even.

If k is odd, then k = 2m+1 and

k^2+k+4 = (2m+1)^2+(2m+1)+4 = 4m^2+4m+1+2m+1+4 = 2(2m^2+3m+3)

That shows k^2+k+4 is even when k is odd.

-------------------------------------

In short, the last section shows that k^2+k+4 is always even for any integer

That then points to 3(k^2+k+4) being a multiple of 6

Which then further points to (k^3+11k) + 3(k^2+k+4) being a multiple of 6

It's a lot of work, but we've shown that (k+1)^3+11(k+1) is a multiple of 6 based on the assumption that k^3+11k is a multiple of 6.

This concludes the inductive step and overall the proof is done by this point.

6 0
3 years ago
Read 2 more answers
Write an equation of the line that passes through (1,2) and is parallel to the line y=-5x+4
valentinak56 [21]

Answer:

The equation of the Parallel line to the given line is  5x+y-7=0

Step-by-step explanation:

<u><em>Explanation:-</em></u>

Given that the line y = - 5x+4 and point (1,2)

The equation of the Parallel line to the given line is  ax+by+k=0

Given straight line  5x + y -4 =0

The equation of the Parallel line to the given line is  5x+y+k=0

This line passes through the point (1,2)

             5x+y+k=0

            5(1)+2+k=0

⇒              7+k=0

                   k =-7

<u>Final answer:-</u>

The equation of the Parallel line to the given line is  5x+y-7=0

7 0
2 years ago
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