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xeze [42]
2 years ago
7

If 2x^2-5x+7 is subtracted from x^2+2x-11, what is the leading coefficient?

Mathematics
1 answer:
guajiro [1.7K]2 years ago
3 0

Answer:

-1

Step-by-step explanation:

2x² - 5x + 7 is subtracted <em>from </em>x² + 2x - 11, so

(x² + 2x - 11) - (2x² - 5x + 7)

x² + 2x - 11 - 2x² + 5x -7 (Distribute the negative)

-x² + 7x - 18 (Combine like terms)

The leading coefficient is the coefficient of the first term, so it is -1.

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Let a1, a2, a3, ... be a sequence of positive integers in arithmetic progression with common difference
Bezzdna [24]

Since a_1,a_2,a_3,\cdots are in arithmetic progression,

a_2 = a_1 + 2

a_3 = a_2 + 2 = a_1 + 2\cdot2

a_4 = a_3+2 = a_1+3\cdot2

\cdots \implies a_n = a_1 + 2(n-1)

and since b_1,b_2,b_3,\cdots are in geometric progression,

b_2 = 2b_1

b_3=2b_2 = 2^2 b_1

b_4=2b_3=2^3b_1

\cdots\implies b_n=2^{n-1}b_1

Recall that

\displaystyle \sum_{k=1}^n 1 = \underbrace{1+1+1+\cdots+1}_{n\,\rm times} = n

\displaystyle \sum_{k=1}^n k = 1 + 2 + 3 + \cdots + n = \frac{n(n+1)}2

It follows that

a_1 + a_2 + \cdots + a_n = \displaystyle \sum_{k=1}^n (a_1 + 2(k-1)) \\\\ ~~~~~~~~ = a_1 \sum_{k=1}^n 1 + 2 \sum_{k=1}^n (k-1) \\\\ ~~~~~~~~ = a_1 n +  n(n-1)

so the left side is

2(a_1+a_2+\cdots+a_n) = 2c n + 2n(n-1) = 2n^2 + 2(c-1)n

Also recall that

\displaystyle \sum_{k=1}^n ar^{k-1} = \frac{a(1-r^n)}{1-r}

so that the right side is

b_1 + b_2 + \cdots + b_n = \displaystyle \sum_{k=1}^n 2^{k-1}b_1 = c(2^n-1)

Solve for c.

2n^2 + 2(c-1)n = c(2^n-1) \implies c = \dfrac{2n^2 - 2n}{2^n - 2n - 1} = \dfrac{2n(n-1)}{2^n - 2n - 1}

Now, the numerator increases more slowly than the denominator, since

\dfrac{d}{dn}(2n(n-1)) = 4n - 2

\dfrac{d}{dn} (2^n-2n-1) = \ln(2)\cdot2^n - 2

and for n\ge5,

2^n > \dfrac4{\ln(2)} n \implies \ln(2)\cdot2^n - 2 > 4n - 2

This means we only need to check if the claim is true for any n\in\{1,2,3,4\}.

n=1 doesn't work, since that makes c=0.

If n=2, then

c = \dfrac{4}{2^2 - 4 - 1} = \dfrac4{-1} = -4 < 0

If n=3, then

c = \dfrac{12}{2^3 - 6 - 1} = 12

If n=4, then

c = \dfrac{24}{2^4 - 8 - 1} = \dfrac{24}7 \not\in\Bbb N

There is only one value for which the claim is true, c=12.

3 0
2 years ago
Can an object be in pure translation as well as in pure rotation simultaneously?
Inessa [10]
  <span>"Pure translational" motion means the object is moving from place to place but isn't rotating at all. Mathematically, that means every point on the object has the same velocity vector as every other point on the object. For example, say you have a box that's sliding along the ground, and say each corner on the box is moving at 15 feet per second at an angle of 26 degrees west of north. Since every point on the box has the same velocity vector, the motion is "pure translational." 

"Pure rotational" motion means the object is rotating, but its position in space isn't changing. Mathematically, that means that every point on the object moves in a circle around some axis (usually through the object's center of mass).

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
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5 0
3 years ago
A stock lost 7 1/8 points on Monday and then another 1 5/8 points on Tuesday. On Wednesday, it gained 13 points. What was the ne
Fudgin [204]

Answer:

It  was gained 4 1/4 points.

Step-by-step explanation:

- 7 1/8 - 1 5/8 + 13 = - 8 6/8 + 13 = - 8 3/4 + 13 = - 8 - 3/4 + 12 + 4/4 =  4 1/4

3 0
3 years ago
Can you please help thanks
Svetach [21]
I'll answer and write my work out -- I will post a link to the picture. It won't be in a diagram but it will tell you the answer. If you are on mobile you should log onto brainly on a computer so you can copy-paste the link into google
4 0
3 years ago
What 6,589 to the nearest thousand
raketka [301]

6,589 to the nearest thousand would be <u>7,000</u> since you have high numbers as 5,8, and 9 that allows you to round up your answer to 7,000.

6 0
3 years ago
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