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3 years ago

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Can someone help with 5 please?

1 answer:

borishaifa [10]3 years ago

4 0

The answer should be 8x^3-36x^2+54x-27
To get this you have to do (2x-3)(2x-3)(2x-3) and distribute through each parenthesis. Hope this helps

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2 yd<br> 3.1 yd <br> find the area

STALIN [3.7K]

Answer:

Area = 6.2 yards²

Step-by-step explanation:

area = bh

area = 2(3.1)

3 0

3 years ago

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I need help with this problem:)

mixer [17]

Answer:

11.5 for first question

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2 years ago

Angle x and y are complementary. Angle x is supplementary to a 128º angle.

tiny-mole [99]

Answer:

x = 52

y = 38

Step-by-step explanation:

Angle x is supplementary to a 128º angle.

Supplementary angles add to 180

x+128 = 180

Subtract 128 from each side

x+128-128 = 180-128

x =52

Angle x and y are complementary.

Complementary angles add to 90

x+y =90

52+y =90

Subtract 52 from each side

52-52 +y = 90-52

y = 38

4 0

3 years ago

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0.85 3/5 0.15 7/10 in order from least to greatest

nydimaria [60]

0.85=0.85

3/5=0.60

0.15=0.15

7/10=0.70

L-G

0.15, 3/5, 7/10, 0.85

7 0

2 years ago

Read 2 more answers

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