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3 years ago

Which sequence can be generated from the formula f(x + 1) = (f(x))?

Mathematics

2 answers:

drek231 [11]3 years ago

7 0

C ,fam im on the cite and everythanng,

melamori03 [73]3 years ago

7 0

Answer:

It is c on edge nudity

Step-by-step explanation:

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While mining, Jason found a large metal bar that weighed 24 ounces. Jason was also able to determine that the bar had 6 ounces o

tino4ka555 [31]

25% 6 divided by 24 will give you the percentage of copper in the bar

5 0

3 years ago

1.64 as a fraction in simplest form

mrs_skeptik [129]

<span>The correct answer is 1 16/25.

Explanation<span>:
This is read as "one and sixty-four one-hundredths." This means the fraction would be 1 64/100.

Both numerator and denominator are even, so we will divide both by 2 and get 1 32/50.

Again, both are even, so we divide by 2 again and get 1 16/25.</span></span>

4 0

3 years ago

Use mathematical induction to prove the statement is true for all positive integers n, or show why it is false:

kondaur [170]

$\text{Proof by induction:}$
$\text{Test that the statement holds or n = 1}$

$LHS = (3 - 2)^{2} = 1$
$RHS = \frac{6 - 4}{2} = \frac{2}{2} = 1 = LHS$
$\text{Thus, the statement holds for the base case.}$

$\text{Assume the statement holds for some arbitrary term, n= k}$
$1^{2} + 4^{2} + 7^{2} + ... + (3k - 2)^{2} = \frac{k(6k^{2} - 3k - 1)}{2}$

$\text{Prove it is true for n = k + 1}$
$RTP: 1^{2} + 4^{2} + 7^{2} + ... + [3(k + 1) - 2]^{2} = \frac{(k + 1)[6(k + 1)^{2} - 3(k + 1) - 1]}{2} = \frac{(k + 1)[6k^{2} + 9k + 2]}{2}$

$LHS = \underbrace{1^{2} + 4^{2} + 7^{2} + ... + (3k - 2)^{2}}_{\frac{k(6k^{2} - 3k - 1)}{2}} + [3(k + 1) - 2]^{2}$
$= \frac{k(6k^{2} - 3k - 1)}{2} + [3(k + 1) - 2]^{2}$
$= \frac{k(6k^{2} - 3k - 1) + 2[3(k + 1) - 2]^{2}}{2}$
$= \frac{k(6k^{2} - 3k - 1) + 2(3k + 1)^{2}}{2}$
$= \frac{k(6k^{2} - 3k - 1) + 18k^{2} + 12k + 2}{2}$
$= \frac{k(6k^{2} - 3k - 1 + 18k + 12) + 2}{2}$
$= \frac{k(6k^{2} + 15k + 11) + 2}{}$
$= \frac{(k + 1)[6k^{2} + 9k + 2]}{2}$
$= \frac{(k + 1)[6(k + 1)^{2} - 3(k + 1) - 1]}{2}$
$= RHS$

Since it is true for n = 1, n = k, and n = k + 1, by the principles of mathematical induction, it is true for all positive values of n.

3 0

3 years ago

What is in simplest form?

viktelen [127]

Answer: 3/5 i think ;-; if its not well then uhhhhh sorry?

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

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Charra [1.4K]

Answer: