Which statement best describes what happened when h is changed from 4 to

If i yeet all my yeet bagels then go to pick them up again and yeetuis steals them all how many yeet knifes do i have to throw a

sukhopar [10]

at least 50 knives, plus a yeet to get to him :>

5 0

3 years ago

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If f(1) = 10 and f '(x) ≥ 3 for 1 ≤ x ≤ 6, how small can f(6) possibly be?

Natalka [10]

Answer:

$f'( x ) = \frac{f(b) - f(a)}{b - a} \\ = \frac{f(6) -f(a)}{6 - 1} \\ 3\leqslant \frac{f(6) - 10}{5} \\ 15\leqslant f(6) - 10 \\ f(6) \geqslant 15 + 10 \\ f(6) \geqslant 25$

6 0

3 years ago

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If you are given a3=2 a5=16, find a100.

ra1l [238]

I suppose $a_n$ denotes the n-th term of some sequence, and we're given the 3rd and 5th terms $a_3=2$ and $a_5=16$ . On this information alone, it's impossible to determine the 100th term $a_{100}$ because there are infinitely many sequences where 2 and 16 are the 3rd and 5th terms.

To get around that, I'll offer two plausible solutions based on different assumptions. So bear in mind that this is not a complete answer, and indeed may not even be applicable.

• Assumption 1: the sequence is arithmetic (a.k.a. linear)

In this case, consecutive terms differ by a constant d, or

$a_n = a_{n-1} + d$

By this relation,

$a_{n-1} = a_{n-2} + d$

and by substitution,

$a_n = (a_{n-2} + d) + d = a_{n-2} + 2d$

We can continue in this fashion to get

$a_n = a_{n-3} + 3d$

$a_n = a_{n-4} + 4d$

and so on, down to writing the n-th term in terms of the first as

$a_n = a_1 + (n-1)d$

Now, with the given known values, we have

$a_3 = a_1 + 2d = 2$

$a_5 = a_1 + 4d = 16$

Eliminate $a_1$ to solve for d :

$(a_1 + 4d) - (a_1 + 2d) = 16 - 2 \implies 2d = 14 \implies d = 7$

Find the first term $a_1$ :

$a_1 + 2\times7 = 2 \implies a_1 = 2 - 14 = -12$

Then the 100th term in the sequence is

$a_{100} = a_1 + 99d = -12 + 99\times7 = \boxed{681}$

• Assumption 2: the sequence is geometric

In this case, the ratio of consecutive terms is a constant r such that

$a_n = r a_{n-1}$

We can solve for $a_n$ in terms of $a_1$ like we did in the arithmetic case.

$a_{n-1} = ra_{n-2} \implies a_n = r\left(ra_{n-2}\right) = r^2 a_{n-2}$

and so on down to

$a_n = r^{n-1} a_1$

Now,

$a_3 = r^2 a_1 = 2$

$a_5 = r^4 a_1 = 16$

Eliminate $a_1$ and solve for r by dividing

$\dfrac{a_5}{a_3} = \dfrac{r^4a_1}{r^2a_1} = \dfrac{16}2 \implies r^2 = 8 \implies r = 2\sqrt2$

Solve for $a_1$ :

$r^2 a_1 = 8a_1 = 2 \implies a_1 = \dfrac14$

Then the 100th term is

$a_{100} = \dfrac{(2\sqrt2)^{99}}4 = \boxed{\dfrac{\sqrt{8^{99}}}4}$

The arithmetic case seems more likely since the final answer is a simple integer, but that's just my opinion...

3 0

2 years ago

How do I move one side to the other: <img src="https://tex.z-dn.net/?f=sin%5E2%5Calpha%20-%20cos%5E2%5Calpha%20%3D%202si

Vera_Pavlovna [14]

9514 1404 393

Explanation:

You can use the addition property of equality to "move" a term from one side of the equation to the other. For example, if you want to move the cos²α term, you can add cos²α to both sides of the equation:

sin²α -cos²α +cos²α = 2sin²α -1 +cos²α

When this is simplified, it becomes ...

sin²α = 2sin²α +cos²α -1 . . . . . . the cos²α term is gone from the left

__

The equation you have is an identity. The left and right sides are equal for any value of α. When you have such an equation as a trig problem, you are often being asked to prove it is true. The way you do that is to make use of other trig identities to transform one side so that it matches the other side.

Here, you can use the trig identity ...

sin²α +cos²α = 1

If you use this to substitute for 1 on the right, you have ...

sin²α - cos²α = 2sin²α - (sin²α +cos²α)

Now, when you collect terms, you get ...

sin²α - cos²α = 2sin²α - sin²α -cos²α . . . eliminate parentheses

sin²α - cos²α = sin²α - cos²α . . . . . proof of your identity

7 0

3 years ago

Guess my profile pictures character for brainliest

Julli [10]

Answer:

hinata hyuge

Step-by-step explanation:

4 0

2 years ago

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Which statement best describes what happened when h is changed from 4 to - 2.