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2 years ago

How is your day? My day is going great.

Mathematics

2 answers:

diamong [38]2 years ago

5 0

Answer:

Im having a good day, im glad yours is good aswell!

Step-by-step explanation:

dalvyx [7]2 years ago

4 0

Answer:

Eh it sucked but im glad you had a good day :)

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Rewrite the expression with rational exponents as a radical expression by extending the properties of Internet exponents

Mekhanik [1.2K]

Answer:

option a

Step-by-step explanation:

y^7/8 / y^1/4

First we will accommodate the fractions

1/4 = 2/8

y^7/8 / y^2/8

Then we will subtract the exponents of the powers since they have the same base and are dividing

7/8 - 2/8 = 5/8

y^5/8

Finally we pass the 8 as radical root

8√y^5

4 0

3 years ago

Evaluate the expression!

Fudgin [204]

Answer:

poop

Step-by-step explanation:

5 0

3 years ago

which transformation causes the described change in the graph of the function y = cos x? the transformation results in a horizon

luda_lava [24]

The transformations that can occur to the graph of the function y = cos x that will exhibit changes would be changes to the angle, or changes to the coefficient. The transformations can be viewed as follows:

y = cos x transforms to y = cos (kx)

k > 1 ; a horizontal shrink occurs
0 < k < 1 ; a horizontal stretch occurs

y = cos x transforms to y = A cos x

|A| > 1 ; a vertical stretch occurs
|A| < 1 ; a vertical shrink occurs

5 0

3 years ago

Prove or disprove (from i=0 to n) sum([2i]^4) <= (4n)^4. If true use induction, else give the smallest value of n that it doe

ddd [48]

Answer:

The statement is true for every n between 0 and 77 and it is false for $n\geq 78$

Step-by-step explanation:

First, observe that, for n=0 and n=1 the statement is true:

For n=0: $\sum^{n}_{i=0} (2i)^4=0 \leq 0=(4n)^4$

For n=1: $\sum^{n}_{i=0} (2i)^4=16 \leq 256=(4n)^4$

From this point we will assume that $n\geq 2$

As we can see, $\sum^{n}_{i=0} (2i)^4=\sum^{n}_{i=0} 16i^4=16\sum^{n}_{i=0} i^4$ and $(4n)^4=256n^4$ . Then,

$\sum^{n}_{i=0} (2i)^4 \leq(4n)^4 \iff \sum^{n}_{i=0} i^4 \leq 16n^4$

Now, we will use the formula for the sum of the first 4th powers:

$\sum^{n}_{i=0} i^4=\frac{n^5}{5} +\frac{n^4}{2} +\frac{n^3}{3}-\frac{n}{30}=\frac{6n^5+15n^4+10n^3-n}{30}$

Therefore:

$\sum^{n}_{i=0} i^4 \leq 16n^4 \iff \frac{6n^5+15n^4+10n^3-n}{30} \leq 16n^4 \\\\ \iff 6n^5+10n^3-n \leq 465n^4 \iff 465n^4-6n^5-10n^3+n\geq 0$

and, because $n \geq 0$ ,

$465n^4-6n^5-10n^3+n\geq 0 \iff n(465n^3-6n^4-10n^2+1)\geq 0 \\\iff 465n^3-6n^4-10n^2+1\geq 0 \iff 465n^3-6n^4-10n^2\geq -1\\\iff n^2(465n-6n^2-10)\geq -1$