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3 years ago

Please help I need to finish this test today

Mathematics

1 answer:

nignag [31]3 years ago

3 0

Answer:

What are the options for the answers?

Step-by-step explanation:

please put the options so I can help you!

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Slove it 3x-5=x+3 :)))))))

lidiya [134]

Answer:

x=4

Slove it 3x-5=x+3

3x-5=x+3

3x-x-5=3

2x-5=3

2x=3+5

2x=8

x=4

5 0

3 years ago

Read 2 more answers

Exercise questions

balu736 [363]

Answer:

a)1

b)1

c)1

Step-by-step explanation:

There's some identity trigonometric equation, which are valid for all angles,and they doesn't depends on the measure of angle!

some of em are follows:. (x is the given angle)

sin(x)^2+cos(x)^2=1
cosec(x)^2=1+cot(x)^2
sec(x)^2=1+tan(x)^2

You can remember these identity, its gonna help alot.

now back to question,. {x is representing angles)

for (a) sin(x)^2+cos(x)^2=1, this is true for all x, dat means that for all the angle given in question(for ,15°,30°,45°,60° and 120°),we will get 1

for(b) ,

cosec(x)^2=1+cot(x)^2

i.e, cosec(x)^2-cot(x)^2=1, again this is true for all x dat means that for all the angle given in question ,we will get 1

for (c),

sec(x)^2=1+tan(x)^2

i.e,sec(x)^2-tan(x)^2=1,again this is true for all x, dat means that for all the angle given in question ,we will get 1

✌️:)

3 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 $n \geq 0$ ,

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

Observe that, because $n \geq 2$ and is an integer,

$n^2(465n-6n^2-10)\geq -1 \iff 465n-6n^2-10 \geq 0 \iff n(465-6n) \geq 10\\\iff 465-6n \geq 0 \iff n \leq \frac{465}{6}=\frac{155}{2}=77.5$

In concusion, the statement is true if and only if n is a non negative integer such that $n\leq 77$

So, 78 is the smallest value of n that does not satisfy the inequality.

Note: If you compute $(4n)^4- \sum^{n}_{i=0} (2i)^4$ for 77 and 78 you will obtain:

$(4n)^4- \sum^{n}_{i=0} (2i)^4=53810064$

$(4n)^4- \sum^{n}_{i=0} (2i)^4=-61754992$

7 0

3 years ago

The function f(x)=x^2 is graphed on the coordinate grid.

Aleksandr [31]

<h3>Answer:</h3>

graphed below.

<h3>Explanation:</h3>

vertex: (0, 0)

other points:

-2 12

-1 3

0 0

1 3

2 12

7 0

2 years ago

Read 2 more answers

PLS HELP WILL GIVE BRANLIEST

BartSMP [9]

The answer is A if I am correct

3 0

3 years ago