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3 years ago

Are the two triangles simliar? How do you know?

Mathematics

1 answer:

lesya [120]3 years ago

7 0

The triangles could be similar by both having right angles, same sides, measurements, ect.

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Please help math experts need your help 15 points !

Alinara [238K]

Answer:

12 = -3 - 3x

Step-by-step explanation:

Same concept as the last question I answered. First, let's derive a formula from the given model. So we have 3 negative x and 3 negative 1 which equal to 12 positive ones, this can be written as:

-3x - 3 = 12

which can be rewritten as

12 = -3 - 3x

7 0

3 years ago

Juan pulled a red card from the deck of regular playing cards. This probability is 2652 or 12 . He puts the card back into the d

kenny6666 [7]

Answer:

Step-by-step explanation:

if he randomly pulls it out it's still the same number of red cards and also the total number of cards as always

can't change

6 0

3 years ago

Read 2 more answers

Which is the same as moving the decimal point 3 places to the left in a decimal

Ludmilka [50]

Multiplying the number by 1,000

5 0

3 years ago

Prove that

Pani-rosa [81]

Let's start from what we know.

$(1)\qquad\sum\limits_{k=1}^n1=\underbrace{1+1+\ldots+1}_{n}=n\cdot 1=n\\\\\\
(2)\qquad\sum\limits_{k=1}^nk=1+2+3+\ldots+n=\dfrac{n(n+1)}{2}\quad\text{(arithmetic series)}\\\\\\
(3)\qquad\sum\limits_{k=1}^nk\ \textgreater \ 0\quad\implies\quad\left|\sum\limits_{k=1}^nk\right|=\sum\limits_{k=1}^nk$

Note that:

$\sum\limits_{k=1}^n(-1)^k\cdot k^2=(-1)^1\cdot1^2+(-1)^2\cdot2^2+(-1)^3\cdot3^2+\dots+(-1)^n\cdot n^2=\\\\\\=-1^2+2^2-3^2+4^2-5^2+\dots\pm n^2$

(sign of last term will be + when n is even and - when n is odd).
Sum is finite so we can split it into two sums, first $S_n^+$ with only positive trems (squares of even numbers) and second $S_n^-$ with negative (squares of odd numbers). So:

$\sum\limits_{k=1}^n(-1)^k\cdot k^2=S_n^+-S_n^-$

And now the proof.

1) n is even.

In this case, both $S_n^+$ and $S_n^-$ have $\dfrac{n}{2}$ terms. For example if n=8 then:

$S_8^+=\underbrace{2^2+4^2+6^2+8^2}_{\frac{8}{2}=4}\qquad\text{(even numbers)}\\\\\\
S_8^-=\underbrace{1^2+3^2+5^2+7^2}_{\frac{8}{2}=4}\qquad\text{(odd numbers)}\\\\\\$

Generally, there will be:

$S_n^+=\sum\limits_{k=1}^\frac{n}{2}(2k)^2\\\\\\S_n^-=\sum\limits_{k=1}^\frac{n}{2}(2k-1)^2\\\\\\$

Now, calculate our sum:

$\left|\sum\limits_{k=1}^n(-1)^k\cdot k^2\right|=\left|S_n^+-S_n^-\right|=
\left|\sum\limits_{k=1}^\frac{n}{2}(2k)^2-\sum\limits_{k=1}^\frac{n}{2}(2k-1)^2\right|=\\\\\\=
\left|\sum\limits_{k=1}^\frac{n}{2}4k^2-\sum\limits_{k=1}^\frac{n}{2}\left(4k^2-4k+1\right)\right|=\\\\\\$

$=\left|4\sum\limits_{k=1}^\frac{n}{2}k^2-4\sum\limits_{k=1}^\frac{n}{2}k^2+4\sum\limits_{k=1}^\frac{n}{2}k-\sum\limits_{k=1}^\frac{n}{2}1\right|=\left|4\sum\limits_{k=1}^\frac{n}{2}k-\sum\limits_{k=1}^\frac{n}{2}1\right|\stackrel{(1),(2)}{=}\\\\\\=
\left|4\dfrac{\frac{n}{2}(\frac{n}{2}+1)}{2}-\dfrac{n}{2}\right|=\left|2\cdot\dfrac{n}{2}\left(\dfrac{n}{2}+1\right)-\dfrac{n}{2}\right|=\left|n\left(\dfrac{n}{2}+1\right)-\dfrac{n}{2}\right|=\\\\\\
$

$=\left|\dfrac{n^2}{2}+n-\dfrac{n}{2}\right|=\left|\dfrac{n^2}{2}+\dfrac{n}{2}\right|=\left|\dfrac{n^2+n}{2}\right|=\left|\dfrac{n(n+1)}{2}\right|\stackrel{(2)}{=}\\\\\\\stackrel{(2)}{=}
\left|\sum\limits_{k=1}^nk\right|\stackrel{(3)}{=}\sum\limits_{k=1}^nk$

So in this case we prove, that:

$\left|\sum\limits_{k=1}^n(-1)^k\cdot k^2\right|=\sum\limits_{k=1}^nk$

2) n is odd.

Here, $S_n^-$ has more terms than $S_n^+$ . For example if n=7 then:

$S_7^-=\underbrace{1^2+3^2+5^2+7^2}_{\frac{n+1}{2}=\frac{7+1}{2}=4}\\\\\\
S_7^+=\underbrace{2^2+4^4+6^2}_{\frac{n+1}{2}-1=\frac{7+1}{2}-1=3}\\\\\\$

So there is $\dfrac{n+1}{2}$ terms in $S_n^-$ , $\dfrac{n+1}{2}-1$ terms in $S_n^+$ and:

$S_n^+=\sum\limits_{k=1}^{\frac{n+1}{2}-1}(2k)^2\\\\\\
S_n^-=\sum\limits_{k=1}^{\frac{n+1}{2}}(2k-1)^2$

Now, we can calculate our sum:

$\left|\sum\limits_{k=1}^n(-1)^k\cdot k^2\right|=\left|S_n^+-S_n^-\right|=
\left|\sum\limits_{k=1}^{\frac{n+1}{2}-1}(2k)^2-\sum\limits_{k=1}^{\frac{n+1}{2}}(2k-1)^2\right|=\\\\\\=
\left|\sum\limits_{k=1}^{\frac{n+1}{2}-1}4k^2-\sum\limits_{k=1}^{\frac{n+1}{2}}\left(4k^2-4k+1\right)\right|=\\\\\\=
\left|\sum\limits_{k=1}^{\frac{n-1}{2}-1}4k^2-\sum\limits_{k=1}^{\frac{n+1}{2}}4k^2+\sum\limits_{k=1}^{\frac{n+1}{2}}4k-\sum\limits_{k=1}^{\frac{n+1}{2}}1\right|=\\\\\\$

$=\left|\sum\limits_{k=1}^{\frac{n-1}{2}-1}4k^2-\sum\limits_{k=1}^{\frac{n+1}{2}-1}4k^2-4\left(\dfrac{n+1}{2}\right)^2+\sum\limits_{k=1}^{\frac{n+1}{2}}4k-\sum\limits_{k=1}^{\frac{n+1}{2}}1\right|=\\\\\\=
\left|-4\left(\dfrac{n+1}{2}\right)^2+4\sum\limits_{k=1}^{\frac{n+1}{2}}k-\sum\limits_{k=1}^{\frac{n+1}{2}}1\right|\stackrel{(1),(2)}{=}\\\\\\
\stackrel{(1),(2)}{=}\left|-4\dfrac{n^2+2n+1}{4}+4\dfrac{\frac{n+1}{2}\left(\frac{n+1}{2}+1\right)}{2}-\dfrac{n+1}{2}\right|=\\\\\\$

$=\left|-n^2-2n-1+2\cdot\dfrac{n+1}{2}\left(\dfrac{n+1}{2}+1\right)-\dfrac{n+1}{2}\right|=\\\\\\=
\left|-n^2-2n-1+(n+1)\left(\dfrac{n+1}{2}+1\right)-\dfrac{n+1}{2}\right|=\\\\\\=
\left|-n^2-2n-1+\dfrac{(n+1)^2}{2}+n+1-\dfrac{n+1}{2}\right|=\\\\\\=
\left|-n^2-n+\dfrac{n^2+2n+1}{2}-\dfrac{n+1}{2}\right|=\\\\\\=
\left|-n^2-n+\dfrac{n^2}{2}+n+\dfrac{1}{2}-\dfrac{n}{2}-\dfrac{1}{2}\right|=\left|-\dfrac{n^2}{2}-\dfrac{n}{2}\right|=\left|-\dfrac{n^2+n}{2}\right|=\\\\\\$

$=\left|-\dfrac{n(n+1)}{2}\right|=|-1|\cdot\left|\dfrac{n(n+1)}{2}\right|=\left|\dfrac{n(n+1)}{2}\right|\stackrel{(2)}{=}\left|\sum\limits_{k=1}^nk\right|\stackrel{(3)}{=}\sum\limits_{k=1}^nk$

We consider all possible n so we prove that:

$\forall_{n\in\mathbb{N}}\quad\left|\sum\limits_{k=1}^n(-1)^k\cdot k^2\right|=\sum\limits_{k=1}^nk$

7 0

3 years ago

Simplify. (3x^2 + 3) - (6x + 4) + (3x^2 - 5x)

r-ruslan [8.4K]

Answer:

6x^2 -11x -1

Step-by-step explanation:

(3x^2 + 3) - (6x + 4) + (3x^2 - 5x)

Distribute the minus sign

(3x^2 + 3) - 6x - 4) + (3x^2 - 5x)

Combine like terms

3x^2 + 3x^2 - 6x - 5x -4+3

6x^2 -11x -1

3 0

3 years ago