All categories

3 years ago

PLEASE HELP ASAP AND THANKS Find x

Mathematics

1 answer:

baherus [9]3 years ago

5 0

Answer:

The value of x is 2.

Step-by-step explanation:

First, you have to make the left side into 1 fraction by making the denorminator the same :

$\frac{x}{x - 2} + \frac{1}{5}$

$= \frac{5x}{5(x - 2)} + \frac{x - 2}{5(x - 2)}$

$= \frac{5x + x - 2}{5(x - 2)}$

$= \frac{6x - 2}{5(x - 2)}$

Then you have to do cross multiplication :

$\frac{6x - 2}{5(x - 2)} = \frac{2}{x - 2}$

$(6x - 2)(x - 2) = 5(2)(x - 2)$

$6 {x}^{2} - 12x - 2x + 4 = 10x - 20$

$6 {x}^{2} - 14x + 4 = 10x - 20$

$6 {x}^{2} - 14x + 4 - 10x + 20 = 0$

$6 {x}^{2} - 24x + 24 = 0$

$6( {x}^{2} - 4x + 4) = 0$

${x}^{2} - 4x + 4 = 0$

${x}^{2} - 2x - 2x + 4 = 0$

$x(x - 2) - 2(x - 2) = 0$

$(x - 2)(x - 2) = 0$

$x = 2$

You might be interested in

A cone is formed from a 300-degree sector of a circle of radius 18 by aligning the two straight sides. what is the result when t

vampirchik [111]

The radius of the completed cone is

... 18×(300°/360°) = 15

The height of the completed cone is found from the Pythagorean theorem:

... 15² + h² = 18²

... h = √99 = 3√11

The volume is given by

... V = (1/3)πr²·h = (1/3)π·15²·3√11

Dividing this value by π gives 225√11 cubic units.

7 0

3 years ago

A department store sold 51 51 shirts one day. All short-sleeved shirts cost $14.00 $ 14.00 each and all long-sleeved shirts cost

Dmitrij [34]

Answer:

28 short-sleeved shirts were sold and 23 long-sleeved shirts were sold.

Step-by-step explanation:

We can solve this question by a system of equations.

I am going to say that:

x is the number of short-sleeved shirts sold.

y is the number of long-sleeved shirts sold.

A department store sold 51 shirts one day.

This means that $x + y = 51$

All short-sleeved shirts cost $14.00 each and all long-sleeved shirts cost $23.00 each. Total receipts for the day were $921.00.

This means that

$14x + 23y = 921$

How many of each kind of shirt were sold?

We have to solve the system of equations.

$x + y = 51$

$x = 51 - y$

--------------

$14x + 23y = 921$

$14(51 - y) + 23y = 921$

$714 - 14y + 23y = 921$

$9y = 207$

$y = \frac{207}{9}$

$y = 23$

--------

$x = 51 - y = 51 - 23 = 28$

28 short-sleeved shirts were sold and 23 long-sleeved shirts were sold.

7 0

3 years ago

HELP!!!!

aniked [119]

Answer:

A. Plane B because it was 9.33 miles away

B. 48 units

Step-by-step explanation:

A. Since the airplanes fly at an angle to the runway, their direction forms a triangle with the runway with their height above the ground as the opposite of the angle and their distance from the airport as the hypotenuse.

So for airplane A with 44° angle of departure,

sin44° = y/h where y = height above the ground and h = distance from airport

So h = y/sin44° = 6/sin44° = 8.64 miles

So for airplane B with 40° angle of departure,

sin40° = y/H where y = height above the ground and H = distance from airport

So H = y/sin40° = 6/sin40° = 9.33 miles

Since airplane B is at 9.33 miles away from the airport whereas airplane A is 8.64 miles from the airport, airplane B is farther away.

B. We know that scale factor = new size/original size

Our scale factor = 4 and original size = 12 units. So,

new size = scale factor original size = 4 × 12 = 48 units.

8 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

Find the value of the expression 2xy^3 if x=3 and y=-2?

Nadusha1986 [10]

So this question basically asks what is 2 x 3 x -2³ and that equals 6 x -8 = -48

6 0

3 years ago