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3 years ago

the length of marshall's rectangular poster is 2 tomes its width. if the perimeter is 24 inches what is the area of the poster

1 answer:

5 0

The answer is 24. There is 2 width sides and 2 length sides. Let's say the width measurement is x. The 2 sides for width equals 2x. The 2 length sides equals 4x because there twice has long has the width which means that each length side is 2x. 2x plus 4x equals 6x. 6x=24 You divided 24 with 6 which gives you 4. That means that both width sides are 4 units long and the length sides are 8 units long. 4+4+8+8=24, 4 times 8=24

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Find the equation of ellipse passing throgh (1,4) and (-3,2)

irinina [24]

Answer:

$\displaystyle \frac{ {3x}^{2} }{ 35 } + \frac{{2y}^{2} }{ 35 } = 1$

Step-by-step explanation:

we want to figure out the ellipse equation which passes through (1,4) and (-3,2)

the standard form of ellipse equation is given by:

$\displaystyle \frac{(x - h {)}^{2} }{ {a}^{2} } + \frac{(y - k {)}^{2} }{ {b}^{2} } = 1$

where:

(h,k) is the centre
a is the horizontal redius
b is the vertical radius

since the centre of the equation is not mentioned, we'd assume it (0,0) therefore our equation will be:

$\displaystyle \frac{ {x}^{2} }{ {a}^{2} } + \frac{{y}^{2} }{ {b}^{2} } = 1$

substituting the value of x and y from the point (1,4),we'd acquire:

$\displaystyle \frac{ 1}{ {a}^{2} } + \frac{16}{ {b}^{2} } = 1$

similarly using the point (-3,2), we'd obtain:

$\displaystyle \frac{ 9}{ {a}^{2} } + \frac{4 }{ {b}^{2} } = 1$

let 1/a² and 1/b² be q and p respectively and transform the equation:

$\displaystyle \begin{cases} q + 16p = 1 \\ 9q + 4p = 1 \end{cases}$

solving the system of linear equation will yield:

$\displaystyle \begin{cases} q = \dfrac{3}{35} \\ \\ p = \dfrac{2}{35} \end{cases}$

substitute back:

$\displaystyle \begin{cases} \dfrac{1}{ {a}^{2} } = \dfrac{3}{35} \\ \\ \dfrac{1}{ {b}^{2} } = \dfrac{2}{35} \end{cases}$

divide both equation by 1 which yields:

$\displaystyle \begin{cases} {a}^{2} = \dfrac{35}{ 3} \\ \\ {b}^{2} = \dfrac{35}{2} \end{cases}$

substitute the value of a² and b² in the ellipse equation , thus:

$\displaystyle \frac{ {x}^{2} }{ \dfrac{35}{3} } + \frac{{y}^{2} }{ \dfrac{35}{2} } = 1$

simplify complex fraction:

$\displaystyle \frac{ {3x}^{2} }{ 35 } + \frac{{2y}^{2} }{ 35 } = 1$

and we're done!

(refer the attachment as well)

8 0

3 years ago

How do u combine like terms ?

Kisachek [45]

We can use #11 as an ex.

7ab and 4 ab

So we combine 7 and 4 (multiply)

=28

Then combine "a". On both sides they are 2 a"s so it would be a^2 and same for b . b^2

Answer: 28a^2b^2.

hope it helps*

8 0

4 years ago

What is the value of X? PLEASE HELP

GREYUIT [131]

Answer:

Step-by-step explanation:

19=x-4

x=23

3 0

3 years ago

Tyshaun buys music online. Each song costs $2 plus a $5 fee for using the artist’s site. He has no more than $40 to spend. Which

HACTEHA [7]

Answer:

I dont see options, but the answer should be: X(2+5)≤40

Step-by-step explanation:

3 0

2 years ago

<img src="https://tex.z-dn.net/?f=27%3D%281%2F3%29x%2F2%5C%5C" id="TexFormula1" title="27=(1/3)x/2\\" alt="27=(1/3)x/2\\" align=

ICE Princess25 [194]

Answer:

x = 162

Step-by-step explanation:

The equation editor lets you properly write exponents, so we have to assume that your equation is supposed to be equivalent to ...

27 = x/6

27·6 = x = 162 . . . . multiply by 6

_____

On the off chance you intend ...

$27=\left(\dfrac{1}{3}\right)^{\dfrac{x}{2}}$

you can solve this by taking logarithms and dividing by the coefficient of x. Using base 3 for the logarithm may be easiest.

$\log_3{(27)}=\dfrac{x}{2}\log_3{(3^{-1})}\\\\3=-\dfrac{x}{2} \qquad\text{simplify}\\\\-6=x \qquad\text{multiply by -2}$

This version of the equation has solution x = -6.

4 0

3 years ago