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3 years ago

The length of a rectangle is 4 m more than the width. The are is 30 m ^2. find the width and length.

Mathematics

1 answer:

zlopas [31]3 years ago

6 0

Answer:

Step-by-step explanation:

AREA FORMULA: LENGTH * WIDTH

L*W=30

L=W+4 thus

(W+4)W=30

W^2+4W=30

W^2+4W-30=0

USING THE QUADRATIC EQUATION WE GET

X=(-B+-SQRT[B^2-4QC])/2A

X=(-4+-SQRT16-4*1*-30])/2*1

X=(-4+-SQRT16+120])/2

X=(-4+-SQRT136)/2

X=(-4+-11.66)/2

X=(-4+11.66)/2

X=7.66/2

X=3.83M IS THE WIDTH.

THUS THE LENGTH IS

L*3.83=30

L=30/3.83

L=7.83M IS THE LENGTH

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Tickets at a particular movie theater have different rates for adults and children. On Friday, the theater sold 4 adult tickets

Thepotemich [5.8K]

Answer: The price for the adult tickets is $12 and the price for the child ticket is $5.

Step-by-step explanation:

Let the price for the adult ticket be x

Let the price for the child ticket be y

According to question, On Friday ,

$4x+7y=83$

and on next day,

$5x+6y=90$

Now, we will use "Substitution Method" to solve the system of equations :

$5x+6y=90\\\\5x=90-6y\\\\x=\frac{90-6y}{5}$

so, we will put the value of x in the first equation i.e.

$4x+7y=83\\\\4(\frac{90-6y}{5})+7y=83\\\\360-24y+35y=83\times 5\\\\35y-24y=415-360\\\\11y=55\\\\y=\frac{55}{11}\\\\y=5$

Now, put the value of y in the equation that is given by

$x=\frac{90-6y}{5}\\\\x=\frac{90-6\times 5}{5}\\\\x=\frac{90-30}{5}\\\\x=\frac{60}{5}\\\\x=\$12$

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3 years ago

Explain how to get that answer!!

ra1l [238]

We need to simplify $\frac{ \sqrt{14x^3} }{ \sqrt{18x} }$

First lets factor $\sqrt{14x^3}$

$\sqrt{14x^3}$ = $\sqrt{14} \sqrt{x^3}$
$\sqrt{14} = \sqrt{2} \sqrt{7}$ by applying the radical rule $\sqrt[n]{ab} = \sqrt[n]{a} \sqrt[n]{b}$
$\sqrt{x^3} = x^{3/2}$ By applying the radical rule $\sqrt[n]{x^m} = x^{m/n}$

So
$\sqrt{14x^3}$ = $\sqrt{14} \sqrt{x^3}$ = $\sqrt{2} \sqrt{7}x^{3/2}$

Now let's factor $\sqrt{18x}$
By applying the radical rule $\sqrt[n]{ab} = \sqrt[n]{a} \sqrt[n]{b}$ ,
$\sqrt{18x} = \sqrt{18} \sqrt{x}$
$\sqrt{18} = \sqrt{2} * 3$

So $\sqrt{18x}$ = $\sqrt{2}*3 \sqrt{x}$

So $\frac{ \sqrt{14x^3} }{ \sqrt{18x} }$ = $\frac{ \sqrt{2} \sqrt{7} x^{3/2} }{ \sqrt{2}*3 \sqrt{x} }$

We know that $\sqrt[n]{x} = x^{1/n}$ so $\sqrt{x} = x^{1/2}$

We now have $\frac{ \sqrt{2} \sqrt{7} x^{3/2} }{ \sqrt{2}*3 \sqrt{x}} = \frac{ \sqrt{2} \sqrt{7} x^{3/2} }{ \sqrt{2}*3x^{1/2}}$

We know that $\frac{x^a}{x^b} = x^{a-b}$
So $\frac{x^{3/2}}{x^{1/2}} = x^{3/2 - 1/2} = x$

We now got $\frac{ \sqrt{2} \sqrt{7} x^{3/2} }{ \sqrt{2}*3x^{1/2}} = \frac{ \sqrt{2} \sqrt{7} x }{ \sqrt{2}*3}
$

We can notice that the numerator and the denominator both got √2 in a multiplication, so we can simplify them, and we get:
$\frac{ \sqrt{2} \sqrt{7} x }{ \sqrt{2}*3} = \frac{ \sqrt{7}x }{3}$

All in All, we get $\frac{ \sqrt{14x^3} }{ \sqrt{18x} }$ = $\frac{ \sqrt{7}x }{3}$

Hope this helps! :D

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