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3 years ago

14

The length of a rectangle is 3 times its width. The rectangle’s width is 4 m. What is the area of the rectangle? Enter your answ

er in the box. m²

2 answers:

Art [367]3 years ago

7 0

Answer:

48m²

Step-by-step explanation:

L= 3*4

W= 4

Area= L*W

A=(3*4)*4

A=12*4

A=48

bekas [8.4K]3 years ago

5 0

Call the length $l$ and the width $w$

We are told:
1. "<span>The length of a rectangle is 3 times its width." i.e.: $l=3w$
2. "</span><span>The rectangle’s width is 4 m." i.e.: $w=4$

So, by substitution we have:
$l=3w$
$l=3(4)$
$l=12m$

Then we can use our area formula:
$A=lw$
$A=(12m)(4m)$
$Area=48m^{2}$ </span>

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What does 2/3x=−1/3x−4 equal to

Misha Larkins [42]

I don’t understand how you worded the question but x=-4 if thats what you are looking for

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3 years ago

Can someone help me pls

iVinArrow [24]

Answer:

$\frac{8}{27}$

Step-by-step explanation:

$\frac{2}{3} \times \frac{2}{ 3} = \frac{4}{9}$

because its cubed, times by 2/3 again.

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hope this helps

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3 years ago

Rick buys remote control cars to resell. He applies a markup of 21%. Enter two expressions that represent the retail price of th

11111nata11111 [884]

The two expressions that represent the retail price of cars is: Retail price = 1.21c and Retail price = c + 0.21c

<em><u>Solution:</u></em>

Given that,

Rick buys remote control cars to resell

He applies a markup of 21%

Let "c" be the original price of remote control cars

To find: Expression for retail price of car

We know that,

Retail price = original price + markup

Here, markup price = 21 % of original price

Markup price = 21 % of c

Therefore, substituting the given values we get,

Retail price = c + 21 % of c

$\text{ Retail price } = c + 21 \% \times c\\\\\text{ Retail price } = c + \frac{21}{100} \times c\\\\\text{ Retail price } = c + 0.21c\\\\$

This can also be expressed as,

$\text{ Retail price } = c + 0.21c = 1.21c$

Thus two expressions that represent the retail price of cars is: Retail price = 1.21c and Retail price = c + 0.21c

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3 years ago

Factoring<br> 3xsquared + 14x +20

lesya692 [45]

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3 years ago

Use the Trapezoidal Rule, the Midpoint Rule, and Simpson's Rule to approximate the given integral with the specified value of n.

Otrada [13]

I guess the "5" is supposed to represent the integral sign?

$I=\displaystyle\int_1^4\ln t\,\mathrm dt$

With $n=10$ subintervals, we split up the domain of integration as

[1, 13/10], [13/10, 8/5], [8/5, 19/10], ... , [37/10, 4]

For each rule, it will help to have a sequence that determines the end points of each subinterval. This is easily, since they form arithmetic sequences. Left endpoints are generated according to

$\ell_i=1+\dfrac{3(i-1)}{10}$

and right endpoints are given by

$r_i=1+\dfrac{3i}{10}$

where $1\le i\le10$ .

a. For the trapezoidal rule, we approximate the area under the curve over each subinterval with the area of a trapezoid with "height" equal to the length of each subinterval, $\dfrac{4-1}{10}=\dfrac3{10}$ , and "bases" equal to the values of $\ln t$ at both endpoints of each subinterval. The area of the trapezoid over the $i$ -th subinterval is

$\dfrac{\ln\ell_i+\ln r_i}2\dfrac3{10}=\dfrac3{20}\ln(ell_ir_i)$

Then the integral is approximately

$I\approx\displaystyle\sum_{i=1}^{10}\frac3{20}\ln(\ell_ir_i)\approx\boxed{2.540}$

b. For the midpoint rule, we take the rectangle over each subinterval with base length equal to the length of each subinterval and height equal to the value of $\ln t$ at the average of the subinterval's endpoints, $\dfrac{\ell_i+r_i}2$ . The area of the rectangle over the $i$ -th subinterval is then

$\ln\left(\dfrac{\ell_i+r_i}2\right)\dfrac3{10}$

so the integral is approximately

$I\approx\displaystyle\sum_{i=1}^{10}\frac3{10}\ln\left(\dfrac{\ell_i+r_i}2\right)\approx\boxed{2.548}$

c. For Simpson's rule, we find a quadratic interpolation of $\ln t$ over each subinterval given by

$P(t_i)=\ln\ell_i\dfrac{(t-m_i)(t-r_i)}{(\ell_i-m_i)(\ell_i-r_i)}+\ln m_i\dfrac{(t-\ell_i)(t-r_i)}{(m_i-\ell_i)(m_i-r_i)}+\ln r_i\dfrac{(t-\ell_i)(t-m_i)}{(r_i-\ell_i)(r_i-m_i)}$

where $m_i$ is the midpoint of the $i$ -th subinterval,

$m_i=\dfrac{\ell_i+r_i}2$

Then the integral $I$ is equal to the sum of the integrals of each interpolation over the corresponding $i$ -th subinterval.

$I\approx\displaystyle\sum_{i=1}^{10}\int_{\ell_i}^{r_i}P(t_i)\,\mathrm dt$

It's easy to show that

$\displaystyle\int_{\ell_i}^{r_i}P(t_i)\,\mathrm dt=\frac{r_i-\ell_i}6(\ln\ell_i+4\ln m_i+\ln r_i)$

so that the value of the overall integral is approximately

$I\approx\displaystyle\sum_{i=1}^{10}\frac{r_i-\ell_i}6(\ln\ell_i+4\ln m_i+\ln r_i)\approx\boxed{2.545}$

4 0

4 years ago