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2 years ago

a rectangle has a width that is 7 more than its length and an area of 30. Find the length and width of the rectangle

Mathematics

1 answer:

just olya [345]2 years ago

5 0

Answer:

length = √ 210/7

width = √ 210

Step-by-step explanation:

Area = 30

length = x

width = 7x

30 = 7x^2

30/7 = x^2

√ 210/7 = x

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Que is on pic.i can't able to type in text.

ad-work [718]

It's not difficult to compute the values of $A$ and $B$ directly:

$A=\displaystyle\int_1^{\sin\theta}\frac{\mathrm dt}{1+t^2}=\tan^{-1}t\bigg|_{t=1}^{t=\sin\theta}$
$A=\tan^{-1}(\sin\theta)-\dfrac\pi4$

$B=\displaystyle\int_1^{\csc\theta}\frac{\mathrm dt}{t(1+t^2)}=\int_1^{\csc\theta}\left(\frac1t-\frac t{1+t^2}\right)\,\mathrm dt$
$B=\left(\ln|t|-\dfrac12\ln|1+t^2|\right)\bigg|_{t=1}^{t=\csc\theta}$
$B=\ln\left|\dfrac{\csc\theta}{\sqrt{1+\csc^2\theta}}\right|+\dfrac12\ln2$

Let's assume $0$ , so that $|\csc\theta|=\csc\theta$ .

Now,

$\Delta=\begin{vmatrix}A&A^2&B\\e^{A+B}&B^2&-1\\1&A^2+B^2&-1\end{vmatrix}$
$\Delta=A\begin{vmatrix}B^2&-1\\A^2+B^2&-1\end{vmatrix}-e^{A+B}\begin{vmatrix}A^2&B\\A^2+B^2&-1\end{vmatrix}+\begin{vmatrix}A^2&B\\B^2&-1\end{vmatrix}$
$\Delta=A(-B^2+A^2+B^2)-e^{A+B}(-A^2-A^2B-B^3)+(-A^2-B^3)$
$\Delta=A^3-A^2-B^3+e^{A+B}(A^2+A^2B+B^3)$

There doesn't seem to be anything interesting about this result... But all that's left to do is plug in $A$ and $B$ .

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2 years ago

Is this equation possibale<br> 3+x=2-3x

Archy [21]

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Yes it is possible

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2 years ago

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daser333 [38]

I think it is: y=0.375x+3.125

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Find b

(5)=0.375(5)+b
(5)=1.875+b
5-1.875=b
b=3.125

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1 year ago

N two years, Linda will be half as old as her mother. Her mother is 38 years old now.

Ksenya-84 [330]

I'm not sure about the formula.
But since 38/2=19
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2 years ago

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Dafna11 [192]

Answer:

x^4 -2x^2 -24

Step-by-step explanation:

(x^2+4)(x^2−6)

FOIL

First: x^2*x^2 = x^4

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Inner: x^2 *4 = 4x^2

Last: 4*-6 = -24

Add them together

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Combine like terms

x^4 -2x^2 -24

This is in standard form since the powers decrease

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2 years ago

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