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3 years ago

The perimeter of a room is 43 ft. The width is 9 1 2ft. What is the length?

2 answers:

7 0

$answer = 12 \: ft \\ solution \\ perimeter = 43 ft\\ breadth = 9 \times \frac{1}{2} = \frac{19}{2 } = 9.5 \: ft\\ now \\ perimeter \: of \: \: rectangle = 2(l + b) \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: or \: 43= 2(l + 9.5) \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: or \: 43 = 2l + 19 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: or \: - 2l = 19 - 43 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: or \: - 2l = - 24 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: or \: l = \frac{ - 24}{ - 2} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: l = 12 \: ft \\ hope \: it \: helps \\ good \: luck \: on \: your \: assignment$

Anettt [7]3 years ago

3 0

Answer:

$length = 12ft$

Step-by-step explanation:

Given that,

$width =9 \frac{1}{2} = 9.5$

Let's find the length

$perimeter \\43 = 2(l + w) \\ 43= 2(l + 9.5) \\ 43 = 2l + 19 \\ 43 - 19 = 2l \\ 24 = 2l \\ \frac{24}{2} = \frac{2l}{2} \\ l = 12$

hope this helps

brainliest appreciated

good luck! have a nice day!

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Answer:

241/3

Trying to find the cube root of 24

Factor 24

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Of course you can also try cubing the choices given one of them will equal 24

Give it a try

I hope this helps

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3 years ago

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Answer:

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Step-by-step explanation:

in the given question,

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Step-by-step explanation:

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3 years ago

Find cos y and tan y if csc y = -√6/2 and cot y >0.

fomenos

Answer:

$\cos y = -\dfrac{\sqrt{3} }{3}$

$\tan y = \sqrt{2}$

Step-by-step explanation:

Recall that

$\boxed{\csc y := \dfrac{1}{\sin y}}$

$\boxed{\cot y := \dfrac{\cos y}{\sin y}}$

We know that

$\csc y = \dfrac{-\sqrt{6} }{2}$

Note that according to the definition of $\csc y$ it is true that both sine and cosine are negative, once $\csc y = \dfrac{-\sqrt{6} }{2}$ . Because $\cot y > 0$ , this conclusion is true. We basically have

$\boxed{(-a)(1/-b)=a/b \text{ such that } a,b\in\mathbb{R}_{\geq 0}}$

Sure it is true $\forall y\in\mathbb{R}$ but perhaps this way is better to understand.

In order to find sine, we can use the definition and manipulate the rational expression.

$\csc y = \dfrac{-\sqrt{6} }{2} = \dfrac{-\sqrt{6} / -\sqrt{6} }{2/-\sqrt{6} } = \dfrac{1 }{-\dfrac{2}{\sqrt{6} } }$

Therefore,

$\sin y =-\dfrac{2}{\sqrt{6} }$

Here I just divided numerator and denominator by $-\sqrt{6}$ .

Now, to find cosine we can use the identity

$\boxed{\sin^2y +\cos ^2y =1}$

Thus,

$\left(-\dfrac{2}{\sqrt{6} }\right)^2 + \cos ^2y =1 \implies \dfrac{4}{6 } +\cos ^2y =1$

$\implies \cos ^2y =1 - \dfrac{4}{6 } \implies \cos ^2y =\dfrac{1}{3 } \implies \cos y = \pm \dfrac{\sqrt{1} }{\sqrt{3} } = \pm \dfrac{\sqrt{1} \sqrt{3} }{3} = \pm \dfrac{\sqrt{3} }{3}$

$\cos y = \pm\dfrac{\sqrt{3} }{3}$

Once we have $\cot y > 0$ , we just consider

$\cos y = -\dfrac{\sqrt{3} }{3}$

FInally, for tangent, just consider

$\boxed{\tan y := \dfrac{\sin y}{\cos y}}$

thus,

$\tan y = \dfrac{\sin y}{\cos y} = \dfrac{-\frac{2}{\sqrt{6} }}{-\frac{\sqrt{3} }{3}} = \dfrac{6}{\sqrt{18} } =\dfrac{6}{3\sqrt{2} } =\dfrac{2}{\sqrt{2} } = \sqrt{2}$

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3 years ago