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saw5 [17]
3 years ago
12

Consider the enlargement of the rectangle. A smaller rectangle with length of x inches and width of three-fifths inch. A larger

rectangle with length of 20 inches and width of 12 inches. Use the proportion to find the missing dimension of the original rectangle. 1. Set up the proportion: StartFraction x over three-fifths EndFraction = StartFraction 20 over 12 EndFraction 2. Use cross products: 12(x) = 20(3 5 ) 3. Simplify: 12x = 4. Divide: x =
Mathematics
2 answers:
Savatey [412]3 years ago
8 0

Answer:

12x= 12

x= 1

Step-by-step explanation:

I got this right

Blizzard [7]3 years ago
5 0

Answer:

its 12 then 1

Step-by-step explanation:

i just did it

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In a right angled triangle, the two sides forming a right angle measure 8cm and 6cm. Area of this triangle is
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<h3>Answer:  24 square cm</h3>

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5 0
3 years ago
Read 2 more answers
Which situation can be modeled by the inequality m &gt; 225? A. The community theater's auditorium can seat at most 225 people.
Georgia [21]

Answer:

<h3>B. The number of people who attended the mathematics conference was more than 225.</h3>

Step-by-step explanation:

Given the inequality m > 225, the inequality means that m is greater than 225. Since the inequality sign does not include equal to sign, then the value of m cannot be less than and equal to 225. The situation that best fits the expression based on the explanation above is "The number of people who attended the mathematics conference was more than 225"

Emphasis should be on 'WAS MORE THAN' according to the statement. The greater sign means that the number of people will always be greater than or more than 225 representing the variable <em>m as the number of people.</em>

6 0
3 years ago
<img src="https://tex.z-dn.net/?f=prove%20that%5C%20%20%5Ctextless%20%5C%20br%20%2F%5C%20%20%5Ctextgreater%20%5C%20%5Cfrac%20%7B
inysia [295]

\large \bigstar \frak{ } \large\underline{\sf{Solution-}}

Consider, LHS

\begin{gathered}\rm \: \dfrac { \tan \theta + \sec \theta - 1 } { \tan \theta - \sec \theta + 1 } \\ \end{gathered}

We know,

\begin{gathered}\boxed{\sf{  \:\rm \: {sec}^{2}x - {tan}^{2}x = 1 \: \: }} \\ \end{gathered}  \\  \\  \text{So, using this identity, we get} \\  \\ \begin{gathered}\rm \: = \:\dfrac { \tan \theta + \sec \theta - ( {sec}^{2}\theta - {tan}^{2}\theta )} { \tan \theta - \sec \theta + 1 } \\ \end{gathered}

We know,

\begin{gathered}\boxed{\sf{  \:\rm \: {x}^{2} - {y}^{2} = (x + y)(x - y) \: \: }} \\ \end{gathered}  \\

So, using this identity, we get

\begin{gathered}\rm \: = \:\dfrac { \tan \theta + \sec \theta - (sec\theta + tan\theta )(sec\theta - tan\theta )} { \tan \theta - \sec \theta + 1 } \\ \end{gathered}

can be rewritten as

\begin{gathered}\rm\:=\:\dfrac {(\sec \theta + tan\theta ) - (sec\theta + tan\theta )(sec\theta -tan\theta )} { \tan \theta - \sec \theta + 1 } \\ \end{gathered} \\  \\  \\\begin{gathered}\rm \: = \:\dfrac {(\sec \theta + tan\theta ) \: \cancel{(1 - sec\theta + tan\theta )}} { \cancel{ \tan \theta - \sec \theta + 1} } \\ \end{gathered} \\  \\  \\\begin{gathered}\rm \: = \:sec\theta + tan\theta \\\end{gathered} \\  \\  \\\begin{gathered}\rm \: = \:\dfrac{1}{cos\theta } + \dfrac{sin\theta }{cos\theta } \\ \end{gathered} \\  \\  \\\begin{gathered}\rm \: = \:\dfrac{1 + sin\theta }{cos\theta } \\ \end{gathered}

<h2>Hence,</h2>

\begin{gathered} \\ \rm\implies \:\boxed{\sf{  \:\rm \: \dfrac { \tan \theta + \sec \theta - 1 } { \tan \theta - \sec \theta + 1 } = \:\dfrac{1 + sin\theta }{cos\theta } \: \: }} \\ \\ \end{gathered}

\rule{190pt}{2pt}

5 0
2 years ago
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