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Elena-2011 [213]
2 years ago
14

This unit, you have learned several formulas and equations. Match the formula or equation with what the answer represents.

Mathematics
1 answer:
Sedbober [7]2 years ago
6 0

Answer:

.

Step-by-step explanation:

.

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Find the area of this,<br> Simplify your answer completely.
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\frac{2}{15}  {cm}^{2}

Step-by-step explanation:

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\frac{1}{5}  \times  \frac{2}{3}  \\  =  \frac{2}{15}  {cm}^{2}

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1 year ago
Please help! If the m = 1/2 and the b = -2, then the linear equation is:
Vsevolod [243]

Answer:

y=1/2x-2

you write it in y=mx+b form

7 0
2 years ago
Read 2 more answers
<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
What's the answer ?<br><br><img src="https://tex.z-dn.net/?f=%20%5Cfrac%7B%20%5Csqrt%7B225%7D%20%7D%7B%20%5Csqrt%7B%7D%2025%7D%2
postnew [5]

Answer:

Hello.....

15/5

Step-by-step explanation:

thanks

6 0
2 years ago
Read 2 more answers
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