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Serga [27]
4 years ago
5

Natasha can type 30 words a minute. How many words can be typed in 7.5 minutes?

Mathematics
2 answers:
Vinvika [58]4 years ago
5 0
225

7*30=210
+15 (1/2 of 30)=225
julia-pushkina [17]4 years ago
3 0
225. I think. However its not clear if total words typed are asked for Natasha or in General.
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What is 13,931 round by 10​
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13,930

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If I can walk 9673.6 meters in 5 hours. How far can I walk in 70 minutes. ( I need the work)
den301095 [7]

First, convert 5 hours into minutes:

5 hours 60 min

------------ * --------------- = 300 min

1 1 hr

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6 0
4 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
3 years ago
Solve the proportions using cross products. Round to the nearest hundredth is necessary. 21miles/49hours = 15miles h hours
SSSSS [86.1K]

\dfrac{21}{49}=\dfrac{21:7}{49:7}=\dfrac{3}{7}\\\\\dfrac{21}{49}=\dfrac{15}{h}\to\dfrac{3}{7}=\dfrac{15}{h}\qquad\text{cross multiply}\\\\3h=(7)(15)\\\\3h=105\qquad\text{divide both sides by 3}\\\\\boxed{h=35}\\\\Answer:\ 35\ hours

8 0
4 years ago
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