All categories

2 years ago

Clara found a nickel, a quarter, two dimes, another quarter, and two pennies in her desk drawer, how much money did clara find?

Mathematics

2 answers:

alisha [4.7K]2 years ago

7 0

Answer:77 cents Clara found 77 cents bc you just need to add up the cents and you have it

adelina 88 [10]2 years ago

4 0

<h3>Answer:</h3>

77 cents

<h3 /><h3>Step-by-step explanation:</h3>

So 2 quarters is 50 cents and you have a nickel: 5 cents, then you have 2 pennies (2 cents), then you have 2 dimes. Now all you have to do is add them all together to get your answer.

5 cents + 25 cents= 30 cents

30 cents + 20 cents= 50 cents

50 cents + 25 cents= 75 cents

75 cents + 2 cents= 77 cents

So Clara would have 77 cents in her desk drawer.

You might be interested in

16. Find the value of y.

devlian [24]

y + (2x -12) = 180 (since they are supplementary angles)

(2x-12) = 96 ( since they are alternate exterior angle)

2x = 96 + 12

2x = 108

x = 108/2

x = 54.

y+( 2x-12)=180

y+(2× 54 -12) = 180

y + (108-12) = 180

y + 96 = 180

y= 180-96

y = 84.

8 0

2 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>

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