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3 years ago

Seven pens and two pencils costs $6.59. seven pens and five cost $7.55. Find the cost of one pencil and the cost of one pen.

1 answer:

5 0

Let one pen be $x$ and one pencil as $y$ , we can produce two equations in terms of $x$ and $y$

$7x+2y=6.59$ ------> Equation 1
$7x+5y=7.55$ ------> Equation 2

We can use either Elimination or Substitution method. Let's try the substitution method

Rearranging equation one to get $7x=6.59-2y$ and substitute into equation 2, gives
$(6.59-2y)+5y=7.55$
$6.59+3y=7.55$
$3y=7.55-6.59$
$3y=0.96$
$y=0.32$

We can use either equation one or equation to work out $x$

$7x+2(0.32)=6.59$
$7x=6.59-0.64$
$x=0.85$

The price of one pen is $0.85
The price of one pencil is $0.32

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2 years ago

Can somebody explain how these would be done? The selected answer is incorrect, and I was told "Nice try...express the product b

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Solution ( Second Attachment ) : - 2.017 + 0.656i

Solution ( First Attachment ) : 16.140 - 5.244i

Step-by-step explanation:

Second Attachment : The quotient of the two expressions would be the following,

$6\left[\cos \left(\frac{2\pi }{5}\right)+i\sin \left(\frac{2\pi \:}{5}\right)\right]$ ÷ $2\sqrt{2}\left[\cos \left(\frac{-\pi }{2}\right)+i\sin \left(\frac{-\pi \:}{2}\right)\right]$

So if we want to determine this expression in standard complex form, we can first convert it into trigonometric form, then apply trivial identities. Either that, or we can straight away apply the following identities and substitute,

( 1 ) cos(x) = sin(π / 2 - x)

( 2 ) sin(x) = cos(π / 2 - x)

If cos(x) = sin(π / 2 - x), then cos(2π / 5) = sin(π / 2 - 2π / 5) = sin(π / 10). Respectively sin(2π / 5) = cos(π / 2 - 2π / 5) = cos(π / 10). Let's simplify sin(π / 10) and cos(π / 10) with two more identities,

( 1 ) $\cos \left(\frac{x}{2}\right)=\sqrt{\frac{1+\cos \left(x\right)}{2}}$

( 2 ) $\sin \left(\frac{x}{2}\right)=\sqrt{\frac{1-\cos \left(x\right)}{2}}$

These two identities makes sin(π / 10) = $\frac{\sqrt{2}\sqrt{3-\sqrt{5}}}{4}$ , and cos(π / 10) = $\frac{\sqrt{2}\sqrt{5+\sqrt{5}}}{4}$ .

Therefore cos(2π / 5) = $\frac{\sqrt{2}\sqrt{3-\sqrt{5}}}{4}$ , and sin(2π / 5) = $\frac{\sqrt{2}\sqrt{5+\sqrt{5}}}{4}$ . Substitute,

$6\left[ \left\frac{\sqrt{2}\sqrt{3-\sqrt{5}}}{4}+i\left\frac{\sqrt{2}\sqrt{5+\sqrt{5}}}{4}\right]$ ÷ $2\sqrt{2}\left[\cos \left(\frac{-\pi }{2}\right)+i\sin \left(\frac{-\pi \:}{2}\right)\right]$

Remember that cos(- π / 2) = 0, and sin(- π / 2) = - 1. Substituting those values,

$6\left[ \left\frac{\sqrt{2}\sqrt{3-\sqrt{5}}}{4}+i\left\frac{\sqrt{2}\sqrt{5+\sqrt{5}}}{4}\right]$ ÷ $2\sqrt{2}\left[0-i\right]$

And now simplify this expression to receive our answer,

$6\left[ \left\frac{\sqrt{2}\sqrt{3-\sqrt{5}}}{4}+i\left\frac{\sqrt{2}\sqrt{5+\sqrt{5}}}{4}\right]$ ÷ $2\sqrt{2}\left[0-i\right]$ = $-\frac{3\sqrt{5+\sqrt{5}}}{4}+\frac{3\sqrt{3-\sqrt{5}}}{4}i$ ,

$-\frac{3\sqrt{5+\sqrt{5}}}{4}$ = $-2.01749\dots$ and $\:\frac{3\sqrt{3-\sqrt{5}}}{4}$ = $0.65552\dots$

= $-2.01749+0.65552i$

As you can see our solution is option c. - 2.01749 was rounded to - 2.017, and 0.65552 was rounded to 0.656.

________________________________________

First Attachment : We know from the previous problem that cos(2π / 5) = $\frac{\sqrt{2}\sqrt{3-\sqrt{5}}}{4}$ , sin(2π / 5) = $\frac{\sqrt{2}\sqrt{5+\sqrt{5}}}{4}$ , cos(- π / 2) = 0, and sin(- π / 2) = - 1. Substituting we receive a simplified expression,

$6\sqrt{5+\sqrt{5}}-6i\sqrt{3-\sqrt{5}}$