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3 years ago

Find the slope of a line parallel to y =

Mathematics

1 answer:

Zigmanuir [339]3 years ago

3 0

Answer:C

Step-by-step explanation:

y=12x is almost the same as y=12x+3 but +3 moves the line 3 up making it parrallel

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Suppose you take $25,000 and deposit some of it into a savings account that pays 2.2% interest. You then take the rest of the mo

FromTheMoon [43]

Answer:

The amount deposited in the savings account is $11,494.12

The amount deposited in the CD account is $13,505.88

Step-by-step explanation:

Firstly, let’s calculate the amount of interest;

Interest = Amount at end of 10 years - amount deposited= 31,648 -25,000 = 6,648

Let the amount deposited in the savings account be $x while the amount deposited in the CD account be $y

Mathematically;

x + y = 25,000 ••••••••••(i)

Now the formula for simple interest is;

PRT/100

where P is the amount deposited, r is the rate and T is the time

On the savings account, amount of interest generated will be;

(x * 2.2 * 10)/100 = 22x/100

On the CD act, amount of interest generated will be;

(y * 3.05 * 10)/100 = 30.5y/100

The addition of both give the total interest;

22x/100 + 30.5y/100 = 6648

Multiply through by 100

22x + 30.5y = 664800 •••••••(ii)

From i, x = 25,000 - y

Substitute this into ii

22(25000-y) + 30.5y = 664800

550,000 -22y + 30.5y = 664,800

8.5y = 664,800-550,000

8.5y = 114,800

y = 114,800/8.5

y = $13,505.88

x = 25,000 -y

x = 25,000 - 13,505.88

x = $11,494.12

3 0

3 years ago

Pls HELP, im stuck(picture included)

Verdich [7]

Answer is x = (355*r^-35)/7 hope led me know if you have any questions

5 0

3 years ago

Read 2 more answers

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

trapecia [35]

Answer:

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,

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