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Vika [28.1K]
2 years ago
14

You have some quarters $0.25 and dimes $0.10 There is a total of 32 coins . You turn them in to the bank and it is worth $5.00.

How many of each coins did you have?
Mathematics
1 answer:
IgorC [24]2 years ago
3 0

$10 bcz you adding on to what you have

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Answer:

The equation is given below as

\frac{\cos2x}{\cos x}=\cos x-\sin x\tan x

Step 1:

We will work on the left-hand side, we will have

\begin{gathered} \cos x-\sin x\tan x \\ \text{recall that,} \\ Quoitent\text{ identity is} \\ \tan x=\frac{\sin x}{\cos x} \end{gathered}

By substituting the identity above, we will have

\begin{gathered} \cos x-\sin x\tan x=\cos x-\frac{\sin x.\sin x}{\cos x}=\cos x-\frac{\sin^2x}{\cos x} \\  \end{gathered}

Here, we will make use of the quotient identity

Step 2:

By writings an expression, we will have

\begin{gathered} \cos x-\sin x\tan x=\cos x-\frac{\sin x.\sin x}{\cos x} \\ \cos x-\sin x\tan x=\frac{\cos^2x-\sin^2x}{\cos x} \end{gathered}

Here, we will use the definition of subtraction

\cos x-\frac{\sin^2x}{\cos x}

Step 3:

We will apply the double number identity given below

\begin{gathered} \cos 2\theta=\cos (\theta+\theta)=\cos ^2\theta-\sin ^2\theta \\ \cos 2x=cos(x+x)=\cos ^2x-\sin ^2x \end{gathered}

By applying this, we will have

\frac{\cos^2x-\sin^2x}{\cos x}=\frac{\cos2x}{\cos x}

Here, we will use the double number identity

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You invested $11,000 in two accounts paying 3% and 7% annual interest, respectively. If the total interest earned for the year w
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Answer:

I invested $ 10,000 at 7% per year, and $ 1,000 at 3% per year.

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Given that I invested $ 11,000 in two accounts paying 3% and 7% annual interest, respectively, if the total interest earned for the year was $ 730, to determine how much was invested at each rate, the following calculation must be performed:

5,000 x 0.07 + 6,000 x 0.03 = 530

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10,000 x 0.07 + 1,000 x 0.03 = 730

Therefore, I invested $ 10,000 at 7% per year, and $ 1,000 at 3% per year.

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