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Basile [38]
3 years ago
11

1) Jim took out a loan for $500. The rate is 8%. How much does Jim owe in interest if he pays it back in 4 1/2 years?

Mathematics
1 answer:
Setler79 [48]3 years ago
8 0

Answer:

ito yung link o /fhytuewfxnopjbfw/ yan na po

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Hello can some one please help me
Artemon [7]
It’s 4/25 because 2/5 times 2/5 the numerator is 2 times 2 which is four and the denominator is 5 times 5 which is 25
8 0
2 years ago
Read 2 more answers
Find the sum of even numbers from 34 to 234 inclusive
valentinak56 [21]
We have 

AP = 34,36,38......,234

a, (first term) = 34

n ( total terms ) =?

d ( common difference ) = 2

so, a +(n-1)d =an

34+(n-1)*2 = 234

(n-1)2 = 234-34

n-1 = 200/2

n = 100+1

n =101

Now sum of the even no.

Sn = n/2( 2a+9n-1)d)
       
       = 101/2 (2*34 +100*2)
 
        = 101/2 * 268
 
          =101 *134
 
         = 13534
7 0
3 years ago
Peter says, "If you subtract 13 from my number and multiply the difference by - 4, the result is -100." What is Peter's number?
HACTEHA [7]

Step-by-step explanation:

peter is still in 1st grade he doesn't understand yet

4 0
3 years ago
Read 2 more answers
Use row operations to solve the system
TiliK225 [7]

Answer:

  (x, y, z) = (1, 12, 15)

Step-by-step explanation:

As with any set of linear equations, there are many possible routes to a solution. We might simplify the notation a bit by writing the coefficients in an augmented matrix. The columns, left to right, represent the coefficients of x, y, and z, in order, and the constant term.

The row operations we'll use are multiplying a row by a value and adding that result to another row, replacing the other row by the sum.

We can make things a little simpler by writing the second equation first. Then the augmented matrix we're starting with is ...

  \left[\begin{array}{ccc|c}4&-1&1&7\\1&1&-1&-2\\1&-3&2&-5\end{array}\right]

Adding the second row to the first, we get ...

  \left[\begin{array}{ccc|c}5&0&0&5\\1&1&-1&-2\\1&-3&2&-5\end{array}\right]

Dividing the first row by 5 gives ...

  \left[\begin{array}{ccc|c}1&0&0&1\\1&1&-1&-2\\1&-3&2&-5\end{array}\right]

Subtracting this from the second row, and again from the third row, we are left with ...

  \left[\begin{array}{ccc|c}1&0&0&1\\0&1&-1&-3\\0&-3&2&-6\end{array}\right]

Multiplying the second row by 3 and adding that to the third row, we get ...

  \left[\begin{array}{ccc|c}1&0&0&1\\0&1&-1&-3\\0&0&-1&-15\end{array}\right]

Subtracting the third row from the second gives ...

  \left[\begin{array}{ccc|c}1&0&0&1\\0&1&0&12\\0&0&-1&-15\end{array}\right]

Finally, multiplying the last row by -1, we have the solution:

  \left[\begin{array}{ccc|c}1&0&0&1\\0&1&0&12\\0&0&1&15\end{array}\right]

This matrix corresponds to the equations ...

  • x = 1
  • y = 12
  • z = 15

_____

The purpose of our choice of row operations is to make the diagonal elements 1 and the off-diagonal elements 0. That is how we end up with the final equations shown.

As we said, there are many ways to go about this. In general, one can ...

  • if necessary, swap rows until the diagonal term of interest is non-zero. If you are doing this using a computer program, generally you want the diagonal term to have the coefficient with the largest magnitude. When doing this by hand, you may want to arrange the rows to avoid fractions when you do the normalizing.
  • divide the row by the coefficient of the diagonal element to "normalize" the diagonal element to a value of 1
  • zero the other elements in that column by multiplying the row just normalized by the element in another row, then subtracting the product. (The 4th matrix shown above shows this for the first column.)
  • proceed to the next diagonal element and repeat the process until all diagonal elements are 1. If you cannot make all diagonal elements 1, then the system of equations does not have a unique solution. If any row becomes all zeros, the system is "dependent" and has infinite solutions. If a row is zeros except for the rightmost column, the system is "inconsistent" and has no solutions.
3 0
3 years ago
Patricia baked some cupcakes for sale. she put half of the cupcakes equally into 6 big boxes and the other half equally into sma
mojhsa [17]

Answer:

The answer is given below

Step-by-step explanation:

a)

Let us assume Patricia baked x number of cakes. She put half of the cupcakes (i.e x/2) equally into 6 big boxes.

6 big boxes contained \frac{x}{2} cakes, therefore 1 big box would contain \frac{x}{2}/6=\frac{x}{12} cakes.

Let us assume she put the other half into 14 small boxes, therefore each small box would contain \frac{x}{2}/14=\frac{x}{28} cakes.

There were 45 cupcakes in 3 big boxes and 8 small boxes altogether. That is:3(\frac{x}{12} )+8(\frac{x}{28})=45\\ 84x+96x=15120\\180x=15120\\x=84

Therefore Patricia baked 84 cup cakes

b)

She sold all the small boxes and collected $189, i.e she sold 14 small box for $189. Each small box = $189/14 = $13.5

6 0
3 years ago
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