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kirza4 [7]
3 years ago
10

Answer as much as you can all of them answers would be very greatly appreciated!

Mathematics
2 answers:
Gnom [1K]3 years ago
7 0
1) quarter- pound can be written as 1/4 pound
100:1/4=100*4=400
400 <span> hamburgers

2) The difference between these temperatures 212-32=180
180

Any number that is more than 32 and less than 212 is between this temperatures.
For example 40, and 136 lie between 32 and 212. 

3) 1000, 100, 10, 1...
We can see that every next number 10 times less than previous number,
so after 1 should be 1/10=0.1
If it is written as fraction 1/10, if it is written as decimal 0.1.</span>
cluponka [151]3 years ago
3 0
1. 400
2.If this is for the halfway point between the two, it is 122 degrees. If it is difference, it is 180 degrees.
3.0.1
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PLZ HELP, 4/5 * 2/3. exsplain how you got it too. i just need an exsplaination. step by step.
slavikrds [6]

Answer:

8/15

Step-by-step explanation:

4x2=8

5x3=15

7 0
2 years ago
Read 2 more answers
Use I = PRT to find the amount of simple interest.
yuradex [85]

Answer:

the simple interest is <em>36.</em>

Step-by-step explanation:

To find the answer we will first<em> divide 3 by 100</em> and we get <em>0.03.</em>  Now <em>multiply 0.03 by $200</em>, which is<em> 6</em>.  Then <em>multiply 6 by 6</em> <em>months</em> and we get 36.  Therefore, <em>36 is our answer. </em>

3 0
3 years ago
If Elisa gets an allowance of $35 per week and spends $10 of it on food and drink, $5 on entertainment and $5 on miscellaneous t
Maslowich

Answer:

15$

Step-by-step explanation:

35-10=25$

25-5=20$

20-5=15$

6 0
3 years ago
Write 60 as the product of two factors
Ilya [14]
Use a factor<span> tree to express </span>60<span> as a </span>product<span> of prime </span>factors<span>. So the prime factorization of </span>60<span> is 2 × 2 × 3 × 5, which can be written as 2 </span>2<span> × 3 × 5.</span>
5 0
3 years ago
Solve the system of equations by finding the reduced row-echelon form of the augmented matrix for the system of equations.
Anuta_ua [19.1K]

Answer:

c

Step-by-step explanation:

First, we can transform this into a matrix. The x coefficients will be the first ones for each row, the y coefficients the second column, etc.

\left[\begin{array}{cccc}1&-2&3&-2\\6&2&2&-48\\1&4&3&-38\end{array}\right]

Next, we can define a reduced row echelon form matrix as follows:

With the leading entry being the first non zero number in the first row, the leading entry in each row must be 1. Next, there must only be 0s above and below the leading entry. After that, the leading entry of a row must be to the left of the leading entry of the next row. Finally, rows with all zeros should be at the bottom of the matrix.

Because there are 3 rows and we want to solve for 3 variables, making the desired matrix of form

\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right] for the first three rows and columns. This would make the equation translate to

x= something

y= something

z = something, making it easy to solve for x, y, and z.

Going back to our matrix,

\left[\begin{array}{cccc}1&-2&3&-2\\6&2&2&-48\\1&4&3&-38\end{array}\right] ,

we can start by removing the nonzero values from the first column for rows 2 and 3 to reach the first column of the desired matrix. We can do this by multiplying the first row by -6 and adding it to the second row, as well as multiplying the first row by -1 and adding it to the third row. This results in

\left[\begin{array}{cccc}1&-2&3&-2\\0&14&-16&-36\\0&6&0&-36\end{array}\right]

as our matrix. * Next, we can reach the second column of our desired matrix by first multiplying the second row by (2/14) and adding it to the first row as well as multiplying the second row by (-6/14) and adding it to the third row. This eliminates the nonzero values from all rows in the second column except for the second row. This results in

\left[\begin{array}{cccc}1&0&10/14&-100/14\\0&14&-16&-36\\0&0&96/14&-288/14\end{array}\right]

After that, to reach the desired second column, we can divide the second row by 14, resulting in

\left[\begin{array}{cccc}1&0&10/14&-100/14\\0&1&-16/14&-36/14\\0&0&96/14&-288/14\end{array}\right]

Finally, to remove the zeros from all rows in the third column outside of the third row, we can multiply the third row by (16/96) and adding it to the second row as well as multiplying the third row by (-10/96) and adding it to the first row. This results in

\left[\begin{array}{cccc}1&0&0&-5\\0&1&0&-6\\0&0&96/14&-288/14\end{array}\right]

We can then divide the third row by -96/14 to reach the desired third column, making the reduced row echelon form of the matrix

\left[\begin{array}{cccc}1&0&0&-5\\0&1&0&-6\\0&0&1&-3\end{array}\right]

Therefore,

x=-5

y=-6

z=-3

* we could also switch the second and third rows here to make the process a little simpler

3 0
3 years ago
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