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bonufazy [111]
3 years ago
10

What is 28 divided by 4?

Mathematics
2 answers:
koban [17]3 years ago
8 0
The answer is seven.
yan [13]3 years ago
6 0

Answer: 7

Step-by-step explanation:

7•4=28 therefor 28\4= 7

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Please help!!<br> Write a matrix representing the system of equations
frozen [14]

Answer:

(4, -1, 3)

Step-by-step explanation:

We have the system of equations:

\left\{        \begin{array}{ll}            x+2y+z =5 \\    2x-y+2z=15\\3x+y-z=8        \end{array}    \right.

We can convert this to a matrix. In order to convert a triple system of equations to matrix, we can use the following format:

\begin{bmatrix}x_1& y_1& z_1&c_1\\x_2 & y_2 & z_2&c_2\\x_3&y_2&z_3&c_3 \end{bmatrix}

Importantly, make sure the coefficients of each variable align vertically, and that each equation aligns horizontally.

In order to solve this matrix and the system, we will have to convert this to the reduced row-echelon form, namely:

\begin{bmatrix}1 & 0& 0&x\\0 & 1 & 0&y\\0&0&1&z \end{bmatrix}

Where the (x, y, z) is our solution set.

Reducing:

With our system, we will have the following matrix:

\begin{bmatrix}1 & 2& 1&5\\2 & -1 & 2&15\\3&1&-1&8 \end{bmatrix}

What we should begin by doing is too see how we can change each row to the reduced-form.

Notice that R₁ and R₂ are rather similar. In fact, we can cancel out the 1s in R₂. To do so, we can add R₂ to -2(R₁). This gives us:

\begin{bmatrix}1 & 2& 1&5\\2+(-2) & -1+(-4) & 2+(-2)&15+(-10) \\3&1&-1&8 \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 2& 1&5\\0 & -5 & 0&5 \\3&1&-1&8 \end{bmatrix}

Now, we can multiply R₂ by -1/5. This yields:

\begin{bmatrix}1 & 2& 1&5\\ -\frac{1}{5}(0) & -\frac{1}{5}(-5) & -\frac{1}{5}(0)& -\frac{1}{5}(5) \\3&1&-1&8 \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\3&1&-1&8 \end{bmatrix}

From here, we can eliminate the 3 in R₃ by adding it to -3(R₁). This yields:

\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\3+(-3)&1+(-6)&-1+(-3)&8+(-15) \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\0&-5&-4&-7 \end{bmatrix}

We can eliminate the -5 in R₃ by adding 5(R₂). This yields:

\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\0+(0)&-5+(5)&-4+(0)&-7+(-5) \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\0&0&-4&-12 \end{bmatrix}

We can now reduce R₃ by multiply it by -1/4:

\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\ -\frac{1}{4}(0)&-\frac{1}{4}(0)&-\frac{1}{4}(-4)&-\frac{1}{4}(-12) \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\0&0&1&3 \end{bmatrix}

Finally, we just have to reduce R₁. Let's eliminate the 2 first. We can do that by adding -2(R₂). So:

\begin{bmatrix}1+(0) & 2+(-2)& 1+(0)&5+(-(-2))\\ 0 & 1 & 0& -1 \\0&0&1&3 \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 0& 1&7\\ 0 & 1 & 0& -1 \\0&0&1&3 \end{bmatrix}

And finally, we can eliminate the second 1 by adding -(R₃):

\begin{bmatrix}1 +(0)& 0+(0)& 1+(-1)&7+(-3)\\ 0 & 1 & 0& -1 \\0&0&1&3 \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 0& 0&4\\ 0 & 1 & 0& -1 \\0&0&1&3 \end{bmatrix}

Therefore, our solution set is (4, -1, 3)

And we're done!

3 0
3 years ago
<img src="https://tex.z-dn.net/?f=%20%7Bx%7D%5E%7B2%7D%20%20%2B%2010%20-%2021%20%3D%20y" id="TexFormula1" title=" {x}^{2} + 10
myrzilka [38]

Answer:

Step-by-step explanation:

5 0
2 years ago
Read 2 more answers
Find the GCF of 27x^2 and 45x^3y^2
Inga [223]

Step-by-step explanation:

27x² = 3*3*3*x*x

45x³y² = 3*3*5*x*x*x*y*y

Now

GCF = 3 * 3 *x * x = 9x²

6 0
3 years ago
Item 8
ludmilkaskok [199]
So, I didn't quite understand your question clearly, but I'll try my best to answer it.

So, the first question is quite simple: Did the runner have a % decrease in his time or an increase.
Well, 7:45 is minor than 5:51. So, his time decreased.

2nd question:
The percent of change. Well, first things first 7:45 has 2 different units, wich are minutes and seconds, I suppose. Same goes for 5:51.
So let's put everything in seconds. multiply both the 7 and 5 for 60(for every minute has 60 seconds.)

That means:
7*60= 420+45 = 465.
5*60=300+51=351.

Now, let's do the math itself:
465 -> 100%
351 -> x%
 Is equal to about  75,5%
How much did it decrease then?
(75,5-100)%= 24,5%, approximately

8 0
3 years ago
A hospital prepares a 100mg supply of technetium-99m which has a half-life of 6 hours. what is the decay factor?
SashulF [63]
For a half- life, the formula is
A(t) = A0(0.5)^( t / t_0.5)
where A(t) is the amount of substance at time t
A0 is the initial amount of subtance
t_0.5 is the time it will reduce its amount to half
t is the time

substitute the given to the formula
A(t) = 100(0.5)^( t/6 )
so the decay factor is 0.5^( t/6 )
6 0
3 years ago
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