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BigorU [14]
2 years ago
14

Consider the following expression and determine which statements are true.

Mathematics
1 answer:
Serhud [2]2 years ago
3 0

Given:

$\frac{7}{r}+2^{3}+\frac{s}{3}+11

To find:

Which statement are true?

Solution:

Option A: The entire expression is a sum.

It is true because it performed addition operation.

Option B: The coefficient of s is 3.

$\frac{7}{r}+2^{3}+\frac{s}{3}+11=\frac{7}{r}+2^{3}+\frac{1}{3}s+11

It is not true because the coefficient of s is \frac{1}{3}.

Option C: The term \frac{7}{r} is a quotient.

If we divide 7 by r, we obtain a quotient.

So it is true.

Option D: The term 2^3 has a variable.

It is not true because it does not contain any variable.

Therefore the entire expression is a sum and the term \frac{7}{r} is a quotient are true statement.

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Write the equation of the line graphed below.
posledela

Answer:

y=2x +3

Step-by-step explanation:

The slope hits the points (-1,2) and (0,2). If you count how many units are right and up, then that's your slope. In this case, the distance between those two points are 1 right and 2 up which is the same thing as 2. The y-intercept is 3 since the slope crosses the y-axis there.

6 0
3 years ago
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
I need help with converting liters, millilters, kiloliters etc? Okay, i am really confused! I don't understand how to get 75.01
Mamont248 [21]
Firstly, understand prefixes :)

giga 1000000000
mega 1000000
kilo 1000
deci 0.1
centi 0.01
milli 0.001
micro 0.000001
nano 0.000000001

6 0
2 years ago
Solve for y<br>7y + 9 - 10y = -18<br>What is y simplify your answer as much as possible.
Tpy6a [65]
Y = 9

7y - 10y = -3y
bring down the 9 = -18 so

-3y + 9 = -18

9 - 9 = 0
- 18 - 9 = - 27

- 3y ÷ - 3 = y
- 27 ÷ -3 = 9

therefore y =9
7 0
3 years ago
The times for running a 5K for a local charity event are Normally distributed, with a mean time of 28 minutes and
liq [111]

Answer:19.12

I’m not sure if this is the answer but I need the points so here.

7 0
2 years ago
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