All categories

3 years ago

What is the parallel slope of y=7

Mathematics

1 answer:

Sergeu [11.5K]3 years ago

3 0

Answer: y = 7 is a horizontal line that has a slope of 0

Step-by-step explanation:

You might be interested in

Write an equation to describe the proportional relationship in this table. Explain your thinking and how you got your answer.

creativ13 [48]

Focus on the values in red outside the table (on the left and right sides). As the output increases by 4, the input increases by 1. This leads to the slope being rise/run = 4/1 = 4. The slope is the number just to the left of the x variable. So that's how they ended up with the equation y = 4x.

You can think of it as y = 4x+0, and note how it's in the form y = mx+b

m = 4 = slope

b = 0 = y intercept

The y intercept is the value of y when x = 0.

--------

Now because x = 0 and y = 0 pair up together, we can find the slope a different way: Pick any row other than the first row. Divide the y value over its paired x value. So let's say we pick on row two and we'd get a slope of y/x = 4/1 = 4.

Or we could pick on row three and get y/x = 8/2 = 4.

Picking on the fourth row gets us the slope to be y/x = 12/3 = 4. We end up with the same slope value each time. Again, this trick only works if (x,y) = (0,0) is on the line. Otherwise, you'll need to use the first method or use the slope formula.

For equations like this, we consider them a direct proportion. As x increases, so does y. As x increases by 1, y increases by 4. This is the same throughout the table.

----------

A real world example of this could be that x represents the number of sodas you buy, and if each soda is $4, then y = 4x represents the total cost of buying x number of sodas.

For instance, if you bought x = 10 sodas, then y = 4*x = 4*10 = 40 dollars is what you spend overall.

7 0

3 years ago

I JUST NEED HELP PLEASE!! AHH!!~∆~

mafiozo [28]

Answer:

Step-by-step explanation:

take theta as reference angle

using sin rule

sin theta=opposite/hypotenuse

sin theta=5/13

sin theta=0.38

theta=sin 23

therefore the value of theta is 23 degree

8 0

3 years ago

Write the equation of the line passing through the two points. Show that this line is perpendicular to the given line.

il63 [147K]

Write the equation of the line passing through the two points. Show that this line is perpendicular to the given line.

(4,1) and (6,-5) y=1/3x-7

(1,-6) and (3,-10) y=1/2x-5

Please help, we learned this today and I’m really confused. How do you do it?

5 0

4 years ago

X2 + 5x - 50 = 0<br><img src="https://tex.z-dn.net/?f=%20%7Bx%7D%5E%7B2%7D%20%20%2B%205x%20-%2050%20%3D%200" id="TexFormula1" ti

love history [14]

Answer:

x = -10

x = 5

Step-by-step explanation:

${x}^{2} + 5x - 50 = 0 \\ (x + 10)(x - 5) = 0 \\ \\ x = - 10 \\ x = 5$

8 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

What is the parallel slope of y=7​

What is the parallel slope of y=7