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Romashka-Z-Leto [24]
2 years ago
8

I’ll mark you brainlist please help

Mathematics
1 answer:
Nina [5.8K]2 years ago
3 0

Answer:

The x is the domain and the y is the range. It's still a function

Step-by-step explanation:

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Factor this polynomial completely x^2-10x+25
san4es73 [151]

<u>x^2-10x+25</u>

=x^2-5x-5x+25

x(x-5)-5(x-5)

<u>=</u><u>(</u><u>x-5)</u><u>(</u><u>x-5)</u>

<u>x=</u><u>5</u><u> </u><u>,</u><u> </u><u>x=</u><u>5</u><u> </u>

hope it's helpful to you

5 0
2 years ago
Read 2 more answers
What would be the surface area of the figure..
valina [46]

Answer:

<h2>           503 cm²</h2>

Step-by-step explanation:

The surface area of the figure is sum of areas of two triangles and three ractangles:

A=2\cdot\frac12\cdot10\cdot8,3+3\cdot10\cdot14=83+420=503\,cm^2

8 0
3 years ago
Order these numbers from least to greatest.<br> 1.401, 1.2011, 1.2, 1.41
ira [324]

Answer:

From least to greatest, the correct order is 1.2, 1.2011, 1.401, and 1.41.

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
Help help I don’t really get this???
insens350 [35]

Answer:

Question 4:  y=\displaystyle -\frac{4}{5}x

Question 5: y=-5x-3

Step-by-step explanation:

Hi there!

Linear equations are typically organized in slope-intercept form: y=mx+b where <em>m</em> is the slope of the line and <em>b</em> is the y-intercept (the y-coordinate of the point where the line crosses the y-axis).

<u>Question 4</u>

<u>1) Determine the slope (</u><u><em>m</em></u><u>)</u>

\displaystyle m=\frac{y_2-y_1}{x_2-x_1} where two points that pass through the line are (x_1,y_1) and (x_2,y_2)

In the graph, two easy-to-identify points on the line are (-5,4) and (5,-4). Plug these into the equation:

\displaystyle m=\frac{-4-4}{5-(-5)}\\\\\displaystyle m=\frac{-4-4}{5+5}\\\\\displaystyle m=\frac{-8}{10}\\\\\displaystyle m=-\frac{4}{5}

Therefore, the slope of the line is \displaystyle -\frac{4}{5}. Plug this into y=mx+b as the slope (<em>m</em>):

y=\displaystyle -\frac{4}{5}x+b

<u>2) Determine the y-intercept (</u><u><em>b</em></u><u>)</u>

On the graph, we can see that the line crosses the y-axis when y is 0. Therefore, the y-intercept (<em>b</em>) is 0. Plug this into y=\displaystyle -\frac{4}{5}x+b:

y=\displaystyle -\frac{4}{5}x+0\\\\y=\displaystyle -\frac{4}{5}x

<u>Question 5</u>

<u>1) Determine the slope (</u><u><em>m</em></u><u>)</u>

\displaystyle m=\frac{y_2-y_1}{x_2-x_1}

Two easy-to-identify points are (-1,2) and (0,-3). Plug these into the equation:

\displaystyle m=\frac{-3-2}{0-(-1)}\\\\\displaystyle m=\frac{-3-2}{0+1}\\\\\displaystyle m=\frac{-5}{1}\\\\m=-5

Therefore, the slope is -5. Plug this into y=mx+b:

y=-5x+b

<u>2) Determine the y-intercept (</u><u><em>b</em></u><u>)</u>

On the graph, we can see that the line crosses the y-axis at the point (0,-3). The y-coordinate of this point is -3. Therefore, the y-intercept (<em>b</em>) is -3. Plug this into y=-5x+b:

y=-5x+(-3)\\y=-5x-3

I hope this helps!

8 0
2 years ago
What is the length side of a triangle that has vertices at (-5, -1), (-5, 5), and (3, -1)?
Rasek [7]

The side lengths of triangle are 6 units, 8 units and 10 units.

<u>SOLUTION: </u>

Given that, we have to find what is the length side of a triangle that has vertices at (-5, -1), (-5, 5), and (3, -1)  

We know that, distance between two points P\left(x_{1}, y_{1}\right) \text { and } Q\left(x_{2}, y_{2}\right) is given by  

P Q=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}

Now,  

\begin{array}{l}{\text { Distance between }(-5,-1) \text { and }(-5,5)=\sqrt{(-5-(-5))^{2}+(5-(-1))^{2}}} \\\\ {\qquad \begin{array}{l}{=\sqrt{(-5-(-5))^{2}+(5-(-1))^{2}}} \\\\ {=\sqrt{(0)^{2}+(5+1)^{2}}=\sqrt{(6)^{2}}=6} \\\\ {=\sqrt{(-5)^{2}+(5+1)^{2}+(5-(-1))^{2}}} \\\\ {=\sqrt{(-8)^{2}+(5+1)^{2}}=\sqrt{64+36}=\sqrt{100}=10} \\\\ {=\sqrt{(3-1)^{2}+(-1-(-1))^{2}}} \\\\ {=\sqrt{(5+3)^{2}+(0)^{2}}=\sqrt{(8)^{2}}=8}\end{array}}\end{array}

8 0
3 years ago
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