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Mars2501 [29]
3 years ago
5

Pleaseeeeeeee someone answer this

Mathematics
1 answer:
Karo-lina-s [1.5K]3 years ago
5 0

Answer:

<u>Ordered pairs of points:</u>

  • (1, 45), (2, 90), (3, 135), (4, 180), (5, 225), (6, 270)

a) Linear

b) Domain is positive integers and is discrete

c) y = 45x

d) Let 500 = 45x, then

x = 500/45 = 11.11

x  is not an integer, so the answer is not

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(3x-5)/(x-5)&gt;=0 <br> How do I get the answers for this problem?
Diano4ka-milaya [45]
\frac{3x-5}{x-5} > 0 

First, note that x is undefined at 5. / x ≠ 5
Second, replace the inequality sign with an equal sign so that we can solve it like a normal equation. / Your problem should look like: \frac{3x-5}{x-5} = 0
Third, multiply both sides by x - 5. / Your problem should look like: 3x - 5 = 0
Forth, add 5 to both sides. / Your problem should look like: 3x = 5
Fifth, divide both sides by 3. / Your problem should look like: x = \frac{5}{3}
Sixth, from the values of x above, we have these 3 intervals to test: 
x < \frac{5}{3}
\frac{5}{3} < x < 5
x > 5
Seventh, pick a test point for each interval. 

1. For the interval x < \frac{5}{3} :
Let's pick x - 0. Then, \frac{3x0-5}{0-5} > 0
After simplifying, we get 1 > 0 which is true.
Keep this interval.

2. For the interval \frac{5}{3} < x < 5:
Let's pick x = 2. Then, \frac{3x2-5}{2-5} > 0
After simplifying, we get -0.3333 > 0, which is false.
Drop this interval.

3. For the interval x > 5:
Let's pick x = 6. Then, \frac{3x6-5}{6-5} > 0
After simplifying, we get 13 > 0, which is ture.
Keep this interval.
Eighth, therefore, x < \frac{5}{3} and x > 5

Answer: x < \frac{5}{3} and x > 5

3 0
4 years ago
30°-60°-90°<br> help me solve this please
dexar [7]

Answer:

the answer to 30-60-90=0

Step-by-step explanation:

90-60=30-30=0

4 0
3 years ago
140cm<br> 3 cm<br> 12 cm<br> 8 cm<br> Find surface area?
Mashcka [7]

Answer:

3cm espero te sirva jajajajaja

8 0
3 years ago
Which of the equations below could be the equation of this parabola?
nirvana33 [79]

Answer:

 y=-4x^2  is the equation of this parabola.

Step-by-step explanation:

Let us consider the equation

y=-4x^2

\mathrm{Domain\:of\:}\:-4x^2\::\quad \begin{bmatrix}\mathrm{Solution:}\:&\:-\infty \:

\mathrm{Range\:of\:}-4x^2:\quad \begin{bmatrix}\mathrm{Solution:}\:&\:f\left(x\right)\le \:0\:\\ \:\mathrm{Interval\:Notation:}&\:(-\infty \:,\:0]\end{bmatrix}

\mathrm{Axis\:interception\:points\:of}\:-4x^2:\quad \mathrm{X\:Intercepts}:\:\left(0,\:0\right),\:\mathrm{Y\:Intercepts}:\:\left(0,\:0\right)

As

\mathrm{The\:vertex\:of\:an\:up-down\:facing\:parabola\:of\:the\:form}\:y=a\left(x-m\right)\left(x-n\right)

\mathrm{is\:the\:average\:of\:the\:zeros}\:x_v=\frac{m+n}{2}

y=-4x^2

\mathrm{The\:parabola\:params\:are:}

a=-4,\:m=0,\:n=0

x_v=\frac{m+n}{2}

x_v=\frac{0+0}{2}

x_v=0

\mathrm{Plug\:in}\:\:x_v=0\:\mathrm{to\:find\:the}\:y_v\:\mathrm{value}

y_v=-4\cdot \:0^2

y_v=0

Therefore, the parabola vertex is

\left(0,\:0\right)

\mathrm{If}\:a

\mathrm{If}\:a>0,\:\mathrm{then\:the\:vertex\:is\:a\:minimum\:value}

a=-4

\mathrm{Maximum}\space\left(0,\:0\right)

so,

\mathrm{Vertex\:of}\:-4x^2:\quad \mathrm{Maximum}\space\left(0,\:0\right)

Therefore,  y=-4x^2  is the equation of this parabola. The graph is also attached.

7 0
3 years ago
How do you graph the equation f(x)=10x+40??
irga5000 [103]
Ok so linear equations come in the form
y = mx + b

B is the y-intercept. In this equation the y intercept is 40. A y- intercept is where the graph crosses the y axis so at the point (0,40) the graph crosses the y axis.

M is the slope which is rise over run so if the slope was 10 (which it is) 10 is equivalent to 10/1 so you move up 10 units for every 1 unit you move across.

So to graph this equation, you would draw your first point at (0,40). For your next point, you would move right one unit, and up 10 units. Draw a point there which would be (1, 50). Hopefully you understand. For going left from the y intercept point, you would move left 1 and down 10.

Actual graph above.

Hope this helps C:

8 0
3 years ago
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