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3 years ago

Determine if the graph is symmetric about the x-axis, the y-axis, or the origin.

Mathematics

1 answer:

iren [92.7K]3 years ago

8 0

Answer:

y-acis

Step-by-step explanation:

the function graph is symmetric about

- y-axis when it is an even function

-the origin ehen it is an even function

A symmetrical graph about the x-axis is not a function graph

f(×) is a even if and only if f(×) =f(×)

f(×) is a odd if and only f(×)=f(×)

We have the function r(0) = 4cos (50)

(only symmetry about the y-acis or about the origin)

Check r(-0)

r(-0) = 4cos (5-0) = 4cos (-50 = 4 cos (50)

Used cos (-× = cos ×

We have r(0). Therefore the graph of r(0) is symmectric about the y-axis.

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The equation y = 1/5x represents a proportional relationship. Explain how you can tell the relationship is proportional from the

andre [41]

If a graph is proportional then the line will go through the origin at point (0, 0). If the equation is proportional then it will be in the form of y=kx with no other operations after. The constant of proportionality is another way to say the slope and in your specific equation the slope would be 1/5.

7 0

3 years ago

Help Me Please ! This is so confusing .‍♀️

Savatey [412]

Answer: -4

Step-by-step explanation:

3-9=-6

-6*5=-30

-30-1=-31

6*-31=-186

-186+2=-184

8-2=6

6*7=42

42+4=46.

-184/46=-4

6 0

3 years ago

Which of the following are coefficient

malfutka [58]

Answer:

2 and -3

Step-by-step explanation:

because a coefficient is a numerical or constant quantity placed before and multiplying the variable in an algebraic expression (e.g. 4 in 4x).

so in 2x-3y+5, 2 and -3 are the coefficients.

4 0

3 years ago

What is the solution of

kobusy [5.1K]

Answer:

Third option: x=0 and x=16

Step-by-step explanation:

$\sqrt{2x+4}-\sqrt{x}=2$

Isolating √(2x+4): Addind √x both sides of the equation:

$\sqrt{2x+4}-\sqrt{x}+\sqrt{x}=2+\sqrt{x}\\ \sqrt{2x+4}=2+\sqrt{x}$

Squaring both sides of the equation:

$(\sqrt{2x+4})^{2}=(2+\sqrt{x})^{2}$

Simplifying on the left side, and applying on the right side the formula:

$(a+b)^{2}=a^{2}+2ab+b^{2}; a=2, b=\sqrt{x}$

$2x+4=(2)^{2}+2(2)(\sqrt{x})+(\sqrt{x})^{2}\\ 2x+4=4+4\sqrt{x}+x$

Isolating the term with √x on the right side of the equation: Subtracting 4 and x from both sides of the equation:

$2x+4-4-x=4+4\sqrt{x}+x-4-x\\ x=4\sqrt{x}$

Squaring both sides of the equation:

$(x)^{2}=(4\sqrt{x})^{2}\\ x^{2}=(4)^{2}(\sqrt{x})^{2}\\ x^{2}=16 x$

This is a quadratic equation. Equaling to zero: Subtract 16x from both sides of the equation:

$x^{2}-16x=16x-16x\\ x^{2}-16x=0$

Factoring: Common factor x:

x (x-16)=0

Two solutions:

1) x=0

2) x-16=0

Solving for x: Adding 16 both sides of the equation:

x-16+16=0+16

x=16

Let's prove the solutions in the orignal equation:

1) x=0:

$\sqrt{2x+4}-\sqrt{x}=2\\ \sqrt{2(0)+4}-\sqrt{0}=2\\ \sqrt{0+4}-0=2\\ \sqrt{4}=2\\ 2=2$

x=0 is a solution

2) x=16

$\sqrt{2x+4}-\sqrt{x}=2\\ \sqrt{2(16)+4}-\sqrt{16}=2\\ \sqrt{32+4}-4=2\\ \sqrt{36}-4=2\\ 6-4=2\\ 2=2$

x=16 is a solution

Then the solutions are x=0 and x=16

5 0

3 years ago

4 pts

Aliun [14]

Answer:

Step-by-step explanation:

So, this is a percentage problem.

Start off by finding how many students 0.28% is:

If 100% = 5780

0.01% = 0.578

Now:

0.01% = 0.578

0.28% = 16.184

The exercise tells you to round for a whole person, so 16.184 turns 17

And that's the answer!

5 0

3 years ago