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2 years ago

Determine if the graph is symmetric about the x-axis, the y-axis, or the origin. r = 4 cos 5θ

Mathematics

1 answer:

Agata [3.3K]2 years ago

6 0

Answer:

It's symmetrical to the origin

Step-by-step explanation:

Use desmos for further graphing

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Can someone please help me with this? I'll mark brainliest for the right answer

lapo4ka [179]

Where’s the question?

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3 years ago

Best way to tell someone no without being mean

m_a_m_a [10]

Answer:

Try to search up euphemisms to help you in an argument where your forced to say no

Step-by-step explanation: In this situation apologize to them and say

I can't do that or no

6 0

2 years ago

If f(x) = x-6 and g(x)= 1/2x (x+3), find g(x) * f(x)

sertanlavr [38]

Answer:

Final answer is $g\left(x\right)\cdot f\left(x\right)=\frac{\left(x-6\right)}{2x\left(x+3\right)}$ .

Step-by-step explanation:

given functions are $f(x)=x-6$ and $g\left(x\right)=\frac{1}{2x\left(x+3\right)}$ .

Now we need to find about what is the value of $g\left(x\right)*f\left(x\right)$ .

$g\left(x\right)*f\left(x\right)$ simply means we need to multiply the value of $f(x)=x-6$ and $g\left(x\right)=\frac{1}{2x\left(x+3\right)}$ .

$g\left(x\right)\cdot f\left(x\right)=\frac{1}{2x\left(x+3\right)}\cdot\left(x-6\right)$

$g\left(x\right)\cdot f\left(x\right)=\frac{\left(x-6\right)}{2x\left(x+3\right)}$

Hence final answer is $g\left(x\right)\cdot f\left(x\right)=\frac{\left(x-6\right)}{2x\left(x+3\right)}$ .

5 0

3 years ago

The figure below shows the circle inside of a square if the radius of the circle Is 8cm find the following by showing the formul

Goryan [66]

A. The circumference of the circle:
C = 2 π r = 2 * 3.14 * 8 = 50.24 cm

B. The area of the circle:
A = π r² = 3.14 * 8² = 200.96 cm²

C. The area of the square:
Let's calculate first the side of the square. It is equal to 2r = 2*8=16 cm

The area of the square is:
16*16 = 256 cm²

Hope that helps.

7 0

3 years ago

Read 2 more answers

What is the inverse of f(x)= cube root of x +2.

Sergeu [11.5K]

$f(x) = \sqrt[3]{x + 2} \\y = \sqrt[3]{x + 2} \\y^{3} = x + 2 \\x^{3} = y + 2 \\x^{3} - 2 = y \\x^{3} - 2 = f^{-1}(x)$

3 0

3 years ago