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IceJOKER [234]
2 years ago
9

Test the function x-y²=0 for symmetries.

Mathematics
1 answer:
mafiozo [28]2 years ago
4 0

For Symmetry About Y-Axis, let's try to replace x with −x, and check if the equations still equal:

-x - y² = 0

The equation changed, so there is no symmetry about Y-Axis.

For Symmetry About X-Axis, let's try to replace y with −y, and check if the equations still equal:

x - (-y)² = 0

x - y² = 0

The equation stayed the same, so there is a symmetry in X-Axis.

Lets try Origin Symmetry Replacing x with −x and y with −y:

-x - (-y)² = 0

-x - y² = 0

The equation changed, so there is no Origin Symmetry.

You might be interested in
Is Garrett's slope correct? If not, identify his error?
dimulka [17.4K]

Answer:

B. No. He should have put the x values in the

denominator and the y values in the numerator.

Step-by-step explanation:

Took the assignment

8 0
3 years ago
Read 2 more answers
Question 2
BabaBlast [244]

Answer:

The equation of the required line is y = x + 5

Step-by-step explanation:

The equation of the given line is y = x - 2

The required line = A line parallel to the given line

The point through which the required line passes = (-3, 2)

The general form of the equation of a straight line, is y = m·x + c

Where;

m = The slope of the line

By comparison, the slope of the given line, m = 1

When two lines are parallel, their slope are equal

Therefore, the slope of the required line = m = 1

The equation of the required line in point and slope form is therefore;

y - 2 = x - (-3) = x + 3

∴y = x + 3 + 2 = x + 5

The equation of the required line is therefore;

y = x + 5.

6 0
3 years ago
Please need help thank you
valentinak56 [21]
Might be 1? Because it is 100 each
5 0
3 years ago
The weight of the chocolate and Hershey Kisses are normally distributed with a mean of 4.5338 G and a standard deviation of 0.10
Salsk061 [2.6K]

For the bell-shaped graph of the normal distribution of weights of Hershey kisses, the area under the curve is 1, the value of the median and mode both is 4.5338 G and the value of variance is 0.0108.

In the given question,

The weight of the chocolate and Hershey Kisses are normally distributed with a mean of 4.5338 G and a standard deviation of 0.1039 G.

We have to find the answer of many question we solve the question one by one.

From the question;

Mean(μ) = 4.5338 G

Standard Deviation(σ) = 0.1039 G

(a) We have to find for the bell-shaped graph of the normal distribution of weights of Hershey kisses what is the area under the curve.

As we know that when the mean is 0 and a standard deviation is 1 then it is known as normal distribution.

So area under the bell shaped curve will be

\int\limits^{\infty}_{-\infty} {f(x)} \, dx= 1

This shows that that the total area of under the curve.

(b) We have to find the median.

In the normal distribution mean, median both are same. So the value of median equal to the value of mean.

As we know that the value of mean is 4.5338 G.

So the value of median is also 4.5338 G.

(c) We have to find the mode.

In the normal distribution mean, mode both are same. So the value of mode equal to the value of mean.

As we know that the value of mean is 4.5338 G.

So the value of mode is also 4.5338 G.

(d) we have to find the value of variance.

The value of variance is equal to the square of standard deviation.

So Variance = (0.1039)^2

Variance = 0.0108

Hence, the value of variance is 0.0108.

To learn more about normally distribution link is here

brainly.com/question/15103234

#SPJ1

3 0
1 year ago
If cosu = 30/34 find sin2u
Tomtit [17]

Answer:

es tal vez otra vez se repite la pregunta

3 0
3 years ago
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