Part a) Write the equation of the circle by completing the square.

Determine whether the statement is true or false. Justify your answers.

marusya05 [52]

Answer:

The graph of y = f(-x) is a reflection of the graph of y = f(x) in the x-axis. ⇒ False

The graph of y = -f(x) is a reflection of the graph of y = f(x) in the y-axis. ⇒ False

Step-by-step explanation:

<em>Let us explain the reflection about the axes</em>

If a graph is reflected about the x-axis, then the y-coordinates of all points on it will opposite in sign

Ex: if a point (2, -3) is on the graph of f(x), and f(x) is reflected about the x-axis, then the point will change to (2, 3)

That means reflection about the x-axis change the sign of y

y = f(x) → reflection about x-axis → y = -f(x)

If a graph is reflected about the y-axis, then the x-coordinates of all points on it will opposite in sign

Ex: if a point (-2, -5) is on the graph of f(x), and f(x) is reflected about the y-axis, then the point will change to (2, -5)

That means reflection about the y-axis change the sign of x

y = f(x) → reflection about y-axis → y = f(-x)

<em>Now let us answer our question</em>

The graph of y = f(-x) is a reflection of the graph of y = f(x) in the x-axis.

It is False because reflection about x-axis change sign of y

The graph of y = -f(x) is a reflection of the graph of y = f(x) in the x-axis

The graph of y = -f(x) is a reflection of the graph of y = f(x) in the y-axis.

It is False because reflection about y-axis change sign of x

The graph of y = f(-x) is a reflection of the graph of y = f(x) in the y-axis

7 0

3 years ago

How do you add 1/15 + 1/6?

Brut [27]

you are going to need to find the common name of common denominator 4/1 over 4 plus 1 over 6 equal to look for the commentator of 15 and 6 what goes into 15 and 6 is going to 15 and 6 the common denominator will be actually 12 to 6 only go into 15 2 times cuz I can't get close to 15 than q2 so you have to common denominator now this is going to be both 1/12 + 1/12 is going to be either 2/24 or 2/12

6 0

3 years ago

Read 2 more answers

The picture below shows a shaded rectangular region inside a large rectangle.

Pachacha [2.7K]

Answer:

the answer is d

Step-by-step explanation:

8 0

3 years ago

Justin’s recipe calls for 5 eggs. If each cupcake ends up with 1/4 of an egg, how many cakecakes will Justin’s recipe make?

ehidna [41]

Answer:

20 cupcakes

Step-by-step explanation:

You multiply 5 times 4 because their are 4/4 in each egg so for 5 eggs their would be 20 cupcakes because 5 times 4 equals 20.

8 0

3 years ago

The operations manager of a manufacturer of television remote controls wants to determine which batteries last the longest in hi

Margarita [4]

Answer:

$t=\frac{107.75-116.75}{\sqrt{\frac{2.75^2}{4}+\frac{9.604^2}{4}}}}=-1.802$

The degrees of freedom are given by:

$df=n_{1}+n_{2}-2=4+4-2=6$

The p value for this case would be given by:

$p_v =2*P(t_{(6)}$

Since the p value is higher than the significance level we have enough evidence to FAIl to reject the null hypothesis and we can conclude that the true mean is not significantly different between the two types of battery

Step-by-step explanation:

Information given

Battery 1 106 111 109 105

Battery 2 125 103 121 118

We can calculate the mean and the deviation with the following formulas"

$\bar X =\frac{\sum_{i=1}^n X_i}{n}$

$s =\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

$\bar X_{1}=107.75$ represent the mean for the Battery 1

$\bar X_{2}=116.75$ represent the mean for the Bettery 2

$s_{1}=2.75$ represent the sample standard deviation for the Battery 1

$s_{2}=9.604$ represent the sample standard deviation for the battery 2

$n_{1}=4$ sample size selected for the Battery 1

$n_{2}=4$ sample size selected for the Battery 2

$\alpha=0.1$ represent the significance level

t would represent the statistic

$p_v$ represent the p value

System of hypothesis

We want to check if the difference in longevity between the two batteries, the system of hypothesis would be:

Null hypothesis: $\mu_{1} = \mu_{2}$

Alternative hypothesis: $\mu_{1} \neq \mu_{2}$

The statistic is given by:

$t=\frac{\bar X_{s1}-\bar X_{2}}{\sqrt{\frac{s^2_{1}}{n_{1}}+\frac{s^2_{2}}{n_{2}}}}$ (1)

The statistic is given by:

$t=\frac{107.75-116.75}{\sqrt{\frac{2.75^2}{4}+\frac{9.604^2}{4}}}}=-1.802$

The degrees of freedom are given by:

$df=n_{1}+n_{2}-2=4+4-2=6$

The p value for this case would be given by:

$p_v =2*P(t_{(6)}$

Since the p value is higher than the significance level we have enough evidence to FAIl to reject the null hypothesis and we can conclude that the true mean is not significantly different between the two types of battery

8 0

3 years ago