Really need help on this graph problem please?

A certain circle can be represented by the following equation. x^2+y^2+18x+14y+105=0 What is the center of the circle? What is t

Gre4nikov [31]

Answer:

See below.

Step-by-step explanation:

Recall the equation for a circle:

$(x-h)^2+(y-k)^2=r^2$ , where (h,k) is the center and r is the radius.

We need to turn the given equation into the above format. We do this by completing the square.

First, group them:

$(x^2+18x)+(y^2+14y)=-105$

For the first section, complete the square for $x^2+18x$ :

$x^2+18x+81-81$

$(x+9)^2-81$

Do the same for the second:

$y^2+14y$

$y^2+14y+49-49$

$(y+7)^2-49$

All together:

$((x+9)^2-81)+((y+7)^2-49)=-105$

$(x+9)^2+(y+7)^2-130=-105$

$(x+9)^2+(y+7)^2=25$

The center is (-9,-7).

The radius is $\sqrt{25}=5$

5 0

3 years ago

Whats the answer to the question? A B C or D?

Yuri [45]

Answer:

a

Step-by-step explanation:

4 0

2 years ago

Sara has a gift box with a volume of 64 in3. The length, width, and height of the box are all y inches. Sara needs to determine

antoniya [11.8K]

Answer:

No, the note papers will not fit the gift box.

Step-by-step explanation:

Given that:

Volume of the gift box = 64 $in^3$

Length, width and height of the box all are equal to $y$ inches.

Dimensions of note papers = 6 inch by 6 inch

To determine:

Whether the note papers will fit the bottom of the box?

Solution:

Here, we need to find the dimensions of the gift box first.

If the dimensions of gift box are greater than or equal to the dimensions of the note papers, only then the papers will fit into the box.

As per the given question statement, the gift box is in the form of a cube.

Formula for volume of a cube is given as:

$Volume = (Side)^3$

Side is given equal to $y$ inches.

$64 = y^3\\\Rightarrow y = 4\ in$

Dimensions of the box are lesser than that of the note papers.

Therefore, <em>the note papers will not fit into the gift box</em>.

6 0

2 years ago

Which function describes the arithmetic sequence shown?

Lynna [10]

f(x)= -2x-3

Step-by-step explanation:

Step 1:

Let the sequence given here is -5, -7, -9, -11, -13 ......

Here the first term (a₁) of sequence is -5

And the common difference between the numbers in the sequence is

d= (-7-(-5)) = -7+5 = -2

Let the number of terms be x

Step 2;

To find the sequence function basic arithmetic sequence formula is

aₙ = a₁ + d( x-1)

Applying the values we get

f(X) = -5 + ((-2)(X-1))

on simplification

f(X) = -5 + (-2X+2)

f(X)= -5+2-2X

f(X)= -3-2X

8 0

3 years ago

Suppose a baker claims that the average bread height is more than 15cm. Several of this customers do not believe him. To persuad

laila [671]

Answer:

$17-2.262\frac{1.9}{\sqrt{10}}=15.641$

$17+2.262\frac{1.9}{\sqrt{10}}=18.359$

So on this case the 95% confidence interval would be given by (15.641;18.359)

And since the lower limit for the confidence interval is higher than 15 we can conclude that at 5% of significance the true mean is higher than 15 cm

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X=17$ represent the sample mean

$\mu$ population mean (variable of interest)

s=1.9 represent the sample standard deviation

n=10 represent the sample size

Confidence interval

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=10-1=9$

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,9)".And we see that $t_{\alpha/2}=2.262$

Now we have everything in order to replace into formula (1):

$17-2.262\frac{1.9}{\sqrt{10}}=15.641$

$17+2.262\frac{1.9}{\sqrt{10}}=18.359$

So on this case the 95% confidence interval would be given by (15.641;18.359)

And since the lower limit for the confidence interval is higher than 15 we can conclude that at 5% of significance the true mean is higher than 15 cm

4 0

3 years ago