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2 years ago

Use the IQR to determine if any of the data in the data set are outliers.

Mathematics

1 answer:

elena-s [515]2 years ago

5 0

Answer:

Step-by-step explanation:

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How would you solve this problem?<br><br> ((2x^12y^3)/(x^10y))^-5

Hatshy [7]

There is no equals sign. Are we looking for x, or y?

Either way, with the -5 exponent outside the parentheses, you can multiply that to each exponent in the fraction to get rid of it.

I'm not sure what you mean by 2x^12y^3. Is it an exponent within another exponent?

7 0

3 years ago

How do you find the Area of a pentagon

Morgarella [4.7K]

Answer:

Area of a regular pentagon = pa/2, where p = the perimeter and a = the apothem. If you don't know the perimeter, calculate it from the side length: p = 5s, where s is the side length.

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers

Write the following quotient in the form a+bi.<br> -7i/2-7i

Helga [31]

Answer:

$\large\boxed{\dfrac{49}{53}-\dfrac{14}{53}i}$

Step-by-step explanation:

$\text{Use}\\\\i=\sqrt{-1}\to i^2=-1\\\\(a-b)(a+b)=a^2-b^2\to(a-bi)(a+bi)=a^2+b^2\\\\\text{distributive property}\ a(b+c)=ab+ac\\---------------------\\\\\dfrac{-7i}{2-7i}=\dfrac{-7i}{2-7i}\cdot\dfrac{2+7i}{2+7i}=\dfrac{-7i(2+7i)}{2^2+7^2}=\dfrac{(-7i)(2)+(-7i)(7i)}{4+49}\\\\=\dfrac{-14i-49i^2}{53}=\dfrac{-14i-49(-1)}{53}=\dfrac{-14i+49}{53}\\\\=\dfrac{49}{53}-\dfrac{14}{53}i$

8 0

3 years ago

Read 2 more answers

Twice the sum of a number and 7 is more than 6 or at most -3. Which compound inequality can represent this relationship

yarga [219]

Answer:

Step-by-step explanation:

6 0

3 years ago

Find the radius of the circle with equation x²+y²+8x+8y+28=0

galina1969 [7]

<h2>Hello!</h2>

The answer is:

Center: (-4,-4)

Radius: 2 units.

To solve the problem, using the given formula of a circle, we need to find its standard equation form which is equal to:

$(x-h)^{2}+(y-k)^{2}=r^{2}$

Where:

"h" and "k" are the coordinates of the center of the circle and "r" is its radius.

So, we need to complete the square for both variable "x" and "y".

The given equation is:

$x^{2}+y^{2}+8x+8y+28=0$

So, solving we have:

$x^{2}+y^{2}+8x+8y=-28$

$(x^2+8x+(\frac{8}{2})^{2})+(y^2+8y+(\frac{8}{2})^{2})=-28+((\frac{8}{2})^{2})++(\frac{8}{2})^{2})\\\\(x^2+8x+16 )+(y^2+8y+16)=-28+16+16\\\\(x^2+4)+(y^2+4)=4$

$(x^2-(-4))+(y^2-(-4))=4$

Now, we have that:

$h=-4\\k=-4\\r=\sqrt{4}=2$

So,

Center: (-4,-4)

Radius: 2 units.

Have a nice day!

Note: I have attached a picture for better understanding.

3 0

2 years ago