All Answers 358483 Answers

Quartile is often a 25% portion of 100%  (1/4.....quartile) ...often spoken of when dealing with statistics and bell curves...

Here is Wikipedia description:

https://en.wikipedia.org/wiki/Interquartile_range      

1:  An equilateral triangle has a side of length 12 inches. What is the area of the triangle, in square inches? Express your answer in simplest radical form.

2:  In triangle ABC, AB = AC = 5 and BC = 6. Let O be the circumcenter of triangle ABC. Find the area of triangle OBC.

3:  The medians AD, BE, and CF of triangle ABC intersect at the centroid G. The line through G that is parallel to BC intersects AB and AC at M and N, respectively. If the area of triangle ABC is 810, then find the area of triangle AMN.

Hello "RainbowPanda" ???!!!.

If the two roots of the quadratic 3x2+5x+k are ( -5+i sqrt(11))/6, ( -5-i sqrt(11) )/6, what is k?

\(\text{Let $ x_1 = \dfrac{ -5+i \sqrt{11} }{6} $ } \\ \text{Let $ x_2 = \dfrac{ -5-i \sqrt{11} }{6} $ } \)

\(\begin{array}{|rcll|} \hline 3x^2+5x+k &=& 0 \quad & | \quad : 3 \\ x^2+\dfrac{5}{3}x+\underbrace{\dfrac{k}{3}}_{=x_1x_2} &=& 0 \\\\ \dfrac{k}{3} &=& x_1x_2 \\\\ \dfrac{k}{3} &=& \left( \dfrac{ -5+i \sqrt{11} }{6} \right) \left( \dfrac{ -5-i \sqrt{11} }{6} \right) \\\\ \dfrac{k}{3} &=& \dfrac{ (-5)^2-(i \sqrt{11})^2 }{36} \\\\ \dfrac{k}{3} &=& \dfrac{25-i^2 \cdot 11 }{36} \quad & | \quad i^2 = -1 \\\\ \dfrac{k}{3} &=& \dfrac{25-(-1) \cdot 11 }{36} \\\\ \dfrac{k}{3} &=& \dfrac{25+11 }{36} \\\\ \dfrac{k}{3} &=& \dfrac{36 }{36} \\\\ \dfrac{k}{3} &=& 1 \\\\ \mathbf{k} & \mathbf{=} & \mathbf{3} \\ \hline \end{array}\)

Thanks CPhill for your acknowledgement........

If the two roots of the quadratic 3x2+5x+k are -5+i sqrt(11)/6, -5-i sqrt(11)/6, what is k?

volume of pyramid  =  (1/3)(area of base)(altitude)

area of base  =  area of rhombus  =  (1/2)(10)(24)  =  120   square units

altitude  =  18

volume of pyramid  =  (1/3)(120)(18)

volume of pyramid  =  (1/3)(2160)

volume of pyramid  =  720   cubic units

You almost had it...maybe just a typo?  

1.

The function is constant when its slope is zero.

When the slope is zero, as the value of  x  changes, the value of  f(x)  doesn't change.

For instance, when  x = -2,  f(x) = 2 .  And when  x = -3, f(x) = 2 .

We changed the value of  x  but the value of  f(x)  did not change.

So the function is constant on the interval  (-4, 0)

2.

f(x)  =  2x³ − x²

To determine whether this function is even, odd, or neither, we need to find  f(-x) .

f(-x)  =  2(-x)³ – (-x)²

Since a negative number times a negative number is a positive number,  (-x)²  =  x²

f(-x)  =  2(-x)³ – x²

A negative number multiplied by itself  3  times is a negative number, so...

f(-x)  =  -2x³ – x²

This is not the same as the original function, and it is not the same as -1 times the original function, so this function is neither.

1.

y  =  f(x)

When  x  =  -8 ,  f(x) = 4  , and  4  is positive, so when  x = -8 , f(x) is positive.

The function is positive at any point above the x-axis.

The function is positive on the intervals   [-8, -6)  and  (-2, 4)

2.

The relative minimum is the y-coordinate of the low point on the graph.

The relative minimum is  -5 .

3.

The function is increasing when its slope is positive.

The function is increasing on the intervals  (0, 1)  and  (2, ∞)

There are a few ways to think about this....

△JKL  is similar to  △MNP  by AA similarity.

So each side of  △MNP  is some scale factor times the corresponding side of  △JKL  .

MN  =  JK * s    where  s  is the scale factor.

30  =  25 * s

30 / 25  =  s

So each side of  △MNP  is  30/25 times the corresponding side of  △JKL  .

MP  =  30/25 * JL

4x - 4  =  30/25 * 20

4x - 4  =  24

I'll let you finish solving for  x . 

Thanks so much!

\(w(x) = \sqrt{x-2} + \sqrt[3]{x-3}~\)

The domain is that which is the most restrictive for both parts of the function

The second term  has a domain of all real numbers because we can take the cube root of all real numbers

However.....the frist term is only defined from  [2, inf)

So.....this is the domain

(4, infinity)

The diagonals of a rhombus bisect each other.......this bisection forms a right triangle with each leg (1/2) the length of its respective diagonal.....the hypotenuse forms one of the equal sides

So.....the perimeter is given by  

4√ [ 9^2 + 6^2  ] =

4 √117  =

4 √ (9 * 13 )  =

12√13 ft

f(x) = x² + 12

f(m) = m² + 12

f(3m) = (3m)² + 12 → 9m² + 12

f(3m) = 3f(m)

9m² + 12 = 3(m² + 12)

9m² + 12 = 3m² + 36

6m² = 24

m² = 4

m = 2  (or -2) 

.

Thanks so much!

Scien not for numerator  256  x 10^-5

Scien Not.for denom         16 x 10^1                16 goes in to 256  16 times

Result =  16 x 10^(-5-1)  = 16 x 10^-6    = .000016

I looked this up:

The properties of the trapezoid are as follows:

The bases are parallel by definition.

Each lower base angle is supplementary to the upper base angle on the same side.

SO angle a + angle d = 180

         angle a = 2d    substitute

              2d +d = 180       d= 60

    a= 2d = 120 degrees                                    This assumes your trapezoid is labeled    A            B

                                                                                                                                          D                      C

Nothing much just asking who's good at percent cause I am

I think he means 95 out of 100. Good job jeff! I had my test the same day you did, and i got a 95% to!

What is that in interval notation?