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Hi Melody,

A square matrix is singular if its determinant is zero (0). (There is no inverse for this matrix.)

Note: This definition applies only if the matrix is square.

GA

--. .-

If I had to, I would do it by matrix elimination.  

Can anyone point out to me why the matrix calculator would not give me an answer?

What is the significance of a matrix being singular?

Very good question.

If we express everything in terms of roots, we get \(P = \sqrt[5]{5} \cdot \sqrt[25]{25} \cdot \sqrt[125]{125} \cdot \dots \)

\(P = \sqrt[5]{5} \cdot \sqrt[5]{\sqrt[5]{25}} \cdot \sqrt[5]{\sqrt[5]{\sqrt[5]{125}}} \cdot \dots\)

Let 5th root of 5 = x.

\(P = x \cdot \sqrt[5]{x^2} \cdot \sqrt[5]{\sqrt[5]{x^3}} \cdot \dots\)

Try to base your solution off of the fact that \(P^{1/5} = x^{1/5} \cdot x^{2/25} \cdot x^{3/125} \cdot \dots\).

If you need more help, just reply.

You are correct that a=5.

\(3 \cdot -4 + 2 \cdot 7 + -5 \cdot -1 + -3 \cdot -1 + 2 \cdot -4 = \boxed{2}\)

The three vertices you specified don't make a rectangle.

Assuming you mean (6,-2):

We can split this region into a circular segment an isosceles triangle. The area of the triangle is 2. I don't have the time to write my process down, but it involved coord bashing and heron's formula. 

The angle of the circular sector is arccos(√15/4), so the degrees value is arccos(√15/4)*180/pi. Divide that by 360, we get that the area is (arccos(√15/4)*180/pi)/360 ths of the whole circle's area, or 16 pi * (arccos(√15/4)*180/pi)/360. Subtract 2, we get \(8 \cdot \text{arccos}\left(\frac{\sqrt15}{4}\right) -2 \text{, or } 4 \pi - 8 \cdot \text{arctan}(\sqrt{15}) - 2\)
It's pretty likely that I'm wrong, but perhaps you can identify my mistake by looking over the steps.

https://www.desmos.com/calculator/uys2utn9vh

g(x)  =  2x + 1     --->     g (g(x) )  =  2( 2x + 1 ) + 1  =  4x + 2 + 1     --->     4x + 3

g( g(x) ) - g(x)  =  [ 4x + 3 ] - [2x + 1 ]  =  4x + 3 - 2x - 1  =  2x + 2

f(3)  =  2(3) + 2  =  6 + 2  =  8

Why's he not online anymore?

Step-by-step explanation

Given

Degree of f(x) = 3

Degree of g(x) = 5

Required

Degree of 2f(x) + 4g(x)

Analyzing both polynomials

f(x)

2f(x) means 2 * f(x)

Since 2 is a constant

Multiplying f(x) by 2 will result in a polynomial with a degree of 3

Hence 2f(x) has a degree of 3

g(x)

4g(x) means 4 * g(x)

Since 4 is also a constant

Multiplying g(x) by 4 will result in a polynomial with a degree of 5

Hence 4g(x) has a degree of 5

Having said that;

When 2 polynomials of different degrees are added together, the degree of the result will be the higher degree of both polynomials;

This means that;

Adding a polynomial of degree 3 and another of degree 5 will result in a polynomial of degree 5.

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From https://web2.0calc.com/questions/rectangle-abcd-is-the-base-of-pyramid-pabcd-if-ab

This is a funny lookin pyramid! The formula for the volume of this pyramid is the same as if "P" were above the center of the base:  volume = (1/3)(area of base)(height)

This helped me verify that.

Draw AC.

Now look at triangle ABC.  From the Pythagorean theorem....

AC2   =   22 + 32   =   4 + 9   =   13

AC is in the same plane as AD and AB, so PA is perpendicular to AC.

Look at triangle PAC.  From the Pythagorean theorem again....

AC2 + PA2  =  52

13  +  PA2  =  25

PA2  =  12

PA  =  √12

PA  =  2√3

And....

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

                               =  (1/3)(  2 * 3  )( PA )

                               =  (1/3)(6)(2√3)

                               =  4√3     cubic units

I just remembered how my math teacher explained that the volume of an oblique cylinder is the same as the volume of a right cylinder (with the same height and base).

Imagine a stack of pennies. If you line the stack straight up, it is a right cylinder. And if you lean the stack over, it is oblique. But the height, area of the base, and the volume stays the same.

And in this case, imagine a stack of square pieces of cardboard that get smaller as you go higher. You can stack the squares up to form a right pyramid, or you can slide them to look like the pyramid in this problem. The volume stays the same. 

According to the problem A is the right angle. 180 - 90 = 90. 90/2 = 45 degrees according to me.

Answer

B = 45 degrees

C = 45 degrees

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I used the first 6 values (n = 2, ... n = 7) to find the values for a, ... f in the equation:

     y  =  ax5 + bx4 + cx3 + dx2 + ex + f

to get these values:

    a  = -7.192460317 x 10-4

    b  =  0.01875

    c  =  -0.1948164682

    d  =  1.01875

    e  =  -2.750892857

    f  =  3.375

If these are correct, they do show that 8/63 is not correct for n = 8.

If AB has length 4 and AC has length 4, in a right triangle, what is the measure of angle B and C in degrees?  

Since it's a right triangle and the two legs are equal, the two acute angles each equals 45^o.  

_.

120

We know that :  m

So, 180-60=120

I didn't understand the problem, so it may be wrong. Can you please post it again with a more understandable question?

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I'll start....you finish

(7+8x)-3(2+6x+2x^2)+8(1+3x+4x^2+12x^3)+9(7-x^2-8x^3+13x^4)

    8x       - 18x                  + 24x                       = _______________

Angle R is also 60 degrees.

Here’s a faster solution method:

\(\large \text{half-life} = \large \dfrac{29} {Log_2(6272) – Log_2(196)} = 5.8 \text { minutes} \)

GA

--. .-

Converting everything to base 10:

49A + 7A + A = 57A

57A + 57B = 228.

This means that A+B=4. Since A and B are positive integers:

(1,3)

(2,2)

(3,1)

Combine like terms. 

-2x^2 + 32x - 25

Factor out a -2.

-2(x^2 - 16x) - 25

Complete the square.

-2(x^2-16x+64) - 25 + ?

To balance the equation, ? = 2x64 = 128.

Thus, our answer is -2(x-8)^2 + 103, so -2-8+103 = 93.

The horizontal asymptote of a rational function with the degree of the numerator and denominator equal is the ratio of the coefficients. 1/1 = 1.

The coefficients don't change the degree. The degree is just 6.

Geogebra is very useful for curve fitting, nice idea also with the matrix solver. Using wolframalpha isn't very intuitive (generic).

I wonder if curve fitting may be useful. I'm using a brute force method, so if there is a better solution please answer!

Let the polynomial be p(x) = ax^5 + bx^4 + cx^3 + dx^2 + ex + f.

We will set up a system of equations now.

2/(2^2-1) = 2/3 = 32a + 16b + 8c + 4d + 2e + f

3/8 = 243a + 81b + 27c + 9d + 3e + f

4/15 = 4^5(a) + 4^4(b) + etc...

You probably see the idea by now. I'm certain this is the tedious method since one wouldn't want to calculate 7^5 by hand, let alone solve a system of equations with monstrous coefficients like these.

Even wolframalpha doesn't accept the equations...

Volume of a tetrahedron =   sqrt (3) / 8   *  h³

Hello, thanks for your correction. I have improved it.

Best regards

  !

At first glance many would say that    p(8)=8/63 which is approx  0.127  but that is not going to be correct.

I solved it graphically, though the answer may not be exact.

Here is the link,  It says the polynomial is of degree 4 not 5.  (Even though I specifically asked for degree 5)

I can't check it easily because the coefficients are rounded to 2 decimal places.