All Answers 358483 Answers

There are four solutions: 5, -5, 5i, and -5i.

Find CM.

Hello Guest!

CM is x.

\(MA=\sqrt{x^2+8^2-2\cdot 8x\cdot cos\ 45^{\circ}}\\ MB=\sqrt{x^2+4^2-2\cdot 4x\cdot cos\ 45^{\circ}}\\ MA+MB=\sqrt{4^2+8^2}=\sqrt{80}\\ \sqrt{x^2+8^2-2\cdot 8x\cdot cos\ 45^{\circ}}+\sqrt{x^2+4^2-2\cdot 4x\cdot cos\ 45^{\circ}}=\sqrt{80}\\ {\color{blue}CM=x=\dfrac{8\sqrt{2}}{3}}=3.771\ (WolframAlpha)\)

  !

Draw a sketch.

You want the perpendicular distance from the chord to the circle. (Which will bisect the chord)

Use Pythatoras's theorem.

The question is not complete

Here's one way:

\(\$3.69=\dfrac{x\cdot 90}{100}\\ x=\dfrac{\$3.69\cdot 100}{90}=\$4.10\)

The price of a gallon of gasoline one year ago was $4.10.

  !

0.4 is mulitipled by something to make 4.  What is the something?

Hello Guest!

0.4 is mulitipled by 10 to make 4.
The something is 10.

  !

Nice work asinus.  And guest too of course. :)

In △ABC△ABC, we have AB=14AB=14, BC=16BC=16, and ∠A=60∘∠A=60∘. Find the sum of all possible values of ACAC.

I made an embarrassing mistake and I apologize.

Here is my correct solution:

\(y = \frac{1}{\sqrt{2}} (x^2 - 13)\\ r^2=x^2+y^2\\ r^2=x^2+\frac{1}{2}(x^2-13)^2\\ \frac{d(r^2)}{dx}=2x+\frac{1}{2}\cdot 2(x^2-13)\cdot 2x=0\\ 2x+2x^3-26x=0\\ 2x^3-24x=0\\ x(x^2-12)=0\)

\(x\in \{0,-\sqrt{12},{\color{blue}\sqrt{12}}\}\\ r^2=x^2+\frac{1}{2}(x^2-13)^2\\ r^2=12+\frac{1}{2}(12-13)^2\\ r^2=12.5\\ \color{blue}r=\sqrt{12.5}=3.5355 \)

This is Guest 2#'s solution.

Sorry again!

  !

If I understand the question properly the team will consist of Grogg and Alex and 3 others from a group of 7

That will be  7C3 ways

PM = 4*sqrt(6)

The area is 13/6*pi

The remainder is 1

BZ = 2*sqrt(11)

How many ways are there to divide a group of 6 friends among the basketball team, the soccer team, and the track team? (Each team could have anywhere from 0 to 6 of the friends on it. Assume the friends are distinguishable.)

Possibilities

0,0,6          1 way to seperate. 3 alternatives for teams  =3

0,1,5         6C1 way to seperate. 3! alternatives for teams  =36

0,2,4         6C2  way to seperate. 3! alternatives for teams  =36

0,3,3         6C3/2  way to seperate. 3! alternatives for teams  =18

1,1,4         6C4  way to seperate. 3! alternatives for teams  =90

1,2,3         6C3*3  way to seperate. 3! alternatives for teams  =360

2,2,2         6C2*4C2/6  way to seperate. 3! alternatives for teams  =90

3+36+36+18+90+360+90 = 633

I could easily have made a mistake.  If you get the correct answer, please share it here.

See https://web2.0calc.com/questions/probability-of-the-game-set-search-it-up-if-you-don

I'd start by considering factors of 24    

plus or minus  1,2,3,4,6,8,12,24     That's a lot  16 in total

Now trial and error to see which one will divide in perfectly.

\(f(x)=x⁴-2x³-13x²+38x-24\\ f(1)=1-2-13+38-24 =39-39=0\;\;great\\ (x-1)\text{ is a factor} \)

Now I would divide f(x) by  x-1    to get the dividend then start the process again.

OR you could just look for another possibility out of the list.

How did you get that? I got -68 since (x-1)=1 and then I plugged that in for x.

5 + what equals -12  

(5)  +  (–17)  =  –12  

As you see above, the answer is negative seventeen.  

_.