All Answers 353121 Answers

"A"  is correct....

If ABC  is equilateral and AB  = 2....then the perimeter    = 6

"D'   i s the correct  choice....

Let  B  =(0,0)   A  = (0,6)   and C  = (8,0)   M  =(0,3)   N  = (4,0)

Let  the slope of MC  = [ 3 -0 ]  / [ 0 - 8]  =  -3/8

And the equation of MC is   y  = -(3/8) x + 3     (1)

Let the slpoe  of AN  = [ 6 - 0 ] / [ 0 - 4 ] = -6/4  = -3/2

And the  equation of An  is   y  = (-3/2)x + 6    (2)

Set (1)  = (2)  to find the x coordinate of P

(-3/8)x + 3  = (-3/2)x + 6

(-3/8 + 3/2)x  = 3

(9/8)x  = 3

24/9 = x  = 8/3

So...the y  coordinate of P  is  (-3/2)(8/3)  + 6 =  -24/6 + 6=  -4 + 6  = 2  = altitude of triangle PNC

The area of triangle APC  = area of triangle ANC  - area of triangle  PNC

Area of ANC  =(1/2) (AB) (NC)  = (1/2)(6)(4)   = 12  units^2

Area  of PNC  = (1/2) (2)(NC) = (1/2) (2) (4)  =4  units^2

So...area of triangle APC =  12 -  4  = 8 units^2

If AK  is an altitude  drawn to  base  BC.....then angles AKB  and AKC  form  right angles

So......using the Pythagorean  Theorem......

√[AK^2 + BK^2] =  AB  = √ [ 6^2 + 8^2  ] =  √100  = 10  units

Ans

√] AK^2 + CK^2 ] =  AC  = √ [ 6^2 + 6^2 ] = √72  = 6√2  units

And  CK + BK  = BC  =  6 + 8  = 14

So...the perimeter is  [ 10 + 6√2 + 14  ] =   24 + 6√2  units

Point \(P\) is inside equilateral triangle \(ABC\) such that the altitudes from \(P\) to \(\overline{AB}, \overline{BC}, \text{and} \ \overline{CA}\) have lengths 5, 6, and 7 respectively. What is the area of triangle \(ABC\). 

Connect \(P\) to \(A,B, \text{and } C\), to form \(\triangle ABP, \triangle ACP, \triangle BCP.\)

\([ABC]=[ABP]+[ACP]+[BCP]=\frac12x(5+6+7).\)

The area of \(\triangle ABC\) can also be represented as \(\frac{\sqrt3}{4}x^2\).

Therefore: \(\frac{\sqrt3}{4}x^2=\frac12x(5+6+7)\Rightarrow x=12\sqrt3\).

Plugging \(x\) into one of our area formulas, \(\frac{\sqrt3}{4}x^2=\frac12x(5+6+7)=108\sqrt3\).

I hope this helped,

Gavin. 

picture 

What  is  " P ??? "

33 + 24  + 8  =  65

Thus....this would be 15 students too many

So...15  play both

33 - 15  = 18  play only hockey

24 - 15  = 9   play only  baseball

18 + 15   = 33

9 + 15  = 24

:

"A certain integer has 4 digits when written in base 8. The same integer has \(d\) digits when written in base 2. What is the sum of all possible values of \(d\)?"

Four-digits base 8 numbers range from 1000₈ - 7777_8.

The range represented into base 2 is 1000000000₂ - 111111111111₂.

Therefore, the minimum is 10 digits, and the maximum is 12. 

I hope this helped,

gavin. 

Ginger, i have to agree with the guest- you are very quick to judge others, and a large portion of your posts is dedicated to bashing other users, some with wrong answers and some with perfectly fine answers. There is no need to attack other users for their answers when their mistakes can be explained to them calmly, but it seems that you just want to cuss them out and to ridicule them.

We all make mistakes, including you, and it's very hypocritical for you to attack others for getting math problems wrong when your posts show that you make mistakes too. The guest gave a good example.

The guest in this thread was wrong- he got the question wrong and gave a wrong solution to the question he thought RBAM was referring to, but there is no need to verbally attack him for that. Keep in mind that many users here are high school students- they're here to learn, and cussing them isn't going to help anyone, its just going to lower their self esteem, and bashing students in the head for failing a challenge or slipping up is not going to encourage them to keep learning.

please take this into consideration

\(\frac{10}{7}\div4=\frac{10}{7}\cdot\frac14=\frac{10}{28}=\frac{5}{7}\)

I remember that post, I was the one who commented on it. Ginger’s twin responses were because of my comments. Though I suppose she might have made similar comments anyway.

 “A home run on an error is still a home run.” LMAO! This is truly a great retort.

What’s your point? What does this have to do with anything here?

Ginger speaks: Yep I screwed up. I neglected the caveat of “at least” in the question.

Ginger Troll speaks: I didn’t post a wrong answer. I correctly answered the wrong question. A home run on an error is still a home run. laugh

GA

GingerAle  Jul 19, 2018
https://web2.0calc.com/questions/help-please_81710#r4

Maybe it was one neg for each comment.

^{Why in the h**l don't you post your own "correct" answer and shut up??!!}

It’s much more beneficial (not forgetting enjoyable) for the forum to give you heII for your, now confirmed, BatShit stupidity, than it is to give a one-off answer to a basic high school (Jr. high) question--that I’ve answered at least once before. You know… give someone a fish and you feed him for a day, stop someone from grazing on BS and you keep him from getting sick. Unless that someone is a dung beetle larva, then it’s nutritious--which is probably why you are full of it.  

^{What the h**l do you think makes you an "expert" in math?}

I don’t think any such thing. It just seems that way to you because you are so poor in the subject. You really can’t see things you don’t understand. Some will latch on to ideas they do understand and erroneously apply them. From this, all sorts of convoluted illogicality arises, and it becomes a way of life.  You understand BS, so that is what you apply, and the way of your life is that of a bullshit bug.

Confirmation of BatShit stupidity:

Divisors of 60 =1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60
Prime "factors" of 60 = 2^2×3×5 (4 prime factors, 3 distinct)
Do you see the difference?? A "divisor" such as 12, or 30 above may have MANY prime "factors". But a "factor", such as 2, 3, 5 above NEVER has a "divisor"!!

I need a babble translator to figure out most of what you wrote.  Though I now understand how you arrived at such an absurd solution. Yep, you should definitely abstain from any and all “math by consensus” votes. 

GA

The function f(n) = 3f(n-2) - 2f(n-1), where f(2) = 3 and f(1) = -1. What is the value of f(5)?

Formula:

\(\begin{array}{|rcll|} \hline f_n &=& -2f_{n-1}+3f_{n-2} \quad & | \quad f_2 = 3 \qquad f_1 = -1 \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline f_3 &=& -2f_{2}+3f_{1} \quad & | \quad f_2 = 3 \qquad f_1 = -1 \\ &=& -2(3)+3(-1) \\ &=& -6-3 \\ \mathbf{f_3} & \mathbf{=} & \mathbf{-9} \\\\ \hline f_4 &=& -2f_{3}+3f_{2} \quad & | \quad f_3 = -9 \qquad f_2 = 3 \\ &=& -2(-9)+3(3) \\ &=& 18+9 \\ \mathbf{f_4} & \mathbf{=} & \mathbf{27} \\\\ \hline f_5 &=& -2f_{4}+3f_{3} \quad & | \quad f_4 = 27 \qquad f_3 = -9 \\ &=& -2(27)+3(-9) \\ &=& -54 -27 \\ \mathbf{f_5} & \mathbf{=} & \mathbf{-81} \\ \hline \end{array}\)

\(\mathbf{f(5) = -81}\)

Formula:

\(\begin{array}{|lrcll|} \hline & f_n &=& -2f_{n-1}+3f_{n-2} \\ \text{ or } \\ & f_n &=& -(-3)^{n-1} \\ & f_5 &=& -(-3)^{5-1} \\ & f_5 &=& -(-3)^{4} \\ & \mathbf{f_5} & \mathbf{=} & \mathbf{-81} \\ \hline \end{array}\)

Formula edited, thank you Alan!

\(\mathbf{f(5) = -81}\)

This forum has many “mathematicians” who could answer this question, but we need a Master Troll to keep the forum’s troll bridge’s structure from deteriorating, and, not to forget, the sewer system free-flowing.

Yep, I care about what Ginger writes and thinks. I wanted her to come back sooner.

33-24=9

50-8=42

33+9=42

ANSWER 9

Find all complex numbers  such that  z^2=2i

Write your solutions in a+bi form, separated by commas. So, "1+2i, 3-i" is an acceptable answer

\(\begin{array}{|lrcll|} \hline & (a+bi)^2 &=& a^2 + 2abi+b^2i^2 \quad | \quad i^2 = -1 \\ & (a+bi)^2 &=& a^2 + 2abi -b^2 \\ & (a+bi)^2 &=& \underbrace{a^2-b^2}_{=0} + \underbrace{2abi}_{=2i} \quad | \quad \text{compare with } 2i \\\\ 1. & a^2-b^2 &=& 0 \\ & a^2 &=& b^2 \\ & \mathbf{a} & \mathbf{=}& \mathbf{b} \\\\ 2. & 2abi &=& 2i \quad | \quad : 2i \\ &\mathbf{ ab } & \mathbf{=}& \mathbf{1 } \quad | \quad a=b \\ &a^2 & = & 1 \\ &a & = & \pm 1 \\ \\ &b &=& a \\ & b &=& \pm 1 \\\\ & && & \mathbf{ab = 1\ ?} \\ & a &=& 1 \quad \text{ and } \quad b = 1 & ab = 1\cdot 1 = 1\ \checkmark \\ \text{or}& a &=& -1 \quad \text{ and } \quad b = -1 & ab = (-1)\cdot (-1) = 1\ \checkmark \\ \text{else}& a &=& -1 \quad \text{ and } \quad b = 1 & ab = (-1)\cdot 1 = -1\ \text{ no solution} \\ & a &=& 1 \quad \text{ and } \quad b = -1 & ab = 1\cdot (-1) = -1\ \text{ no solution} \\ \hline \end{array}\)

The answer is \(\mathbf{"1+i, -1-i"}\)

The least multiple of 72 that has exactly 16 prime factors is 72*2¹¹, not 72*2⁵*3⁶

EDIT: my bad, its 2¹¹, not 2¹³

How many positive multiples of 7 that are less than 1000 end with the digit 3?

\(\begin{array}{|r|r|r|} \hline & n & 7n \\ \hline 1. &9& 63\\ 2. &19& 133\\ 3. &29& 203\\ 4.&39& 273\\ 5.&49& 343\\ 6.&59& 413\\ 7.&69& 483\\ 8.&79& 553\\ 9.&89& 623\\ 10.&99& 693\\ 11.&109& 763\\ 12.&119& 833\\ 13.&129& 903\\ 14.&139& 973\\ \hline \end{array}\)

14 positive multiples of 7 that are less than 1000 end with the digit 3.

"Your prime factor answer is wrong." Why in the h**l don't you post your own "correct" answer and shut up??!! You "genetically enhanced chimp" should stick to your monkey business! Quite clearly, this is out of your league! I looked briefly at your profile. What the h**l do you think makes you an "expert" in math? I clearly see that you are seeking a confrontation with anybody and everybody. What the h**l is the matter with you? Are you a "normal" being or do have "mental problems"? If the latter, then seek professional help. I suggest that if you wish to reform the world, then start with yourself!!. God knows you needed it badly!.
Divisors of 60 =1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60
Prime "factors" of 60 = 2^2×3×5 (4 prime factors, 3 distinct)
Do you see the difference?? A "divisor" such as 12, or 30 above may have MANY prime "factors". But a "factor", such as 2, 3, 5 above NEVER has a "divisor"!! YOU confuse the two, you "Genetically Enhanced Chimp"!.

i got it it's ok