All Answers 352916 Answers

1)  If this were a right triangle...it would have to be that

8^2 + 9^2  = 12^2      but

64 + 81  >  144

145 > 144

Thus....angle ABC isn't quite large enough to be 90°

And the other two angles are definitely < 90° each....so this is an acute triangle

2) In isosceles triangle APS, we have  angle P is equal to 54 degrees and AP = AS. Which is larger, AP or PS?

                                                       A

                                           P                   S

Angle  APS  = Angle  ASP......So....angle  PAS =  [ 180  - 2*54 ] =    180 - 108  = 72°

And in a triangle.....the greater side is opposite the greater angle......and angle PAS > angle ASP.....so ...... PS > AP

3)In actute triangle ABC, we know  AB = 7, BC = 8 and that CA is the shortest side. What is the smallest possible integer value of CA?

Since this is an acute triangle, we must have that

7^2  + CA^2 >  8^2

49 + CA^2  > 64

CA^2 > 15

CA > √15  ⇒    CA > ≈ 3.8

So....the least integer value for CA  is  4

4)In obtuse triangle ABC, we know  AB = 7, BC = 8 and that CA is the longest side. What is the smallest possible integer value of CA?

Since the triangle is obtuse and CA is the longlest side, we must have that

AB^2 + BC^2  <  CA^2

7^2  + 8^2  < CA^2

49 + 64 < CA^2

113 < CA^2

√113  < CA

CA > √113 ≈ 10.6

So ......   the smallest possible integer value for CA  is 11

5)Two diagonals of a parallelogram have lengths 6 and 8. What is the largest possible length of the shortest side of the parallelogram?

The diagonals will bisect each other......

So we have 4 triangles that have two sides of 3 and 4

And....by the triangular inequality, we have that

Longest side + Intermediate side > Shortest side

4 + 3  >  Shortest side

7 > Shortest side

Shortest side < 7

So......the largest possible value for the shortest side must be < 3

6)Two sides of an acute triangle are 8 and 15. How many possible lengths are there for the third side if it is an integer?

Longest possible side length  

8 + 15  >  Third side length

23 > Third side length

Third side length < 23

Shortest possible side length

8 + Shortest side length > 15

Shortest side > 7

So.....the possible integer side lengths are

8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18 19,, 20 21, 22

So.....15 integer lengths are possible

Don't know how to do 7....maybe someone else has a solution......!!!

x = yz + y

Make y the subject.....note that the right side can be factored as

x  = y (z + 1)       now....divide both sides by (z + 1)   and we have that

x / ( z + 1)   =  y

a)  Here, we are trying to find angle PQR.....we can use the Law of Cosines for this...so we have...

3^2  =  6^2 + 4^2  - 2(6)(4)cosQPR    rearrange as

[ 3^2 - 6^2 - 4^2 ] / [ - 2(6)(4)]  =  cos QPR

Using the cosine inverse (arccos), we have that

arccos ( [ 3^2 - 6^2 - 4^2 ] / [ - 2(6)(4)] )  = QPR ≈  26.38°

So....the bearing of R from Q  =  [180 - 26.38]°  ≈  153.62°  = 154°

b) This one is easier...here we are trying to find angle QPR.....we can still use the Law of Cosines

4^2 =  6^2  +  3^2  - 2(6)(3)cos QPR   rearrange

([ 4^2 - 6^2 - 3^2 ] / [ -2(6)(3)] )  =  cosQPR

Using the cosine inverse (arccos), we have the bearing of R from P  is

arccos ([ 4^2 - 6^2 - 3^2 ] / [ -2(6)(3)] )  = QPR  ≈  36.34° = 36°

Here's a pic :

y = 1/x    Cross multiply

xy = 1     Divide both sides by y

x = 1/y

We can use the Law of Cosines to solve this

We have that

96.8^2  = 130^2  + 56^2  - 2(130)(56)cos(ABC)       rearrange as

[ 96.8^2 - 130^2 - 56^2 ] / [ -2(130)(56)]  =  cos(ABC)

Using the cosine inverse (arccos), we have

arccos ( [ 96.8^2 - 130^2 - 56^2 ] / [ -2(130)(56)] )  = ABC ≈  42.9°

maybe around 2 hours

y = 5(x + 2)

y = 5x + 10   subtract 10 from both sides

y - 10 = 5x    divide both sides by 5

[y - 10] / 5 =x, or:

x = [y - 10] / 5

You could also do this:

y = 5(x + 2)    divide both sides by 5

y/5 = x + 2     subtract 2 from both sides

y/5 - 2 = x, or:

x =y/5 - 2  Note: Even though they look different, but in fact they are exactly the same.

Answered here:  https://web2.0calc.com/questions/help_16307

Please don't vote your answer down EP, it was just a small oversight.

I am the one who gave you that point and I thought you deserved it.

Now you took it away 

Agree !   Melody is correct.... !      I was thinking about  a^2   and b^4   both being POSITIVE results...but that does not tell us if a or b are positive themselves !

2)

Here's a pic of the way I set this up :

The radius  of the larger  hexagon can be found as follows :

Area   =  1            ....so....

3 r^2 sin (60)  =  1

3r^2 * √3/2 = 1

r^2  =    2  / √27

r  =  √  [ 2 / √27 ]

The y coordinate of  B  is      r * sin (60°) =  √  [ 2 / √27 ] * [ √3 / 2 ]  

And the y coordinate  of F  is  r * sin (-60°)   - √  [ 2 / √27 ] * [ √3 / 2 ]

And the distance between them, BF,  is just    2 * √  [ 2 / √27 ] * [ √3 / 2 ]    =  

√  [ 2 / √27 ] *  √3 =

√ [ 6 / √27]

Note that because they span equal arcs, chords AC and BF are equal

And by SSS, triangle CBA is congruent to triangle BAF  and both are isosceles

And angle ACB  = angle BAC =  angle BAF = angle FBA

So  angle FBA  =  angle BAC....so BH  = HA

And angle CBA  = angle BAE  =  90°

So...subtracting equal angles , then angle HBI  = angle HAM

And angle BHI  = angle AHM

So...by  SAS, triangle BHI  is congruent to triangle AHM

But  since HI  is a transversal that cuts parallels  DB  and AE....then angle HAM  = angle HIB

But angle HAM  =  angle HBI......so  angle HBI  = angle HIB

And angle HIB  =  angle HMA.....so angle HAM  = angle HMA

So  HM   = AH      and  HI  = HB

So.....HM = HI = AH = HB

So .....since triangles BHI and AHM are congruent.....then HM  = HB

And by similar logic, we can prove that HM  = MF

So..... HB  = HM = HF

So....side HM of hexagon HIJKLM  is 1/3  of BF

So....the area of the hexagon is

3 * [ 6 / √27] / 9 * √3 / 2  =

√3 / √27  =

√3 /  [ 3 √3]  =

1/3 units^2

3 and 4 answered here :

https://web2.0calc.com/questions/math-help-please_16

5)

\($\dfrac {1}{1\cdot 3} + \dfrac {1}{3\cdot 5} + \dfrac {1}{5\cdot 7} + \dfrac {1}{7\cdot 9} + \cdots + \dfrac {1}{199\cdot 201}$\)

Note that  we can write      1 /  [  n (n + 2) ]  as     (1/2)  [ 1/n  -  1 / (n + 2)]

So  we can write

(1/2) [  (1/1  - 1/3)  + ( 1/3 - 1/5) + ( 1/5 - 1/7) + ( 1/7 - 1/9)  + .....+ (1/197 - 1/199) + (1/199 - 1/201) ]  =

(1/2) (1/1  -  1/201)  =

(1/2) ( 201 - 1 ) / 201  =

(1/2) (200) / 201  =

100 / 201

1)

(1 - 1/2) * (1 - 1/3) * (1- 1/4) * (1 - 1/5) *...........* ( 1 - 1/2007) * (1 - 1/2008) * (1 -1/2009)  = 

(1/2) (2/3) (3/4) ( 4/5)  *...........*   ( 2006/2007) (2007/2008) (2008/ 2009) =

(1) (2/2) (3/3) (4/ 4) (5/5) *.........* ( 2006/2006) (2007/ 2007) (008/ 2008) (1/2009) =

(1) (1) (1) *...... *(1/2009)  =

(1 / 2009)

Here's the first one:

See the way I've set it up

angle ABC  =  120°  ......so the area of triangle ABC  =  (1/2)*2^2 sin (120) =  √3 units^2

And ACDE forms an isosceles trapezoid

Let the bottom base be  DE  = 2

A  = (0,2)    and C  =  (-√ (3) , -1)    E  = (2, 0)    D =  ( 1, -√3)

And  AC forms the top base  = √  [   3 + 9  }  =  √12 units

Call G the mid-point of  AC  =  (-√3/2 , 1/2 )

And  F  is the midpoint  of  DE  =  ( 3/2, -√3/2 )

And FG  is the height of the trapezoid  =  √ [  ( 3 + √3)^2  + (1 + √3)^2 ]  /  2  = 

√ [  4  ( 4  + 2√3) ]/ 2  =  √ (4 + 2√3)  =  √[ 1 + √3 ]^2  =  1 + √3 units

So.....the area of the trapezoid  is

(1/2) (height)  (sum of the bases)  =

(1/2) ( 1 + √3)  [ 2  + √12 ]  =

(1/2) (1 + √3) [ 2 + 2√3 ]  =

( 1 + √3)^2  =   [ 4 + 2√3 ] units^2

So.....the total area is  

[ √3  + 4 + 2√3 ]    =    4 + 3√3  units^2

Find the least positive four-digit solution r of the congruence .

\(r^2 + 4r + 4 \equiv r^2 + 2r + 1 \pmod{55}\)

 It can be written as follows: (r + 2)^2 ≡(r + 1)^2 mod 55.  ( just as guest said )

 (r + 2)^2 ≡(r + 1)^2 mod 55.

Let x=r+1

\((x+1)^2\equiv x^2 \pmod{55}\\ x^2+2x+1\equiv x^2 \pmod{55}\\ 2x+1\equiv 0 \pmod{55}\\ 2x+1=55k \qquad k\in Z\\ 2x=55k-1 \qquad k\in Z\\ \text{(x-1) must have 4 digits}\\ \text{ k=37 tis the smallest, k needs to be odd}\\ Try\;\; k=37, 39,41,43,.....\\ 2x+1= 2035, 2145, 2255, 2365,\dots\\ 2x=2034, 2144, 2254, 2364, \dots \\ x=1017,1072, 1127, 1182 \dots\\ r=1016, 1071, 1126, 1181 \dots (1016+55c) \qquad c\in Z \)

check

mod(1017^2,55) = 14

mod(1018^2,55) = 14

mod(1072^2,55) = 14

mod(1073^2,55) = 14 

Well the first 2 work anyway , so it looks ok 

thats what i want to know

If tea has twice as much caffeine as coffee per volume,  then 160 cups of tea must have twice the amount of caffeine as 40 cups of coffee. So.....80 cups of tea must have the same amount of caffeine as 40 cups of coffee. So....one cup of coffee must have twice the caffeine of one cup of tea.

One cup of tea has 1/2 the caffeine as a cup of coffee. So.......since a 12-ounce can of cola has 1/4 the caffeine of a cup of coffee, a cup of tea must have twice the caffeine of a 12-ounce cola.

If a pound of tea has twice as much caffeine as a pound of coffee,

and if a pound of tea is enough to make 160 cups of tea,

and if a pound of coffee to make 40 cups,

How much caffeine does one cup of tea have compared to one cup of coffee?

Let’s see:

160cups of tea = 40cups of coffee (by weight)

160cups of tea= 80cups of coffee (by amount of caffeine)

so

1cup of tea= ½ cup of coffee (by amount of caffeine)

and if a 12-ounce can of cola has about one-fourth the caffeine as a cup of coffee.

How much caffeine does one cup of tea have compared to a 12-ounce can of cola? 

4 can cola = 1 cup coffee   (by amount of caffeine)

2 can cola = 1/2 cup coffee   (by amount of caffeine)

So

2 can cola = 1 cup tea   (by amount of caffeine)

So 1 cup of tea has twice as much caffeine as one can of cola.

Chris got the same answer.    

Given that  \(xy = \dfrac32 \)   and both x and y are nonnegative real numbers, find the minimum value of 

\(10x + \dfrac{3y}5\)

\(y=\frac{3}{2x}\)

so

\(​​​​z=10x+\frac{3y}{5}\\ ​​​​z=10x+\frac{3}{5}\times \frac{3}{2x}\\ ​​​​z=10x+\frac{9}{10x}\\ ​​​​z=10x+0.9x^{-1}\\ \frac{dz}{dx}=10-0.9x^{-2}\\ \frac{d^2z}{dx^2}=1.8x^{-3}>0\qquad \text{So any stationary point will be a minimum.}\\ \text{Find stat point}\\ 10-0.9x^{-2}=0\\ 10=0.9x^{-2}\\ 10=\frac{9}{10x^2}\\ 100x^2=9\\ x^2=\frac{9}{100}\\ x=\frac{3}{10}\qquad \text{Since x>0}\\ so\\ ​​​​z=10x+0.9x^{-1}\\ min\;\;value\\ =10*\frac{3}{10}+\frac{9*10}{10*3}\\ =3+3\\ =6 \)

check:

Here is the graph:

1)

2, 1/4, 16, 1/256, 65,536........etc.

Sqrt(65,536) = 256

3)

[a*(a+1)] / 2 =400, solve for a

a = ~ 27, so that:

1+2+3+.........+27 = 378

4)

2^1993 + 3^1993 =2^1 + 3^1 =5 - last digit !!.

Are you sure your congruence is written correctly? With a modulus of 55, there doesn't appear to be a solution !!. It can be written as follows: (r + 2)^2 ≡(r + 1)^2 mod 55.

6x³ - 12x² - 48x   =   0

6x( x² - 2x - 8 )   =   0

6x( x - 4 )( x + 2 )  =  0

6x  =  0     or     x - 4  =  0     or     x + 2  =  0

 x  =  0                 x  =  4                  x  =  -2

64x¹² + 27y³   =   ( 4x⁴ )³  +  ( 3y )³

This is a sum of cubes.   a³ + b³   =   (a + b)(a² - ab + b²)

( 4x⁴ )³  +  ( 3y )³

=   ( 4x⁴ + 3y )( (4x⁴)² - (4x⁴)(3y) + (3y)² )

=   ( 4x⁴ + 3y )( 16x⁸ - 12x⁴y + 9y² )