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For all 30° 60° 90° right triangles if x = the shortest side's length the hypotenuse would be 2x

So the hypotenuse of the first triangle would be \(2 \times 1 = 2\) 

We could see that the second triangle is also 30° 60° 90°

2 is the length of the shortest side in the second triangle so \(x = 2 \times 2\)

\(x = 4\)

John flew a kite. He started flying the kite at 2:20 pm and ended at 3:14 pm. How many minutes did John fly a kite?    

From 2:20 pm to 3:14 pm is just 6 minutes short of an hour. 

An hour has 60 minutes, so 60 minus 6 equals 54 minutes.  

_.

There are two positive integers c for which the equation
5x^2+11x+c=4x^2 + 3x
has rational solutions. What is the product of those two values of c?     

Start with                                           5x² + 11x + c = 4x² + 3x  

Subtract (4x² + 3x) from both sides      x² + 8x + c  =  0   

If c = 16, then the quadratic will factor to (x + 4)(x + 4)   

If c = 15, then the quadratic will factor to (x + 5)(x + 3)  

If c = 12, then the quadratic will factor to (x + 6)(x + 2)  

If c =  7, then the quadratic will factor to  (x + 7)(x + 1)  

The problem says two.  If that means ONLY two, then I'm misunderstanding the problem.     

_.

54 minutes

Nevermind thanks for all the help!

Find three consecutive odd integers such that the sum of the first, two times the second and three times the third is 34.     

Call the 1^st integer (x)  

Call the 2^nd integer (x + 2)  

Call the 3^rd integer (x + 4)  

Given                                                  (x) + 2(x + 2) + 3(x + 4)  =  34  

Now, just multiply it out                         x + 2x + 4 + 3x + 12  =  34   

Combine like terms                              6x + 16  =  34  

Subtract 16 from both sides                6x  =  18  

Divide both sides by 6                          x  =  3  

The 1^st integer is 3, the 2^nd integer is 5, and the 3^rd integer is 7  

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It may be correct by you have not learned anything from this answer only reply.

I have been playing with this but I cannot quite get it to work.

I do not really know what I am doing.

If Alan or Heureka or some other talented mathematician could show us both that would be great.

Four positive integers A, B, C and D have a sum of 36. If A + 2 = B - 2 = C*2 = D*2, what is the value of the product ABCD?

Unfortunately, some of their answers were wrong!  See https://web2.0calc.com/questions/help_74185#r2

See the following plot:

The first equation allows you to calculate \(\theta\)

Then you can calculate \(\beta\) from the second equation

Then you can calculate \(\alpha\) from the third equation

Then you can calculate \(\sin \alpha\)

THX, it's correct

What is the coefficient of x^2 when -5x^3 - 5x^2 - 7x + 1 is multiplied by -8x^2 + 6x + 2 and the like terms are combined?

Hello Guest!

\(( -5x^3 - 5x^2 - 7x + 1)(-8x^2 + 6x + 2 )\\ =40x^5+10x^4+16x^3-60x^2-8x+2\)

\(The\ coefficient\ of\ x^2\ is\ -60\)

  !

The parabola y=ax^2+bx+c is graphed below. Find a+b+c.

Hello Guest!

\(y= (x-1)^2-4\\ \color{blue}y=x^2-2x-3\)

a = 1,   b = - 2,   c = -3     a + b + c = - 4

  !

Find the vertex of the graph of the equation y=-2x^2+18x-15.

Hello Guest!

\(y=-2x^2+18x-15.\\ \frac{dy}{dx}=-4x+18=0\\ -4x=-18\\ \color{blue}x=4.5\)

\(y=-2x^2+18x-15\\ y=-2\cdot 4.5^2+18\cdot 4.5-15\\ \color{blue}y=25.5\)

V(4.5, 25.5)

  !

Ok, now I answered it :)

I agree, this is a really confusing question.

Let                             Melville start with M                and      Danny start with D

Melville give danny 2/5 of his so

afer that                      Melville has 3/5M                  and      Danny has    D+2/5M

Now Danny gives Melville half of what he has

so now        Melville has          3/5M+1/2(D+2/5M)   and      Danny has    1/2(D+2/5M)

Now Melville has 3 times more than Danny

\(\frac{3M}{5}+\frac{1}{2}(D+\frac{2M}{5})=3(\frac{1}{2}(D+\frac{2M}{5})\\ 10*[\frac{3M}{5}+\frac{1}{2}(D+\frac{2M}{5})]=10*[3(\frac{1}{2}(D+\frac{2M}{5}))]\\ 6M+5D+2M=15D+6M\\ 8M+5D=15D+6M\\ M=5D \)

Melville gave Danny    \(\frac{2M}{5}=\frac{2*5D}{5}=2D\)

Danny gave Melville    \(0.5(D+\frac{2M}{5})=\frac{5D+2M}{10}=\frac{5D+2*5D}{10}=\frac{15D}{10}=\frac{3D}{2}\)

We are told that Melvin gave Danny 28 more water guns than what Danny gave to Melvin (though it is worded badly)

\(\frac{3D}{2}+28=2D\\ 3D+56=4D\\ D=56\)

So Danny started with 56 water guns.

I leave it to you to check.

LaTex

\frac{3M}{5}+\frac{1}{2}(D+\frac{2M}{5})=3(\frac{1}{2}(D+\frac{2M}{5})\\

10*[\frac{3M}{5}+\frac{1}{2}(D+\frac{2M}{5})]=10*[3(\frac{1}{2}(D+\frac{2M}{5}))]\\
6M+5D+2M=15D+6M\\
8M+5D=15D+6M\\
M=5D

The maximum area of the rectangle is 2e.

By the midsegment theorem, AB = 16, and PQ = 1/2(AB)

Area = 1/2(AB * CD)

r = 1/2√305

or

r = sqrt(8.5² + 2²)

r = 8.732124598

Let's see

angle BMC +angle A = 201

consider triangle CMA

2

2A=201-A

3A=201

A=67 degrees

so

angle CMA= 180-2*67 = 46degrees

so angle B = half of 46 = 23 degrees.

I have not carefully checked this.

This question is nonsense.

Answerers have changed the question to fit in with their solutions but the simple fact is that this question is total nonsense.

The angle sum of a triangle is 180 degrees therefore 2 internal angles cannot add up to more than that.