+0  
 
+1
141
6
avatar+211 

 

Help please

Rollingblade  May 12, 2018
 #1
avatar+970 
+2

3. 

 

Since the sum of two absolute value equation is equal to 0, each of the absolute value equations is equal to zero. 

 

This is because \(|x|\ge0\).

 

We can write the systems of equation: 

 

\(x + y + 7=0\)

 

\(2x - y + 2 =0\)

 

Solve by adding:

 

\(x+2x+y-y+7+2=0 \\ 3x = -9 \\ x=-3\)

 

\(y+7-3=0 \\ y=-4\)


\( \boxed{(-3,-4)}\)

 

I hope this helped,

 

Gavin

GYanggg  May 12, 2018
edited by GYanggg  May 12, 2018
 #2
avatar+970 
+2

1. 

 

If we set the other 4 test scores as high as possible, the 5th one must be the lowerst possible score. 

 

We can assign the lowest possible score as x. 

 

Then we have the equation: 

 

\(\frac{4\cdot100+x}{5}=88\\ 400+x=440 \\ x=40\)

 

The lowest possible score is 40. 

 

I hope this helped

 

Gavin

GYanggg  May 12, 2018
 #3
avatar+970 
+2

2. 

 

Solve by graphing. 

 

https://www.desmos.com/calculator/x1mgab6pen

 

We get \(5^x+12^x=y\)

 

Y-intecept = 2. 

 

Only real solution is 2. 

 

I hope this helped,

 

gavin

GYanggg  May 12, 2018
 #4
avatar+970 
+2

4. Expand

 

 

\(ay^3+by^2+cy+d=3x^3-8x^2+7\\ \ \\ a(x-2)^3+b(x-2)^2+c(x-2)+d=3x^3-8x^2+7\\ \ \\ ax^3−6ax^2+12ax−8a+bx^2−4bx+4b+cx-2c+d=3x^3-8x^2+7\\ \ \\ ax^3-6ax^2+bx^2+12ax-4bx+cx-8a+4b-2c+d=3x^3-8x^2+7 \)

 

We get:

 

\(ax^3=3x^3 \\ bx^2-6ax^2=-8x^2\\ 12ax-4bx+cx=0\\ 4b-8a-2c+d=7\)

 

We get the systems:

 

\(a=3\\ b-6a=-8\\ 12a-4b+c=0\\ 4b-8a-2c+d=7\)

 

\(a=3\\ b-6\cdot3=-8\Rightarrow b=10\\ 12\cdot3-4\cdot10+c=0\Rightarrow 36-40+c=0 \Rightarrow c=4\\ 4\cdot10-8\cdot3-2\cdot4+d=7\Rightarrow40-24-8+d=7 \Rightarrow d=-1\)

 

\(a+b+c+d=3+10+4-1=\boxed{16}\)

 

I hope this helped,

Gavin

GYanggg  May 12, 2018
 #5
avatar+94105 
+2

I normally will not answer questions when there are several asked on the same post but I am making an exception here because a much respected member has asked me too.  (He wants to see how it can be answered for himself)

 

From a point interior to a regular hexagon, perpendiculars drawn to the sides on the hexagon have lengths, in order of size of 3, 6, 7.5, 13.5, 15 and 18. How long is the side of this hexagon?

 

 

A hexagon has got 6 sides. Opposite sides are parallel. So the perpendicular from a point to one side will also be the perpendicular from the point to the opposite sides.

If you look at the given lengths you can see that 

3+18=21

6+15=21

7.5+13.5=21

So 21 is the perpendicular distance between the parallel sides.

So the centre of the hexagon is 10.5 perpendicular units from each side.

 

The angle sum of a hexagon is 180*(6-2) = 720 degrees

Each internal angle is 720 / 6 = 120 degrees.

Half of each internal angle is 60 degrees.

So this is what we have.

 

 

 

\(tan60^\circ=\frac{10.5}{x}\\ \sqrt3=\frac{10.5}{x}\\ x=\frac{10.5}{\sqrt3}\\ x=\frac{21}{2\sqrt3}\\ x=\frac{21\sqrt3}{2*3}\\ x=\frac{7\sqrt3}{2}\\ \text {the side length is 2 * x}\\ Side\;length\;=7\sqrt3\;units \\ Side\;length\;\approx 12.124\;units \\ \)

 

Here is the pic with the original figures. This pic was not needed to answer this question but it may help you to understand better.

 

Melody  May 13, 2018
edited by Melody  May 13, 2018
 #6
avatar+3446 
+1

2. We think of a Pythagorean triplet, here:

a^2+b^2=c^2

So, 5^2+12^2=13^2

So, the answer is \(\boxed{2}\)

smileysmiley

tertre  May 13, 2018

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