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<span><span> 5x2-23x=-26</span> </span>Two solutions were found :<span><span> x = 2
</span><span> x = 13/5 = 2.600
</span></span>
Rearrange:
Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :
5*x^2-23*x-(-26)=0
Step by step solution :Skip Ad
<span> Step 1 :</span><span>Equation at the end of step 1 :</span><span> (5x2 - 23x) - -26 = 0
</span><span> Step 2 :</span>Trying to factor by splitting the middle term
<span> 2.1 </span> Factoring <span> 5x2-23x+26</span>
The first term is, <span> <span>5x2</span> </span> its coefficient is <span> 5 </span>.
The middle term is, <span> -23x </span> its coefficient is <span> -23 </span>.
The last term, "the constant", is <span> +26 </span>
Step-1 : Multiply the coefficient of the first term by the constant <span> 5 • 26 = 130</span>
Step-2 : Find two factors of 130 whose sum equals the coefficient of the middle term, which is <span> -23 </span>.
<span><span>
-130
+
-1
=
-131
<span>
</span>
</span>
<span>
-65
+
-2
=
-67
<span>
</span>
</span>
<span>
-26
+
-5
=
-31
<span>
</span>
</span>
<span>
-13
+
-10
=
-23
That's it</span></span>
Answer:
Step-by-step explanation:
They wanted to complete the square so they took the thing in front of x and divided by 2 then squared. Whatever you add in, you must take out.
Now we are read to write that one part (the first three terms together) as a square:
I don't see this but what happens if we find a common denominator for those 2 terms after the square. (b/2a)^2=b^2/4a^2 so we need to multiply that one fraction by 4a/4a.
They put it in ( )
I'm going to go ahead and combine those fractions now:
I'm going to factor out a -1 in the second term ( the one in the second ( ) ):
Now I'm going to add (b^2-4ac)/(4a^2) on both sides:
I'm going to square root both sides to rid of the square on the x+b/(2a) part:
Now subtract b/(2a) on both sides:
Combine the fractions (they have the same denominator):
Let
x---------------> distance from people living to the city center
we Know that
Zone 1 covers people living within three miles of the city center
Zone 1 ------------> [x < 3 miles]
Zone 2 covers those between three and seven miles from the center
Zone 2 ------------> [ 3 <= x < = 7 miles]
Zone 3 covers those over seven miles from the center
Zone 3 ------------> [ x > 7 miles]
<span>calculate the distance between two points to find the value of x
</span>
case A) point (0,0) point (3,4)
x=√[(y2-y1)² +(x2-x1)²]----------> √[(4-0)² +(3-0)²]------> √[16+9]
x=√25-------------> x=5 miles
the answer Part A)
people living in (3,4)
x=5 miles -------------> covers Zone 2 [ 3 < =x <= 7 miles]
case B) point (0,0) point (6,5)
x=√[(y2-y1)² +(x2-x1)²]----------> √[(5-0)² +(6-0)²]------> √[25+36]
x=√61-------------> x=7.81 miles
the answer Part B)
people living in (6,5)
x=7.81 miles -------------> covers Zone 3 [ x > 7 miles]
case C) point (0,0) point (1,2)
x=√[(y2-y1)² +(x2-x1)²]----------> √[(2-0)² +(1-0)²]------> √[4+1]
x=√5-------------> x=2.23 miles
the answer Part C)
people living in (1,2)
x=2.23 miles -------------> covers Zone 1 [ x < 3 miles]
case D) point (0,0) point (0,3)
x=√[(y2-y1)² +(x2-x1)²]----------> √[(3-0)² +(0-0)²]------> √[9]
x=√9-------------> x=3 miles
the answer Part D)
people living in (0,3)
x=3 miles -------------> covers Zone 2 [ 3 < =x <= 7 miles]
case E) point (0,0) point (1,6)
x=√[(y2-y1)² +(x2-x1)²]----------> √[(6-0)² +(1-0)²]------> √[36+1]
x=√37-------------> x=6.08 miles
the answer Part E)
people living in (1,6)
x=6.08 miles -------------> covers Zone 2 [ 3 < = x <= 7 miles]
Yes. If the diagonals bisect the angles, the quadrilateral is always a parallelogram, specifically, a rhombus.
Consider quadrilateral ABCD. If diagonal AC bisects angles A and C, then ΔACB is congruent to ΔACD (ASA). Hence AB=AD and BC=CD (CPCTC).
Likewise, if diagonal BD bisects angles B and D, triangles BDA and BDC are congruent, thus AB=BC and AD=CD. (CPCTC again). Now, we have AB=BC=CD=AD, so the figure is a rhombus, hence a parallelogram.
