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JulsSmile [24]
3 years ago
15

The park behind James's house used to have a bike trail that was 12.347 kilometers long. Last year, the trail was reduced to 9.4

32 kilometers to make space for a meditation garden. What is the difference in the lengths of the old and the new bike trails?
Mathematics
2 answers:
aksik [14]3 years ago
6 0
12.347 - 9.432 = 2.915 kilometers
Alina [70]3 years ago
5 0

Answer:

The difference in the lengths of the old and the new bike trails is 2.915 kilometers.

Step-by-step explanation:

The park behind James's house used to have a bike trail that was 12.347 kilometers long.

Last year, the trail was reduced to 9.432 kilometers to make space for a meditation garden.

So, the difference in lengths is = 12.347-9.432=2.915 kilometers

Therefore, the difference in the lengths of the old and the new bike trails is 2.915 kilometers.

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Solve for equation √-5p=√24-p
blsea [12.9K]

Answer:


Step-by-step explanation:

Algebra Calculator | Latest | Discuss | About | Help | Translation

25,219,829 solved | 850 online

p2-5p-24=0  

Two solutions were found :

    p = 8

    p = -3  

Reformatting the input :

Changes made to your input should not affect the solution:

(1): "p2"   was replaced by   "p^2".  

Step by step solution :

Step  1  :

Trying to factor by splitting the middle term

1.1     Factoring  p2-5p-24  

The first term is,  p2  its coefficient is  1 .

The middle term is,  -5p  its coefficient is  -5 .

The last term, "the constant", is  -24  

Step-1 : Multiply the coefficient of the first term by the constant   1 • -24 = -24  

Step-2 : Find two factors of  -24  whose sum equals the coefficient of the middle term, which is   -5 .

      -24     +     1     =     -23  

      -12     +     2     =     -10  

      -8     +     3     =     -5     That's it

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above,  -8  and  3  

                    p2 - 8p + 3p - 24

Step-4 : Add up the first 2 terms, pulling out like factors :

                   p • (p-8)

             Add up the last 2 terms, pulling out common factors :

                   3 • (p-8)

Step-5 : Add up the four terms of step 4 :

                   (p+3)  •  (p-8)

            Which is the desired factorization

Equation at the end of step  1  :

 (p + 3) • (p - 8)  = 0  

Step  2  :

Theory - Roots of a product :

2.1    A product of several terms equals zero.  

When a product of two or more terms equals zero, then at least one of the terms must be zero.  

We shall now solve each term = 0 separately  

In other words, we are going to solve as many equations as there are terms in the product  

Any solution of term = 0 solves product = 0 as well.

Solving a Single Variable Equation :

2.2      Solve  :    p+3 = 0  

Subtract  3  from both sides of the equation :  

                     p = -3

Solving a Single Variable Equation :

2.3      Solve  :    p-8 = 0  

Add  8  to both sides of the equation :  

                     p = 8

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3 years ago
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Answer:

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5 0
3 years ago
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Need help with the blanks
Crazy boy [7]
Answers: 
33. Angle R is 68 degrees
35. The fraction 21/2 or the decimal 10.5
36. Triangle ACG
37. Segment AB
38. The values are x = 6; y = 2
40. The value of x is x = 29
41. C) 108 degrees
42. The value of x is x = 70
43. The segment WY is 24 units long
------------------------------------------------------
Work Shown:
Problem 33) 
RS = ST, means that the vertex angle is at angle S
Angle S = 44
Angle R = x, angle T = x are the base angles
R+S+T = 180
x+44+x = 180
2x+44 = 180
2x+44-44 = 180-44
2x = 136
2x/2 = 136/2
x = 68
So angle R is 68 degrees
-----------------
Problem 35) 
Angle A = angle H
Angle B = angle I
Angle C = angle J
A = 97
B = 4x+4
C = J = 37
A+B+C = 180
97+4x+4+37 = 180
4x+138 = 180
4x+138-138 = 180-138
4x = 42
4x/4 = 42/4
x = 21/2
x = 10.5
-----------------
Problem 36) 
GD is the median of triangle ACG. It stretches from the vertex G to point D. Point D is the midpoint of segment AC
-----------------
Problem 37)
Segment AB is an altitude of triangle ACG. It is perpendicular to line CG (extend out segment CG) and it goes through vertex A.
-----------------
Problem 38) 
triangle LMN = triangle PQR
LM = PQ
MN = QR
LN = PR
Since LM = PQ, we can say 2x+3 = 5x-15. Let's solve for x
2x+3 = 5x-15
2x-5x = -15-3
-3x = -18
x = -18/(-3)
x = 6
Similarly, MN = QR, so 9 = 3y+3
Solve for y
9 = 3y+3
3y+3 = 9
3y+3-3 = 9-3
3y = 6
3y/3 = 6/3
y = 2
-----------------
Problem 40) 
The remote interior angles (2x and 21) add up to the exterior angle (3x-8)
2x+21 = 3x-8
2x-3x = -8-21
-x = -29
x = 29
-----------------
Problem 41) 
For any quadrilateral, the four angles always add to 360 degrees
J+K+L+M = 360
3x+45+2x+45 = 360
5x+90 = 360
5x+90-90 = 360-90
5x = 270
5x/5 = 270/5
x = 54
Use this to find L
L = 2x
L = 2*54
L = 108
-----------------
Problem 42) 
The adjacent or consecutive angles are supplementary. They add to 180 degrees
K+N = 180
2x+40 = 180
2x+40-40 = 180-40
2x = 140
2x/2 = 140/2
x = 70
-----------------
Problem 43) 
All sides of the rhombus are congruent, so WX = WZ.
Triangle WPZ is a right triangle (right angle at point P).
Use the pythagorean theorem to find PW
a^2+b^2 = c^2
(PW)^2+(PZ)^2 = (WZ)^2
(PW)^2+256 = 400
(PW)^2+256-256 = 400-256
(PW)^2 = 144
PW = sqrt(144)
PW = 12
WY = 2*PW
WY = 2*12
WY = 24
3 0
2 years ago
Mr. Shaw wants to replace the flooring his family room. The floor has an area of 262.8 square feet. If the room is 18 feet long,
iVinArrow [24]

Answer:

14.6

Step-by-step explanation:

The area of a rectangle is area=length*width

so we can substitute the numbers here, 262.8=18w

divide 262.8 by 18 to isolate the variable.

You get 14.6.

The width is 14.6.

5 0
3 years ago
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