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Oduvanchick [21]
3 years ago
10

PLZ HELP ME!!! I NEED HELP!!!

Mathematics
1 answer:
mariarad [96]3 years ago
6 0

Answer:

The answer is 1

Step-by-step explanation:

i) \dfrac{(5\frac{4}{45} -4 \frac{1}{6})\div5\frac{8}{15}   }{(4\frac{2}{3} +0.75)\times3\frac{9}{13}}\times34\frac{2}{7} + \frac{0.3\div0.01}{70} + \frac{2}{7}

ii)\dfrac{(\frac{229}{45} -\frac{25}{6})\div\frac{83}{15}   }{(\frac{14}{3} +\frac{3}{4})\times\frac{48}{13}}\times\frac{240}{7} + \frac{30}{70} + \frac{2}{7}

<h3 />

iii)(\dfrac{(\frac{458}{90} -\frac{375}{90})\times\frac{15}{83}   }{(\frac{56}{12} +\frac{9}{12})\times\frac{48}{13}}\times\frac{240}{7}) + \frac{30}{70} + \frac{2}{7}

<h3 />

iv) (\dfrac{(\frac{83}{90})\times\frac{15}{83}   }{(\dfrac{65}{12} )\times\dfrac{48}{13}}\times\dfrac{240}{7}) + \dfrac{30}{70} + \dfrac{2}{7}

<h3 />

v)(\dfrac{(\frac{1}{6})   }{(5\times4)}\times\dfrac{240}{7}) + \dfrac{30}{70} + \dfrac{20}{70}

<h3 />

vi)(\dfrac{1}{(6\times5\times4)}\times\dfrac{240}{7}) + \dfrac{50}{70}

<h3 />

vii)(\dfrac{1}{120}\times\dfrac{240}{7}) + \dfrac{50}{70}

viii)(\dfrac{2}{7}) + \dfrac{50}{70}

ix)\dfrac{20}{70} +\dfrac{50}{70}

x)= \dfrac{70}{70} \hspace{0.2cm} = \hspace{0.2cm} 1

<h3 />
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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
3 years ago
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