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OverLord2011 [107]
3 years ago
9

The product of two consecutive integers is 72. The equation x(x + 1) = 72 represents the situation, where x represents the small

er integer. Which equation can be factored and solved for the smaller integer?
Mathematics
3 answers:
FrozenT [24]3 years ago
5 0
That would be:-

x^2 + x - 72 = 0

(x + 9)(x - 8) = 0

smallest number is 8
Sergeeva-Olga [200]3 years ago
5 0
<span>x(x + 1) = 72 
1) x²+x-72=0 - This equation can be factored.
2) (x+9)(x-8)=0
x=-9, x=8
3) If x=8, x+1=9
First number = 8
Second number = 9
4) If x=-9, x+1=-9+1=-8
First number = -9
Second number = -8

</span>
XxMandy_WolfxX2 years ago
0 0

The answer is A. on edge

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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
In a experiment, the probability that event A occurs is 5/9, the probability that event B occurs is 5/7, and the probability tha
Ymorist [56]

Answer:

52.5% probability that A occurs given B occurs

Step-by-step explanation:

Suppose we have two events, A and B, the conditional probability formula is:

P(A|B) = \frac{P(A \cap B)}{P(B)}

In which

P(A|B) is the probability of A happening given that B happened.

P(A \cap B) is the probability of both A and B happening.

P(B) is the probability of B happening.

In this problem, we have that:

P(A \cap B) = \frac{3}{8}, P(B) = \frac{5}{7}

So

P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{\frac{3}{8}}{\frac{5}{7}} = 0.525

52.5% probability that A occurs given B occurs

7 0
3 years ago
Find the area and the circumference of the circle with the radius 6yd.
fgiga [73]

Answer:

<h2>A = 36π yd² ≈ 113.04 yd²</h2><h2>C = 12π yd ≈ 37.68 yd</h2>

Step-by-step explanation:

The formula of an area of a circle:

A=\pi r^2

The formula of a circumference of a circle:

C=2\pi r

<em>r</em><em> - radius</em>

<em />

We have <em>r = 6yd</em>.

Substitute:

A=\pi(6^2)=36\pi\ yd^2

C=2\pi(6)=12\pi\ yd

If you want round the answers, then use <em>π ≈ 3.14</em>

A\approx(36)(3.14)=113.04\ yd^2

C\approx(12)(3.14)=37.68\ yd

3 0
3 years ago
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