The <u>probability</u> that a point <u>chosen at random</u> in the triangle is also in the blue square can be calculated using <u>geometrical definition of the probability</u>:
1. Find the total area of the triangle:
2. Find the desired area of the square:
Then the probability is
Answer: correct choice is B
pH = f(x) = -log₁₀x
1. Graphs
I used Excel to calculate the pH values and draw the graphs (see the Figure).
f(x) and f(x) +1 are plotted against the left-hand axis, while f(x+ 1) is plotted against the expanded right-hand axis.
The points at which pH = 0 and pH = 1 are indicated by the large red dots.
2. x = 0.5
When x = 0.5, pH ≈0.30. The point is indicated by the red diamond.
3. Transformations
(a) ƒ(x) = -log(x) + 1
This function has no y-intercept, because log(0) is undefined.
(b) ƒ(x +1) = -log(x + 1)
f(0) = -log(0 + 1) = -log(1) = 0
This function has a y-intercept at (0,0).
hope this helps please mark me brainliest!
Answer:
Step-by-step explanation:
using the formula below, the central angle is 20° (bc 360-180-160=20)
radius is 14 (28/2=14)
Step-by-step explanation:
A. this is a geometric sequence.
according to y-coordinates =>
3, 6, 12, 24
have a T1 = 3 and ratio = 6/3 = 2
B. the formula = Tn = r. T(n-1)
T5 = 2. T4
= 2. 24 = 48 minutes
C. the formula : Tn = T1 . r^(n-1)
T9 = 3. 2^(9-1)
= 3. 2^8
= 3. 256
= 768 munutes
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
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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
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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
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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
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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.
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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
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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
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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
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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
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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
