The most cookies Heidy can make is 36 cookies.
<h3>How to calculate how many cookies can Heidy make?</h3>
To know how many cookies Heidy can make, you have to take into account the following information:
12 cookies need the following ingredients:
- 125g butter
- 200g flour
- 50g sugar
In the case in which Heidy has more ingredients, we must carry out the following operations:
Divide the quantities, in the reference quantity we have:
- 500g of butter ÷ 125g of butter = 4
- 700g flour ÷ 200g flour = 3.5
- 250g of sugar ÷ 50g of sugar = 5
According to the above, we must take into account the lowest value of all because if that ingredient is enough, we can infer that the rest of the ingredients also.
So the number of cookies Heidy can make are:
12 × 3.5 = 42
Learn more about ingredients in: brainly.com/question/26532763
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
Answer:
The correct answer is:
Between 600 and 700 years (B)
Step-by-step explanation:
At a constant decay rate, the half-life of a radioactive substance is the time taken for the substance to decay to half of its original mass. The formula for radioactive exponential decay is given by:
First, let us calculate the decay constant (k)
Next, let us calculate the half-life as follows:
Therefore the half-life is between 600 and 700 years
We are looking to figure out the size of m<CAB
Since line AB is parallel to the line CD, m<CAB corresponds to m<ECD which means the size of the angles equals
m<ECD can be found by using the fact that angles in a triangle add up to 180°,
hence, 180°-58°-43°=79°
The size of m<CAB is 79°
The graph that represents the function is graph b. Graph b is the only graph the appropriately represents the y intercept (the value of y when x = 0).
