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3 years ago

Help Please? GEOMETRY Use the word bank to help you with possible answers (there are extra options that are not to be used)

Mathematics

1 answer:

ipn [44]3 years ago

8 0

Short Answers:

Answer for part A: Definition of perpendicular
Answer for part B: Right Angle Congruence Theorem
Answer for part C: Reflexive Property of Congruence
Answer for part D: Definition of Midpoint
Answer for part E: $\triangle SXR \cong \triangle TXR$
Answer for part F: CPCTC

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Explanations:

Part A:

We are given that $\overline{RX} \perp \overline{ST}$ which means, in english, "line segment RX is perpendicular to line segment ST"

By the very definition of perpendicular, this means that the two line segments form a right angle. This is visually shown as the red square angle marker for angle RXT. Angle RXS is also a right angle as well.

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Part B:

The Right Angle Congruence Theorem (aka Right Angle Theorem) is the idea that if we have two right angles, then we know that they are both 90 degrees so they must be congruent to one another.

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Part C:

Any line segment is congruent to itself. This is because any line segment will have the same length as itself. It seems silly to even mention something so trivial but it helps establish what we need for the proof.

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Part D:

We are given "X is the midpoint of segment ST" so by definition, X is in the very exact middle of ST. Midpoints cut segments exactly in half. SX is one half while TX is the other half. The two halves are congruent which is why SX = TX

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Part E:

Writing $\triangle SXR \cong \triangle TXR$ means "triangle SXR is congruent to triangle TXR". These two triangles are the smaller triangles that form when you draw in segment RX

Side Note: SAS stands for "side angle side". The angle must be between the two sides. The pairing RX and RX forms one of the 'S' letters (see part C), while the pairing SX and TX forms the other 'S' (see part D). The angles between the sides are RXS and RXT (see part B).

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Part F:

CPCTC stands for "Corresponding Parts of Congruent Triangles are Congruent"

It means that if we have two congruent triangles, then the corresponding parts are congruent. Back in part E, we proved the triangles congruent. For this part, we look at the pieces RS and RT (which correspond to one another; they are the hypotenuse of each triangle). They are proven congruent by CPCTC

If CPCTC is an odd concept to think about, then try thinking about something like this: you have two houses which are completely identical in every way. We can say that those two houses are congruent. If the houses are identical, then surely every piece that makes up the house is identical to its corresponding piece to the other house. For example, the front door to each house is both the same size, shape, color, made of the same material, designed in the same pattern, etc. So the two doors are congruent as well.

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You play the following game against your friend. You have 2 urns and 4 balls One of the balls is black and the other 3 are white

Rom4ik [11]

Answer:

Part a: The case in such a way that the chances are minimized so the case is where all the four balls are in 1 of the urns the probability of her winning is least as 0.125.

Part b: The case in such a way that the chances are maximized so the case where the black ball is in one of the urns and the remaining 3 white balls in the second urn than, the probability of her winning is maximum as 0.5.

Part c: The minimum and maximum probabilities of winning for n number of balls are such that

when all the n balls are placed in one of the urns the probability of the winning will be least as 1/2n

when the black ball is placed in one of the urns and the n-1 white balls are placed in the second urn the probability is maximum, as 0.5

Step-by-step explanation:

Let us suppose there are two urns A and A'. The event of selecting a urn is given as A thus the probability of this is given as

P(A)=P(A')=0.5

Now the probability of finding the black ball is given as

P(B)=P(B∩A)+P(P(B∩A')

P(B)=(P(B|A)P(A))+(P(B|A')P(A'))

Now there can be four cases as follows

Case 1: When all the four balls are in urn A and no ball is in urn A'

P(B|A)=0.25 and P(B|A')=0 So the probability of black ball is given as

P(B)=(0.25*0.5)+(0*0.5)

P(B)=0.125;

Case 2: When the black ball is in urn A and 3 white balls are in urn A'

P(B|A)=1.0 and P(B|A')=0 So the probability of black ball is given as

P(B)=(1*0.5)+(0*0.5)

P(B)=0.5;

Case 3: When there is 1 black ball and 1 white ball in urn A and 2 white balls are in urn A'

P(B|A)=0.5 and P(B|A')=0 So the probability of black ball is given as

P(B)=(0.5*0.5)+(0*0.5)

P(B)=0.25;

Case 4: When there is 1 black ball and 2 white balls in urn A and 1 white ball are in urn A'

P(B|A)=0.33 and P(B|A')=0 So the probability of black ball is given as

P(B)=(0.33*0.5)+(0*0.5)

P(B)=0.165;

Part a:

As it says the case in such a way that the chances are minimized so the case is case 1 where all the four balls are in 1 of the urns the probability of her winning is least as 0.125.

Part b:

As it says the case in such a way that the chances are maximized so the case is case 2 where the black ball is in one of the urns and the remaining 3 white balls in the second urn than, the probability of her winning is maximum as 0.5.

Part c:

The minimum and maximum probabilities of winning for n number of balls are such that

when all the n balls are placed in one of the urns the probability of the winning will be least given as

P(B|A)=1/n and P(B|A')=0 So the probability of black ball is given as

P(B)=(1/n*1/2)+(0*0.5)

P(B)=1/2n;

when the black ball is placed in one of the urns and the n-1 white balls are placed in the second urn the probability is maximum, equal to calculated above and is given as

P(B|A)=1/1 and P(B|A')=0 So the probability of black ball is given as

P(B)=(1/1*1/2)+(0*0.5)

P(B)=0.5;

5 0

3 years ago

Consider the function f(x) = x2 + 2x – 35. If one of the zeros of the

Whitepunk [10]

If one zeros is x= -7 then the other zero is x=5

Step-by-step explanation:

We are given the function: $f(x) = x^2 + 2x-35$