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NISA [10]
3 years ago
14

Triangle ABC has been dilated to form triangle A'B'C'. If sides AB and A'B' are proportional, what is the least amount of additi

onal information needed to determine if the two triangles are similar? Angles B and B' are congruent, and angles C and C' are congruent. Segments AC and A'C' are congruent, and segments BC and B'C' are congruent. Angle C=C', angle B=B', and segments BC and B'C' are congruent. Segment BC=B'C', segment AC=A'C', and angles B and B' are congruent.
Mathematics
2 answers:
snow_tiger [21]3 years ago
4 0

Answer:

Its Angles B and B' are congruent, and angles C and C' are congruent.

Ede4ka [16]3 years ago
3 0
When considering similar triangles, we need congruent angles and proportional sides.

Hence
"Angles B and B' are congruent, and angles C and C' are congruent." is sufficient to prove similarity of two triangles.

"Segments AC and A'C' are congruent, and segments BC and B'C' are congruent." does not prove anything because we know nothing about the angles.

"Angle C=C', angle B=B', and segments BC and B'C' are congruent." would prove ABC is congruent to A'B'C' if and only if AB is congruent to A'B' (not just proportional).

"<span>Segment BC=B'C', segment AC=A'C', and angles B and B' are congruent</span>" is not sufficient to prove similarity nor congruence because SSA is not generally sufficient.

To conclude, the first option is sufficient to prove similarity (AAA)

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What is the value of -17 - (-1) + 2/3 + -1/2?
netineya [11]

Answer:

-95/6 or -15.83 repeating

Step-by-step explanation:

6 0
3 years ago
Stefanie bought a package of pencils for $1.75 and some erasers that cost $0.25 each. She paid a total of $4.25 for these items,
melamori03 [73]

Answer:

Stefanie bought 10 erasers

Step-by-step explanation:

We can establish the following equation:

TP = NP*PP + NE*EP

Where:

TP = Total price; (4.25)

NP = Number of pencil packages; (1)

PP = Pencil Package's Price; (1.75)

NE = Number of erasers; (What we are looking for)

EP = Erasers Price; (0.25);

Then:

4.25 = 1*1.75 + NE*0.25;

4.25 - 1.75 = NE*0.25;

2.5/0.25 = NE;

10 = NE.

5 0
3 years ago
Let a triangle have sides of 137 in, 127 in,
Margarita [4]

Answer:

I think first add the side 137+ 127 +118=382 / 382cm

8 0
2 years ago
Identify the x-intercept and y-intercept of the equation below.
Lady_Fox [76]
Divide by 21 to put the equation in intercept form.
  x/(21/9) + y/(-21/7) = 1
  x/(7/3) + y/(-3) = 1

The x-intercept is (7/3, 0)
The y-intercept is (0, -3)

The 3rd choice is appropriate.

8 0
3 years ago
What is a quick and easy way to remember explicit and recursive formulas?
Oliga [24]
I always found derivation to be helpful in remembering. Since this question is tagged as at the middle school level, I assume you've only learned about arithmetic and geometric sequences.

First, remember what these names mean. An arithmetic sequence is a sequence in which consecutive terms are increased by a fixed amount; in other words, it is an additive sequence. If a_n is the nth term in the sequence, then the next term a_{n+1} is a fixed constant (the common difference d) added to the previous term. As a recursive formula, that's

a_{n+1}=a_n+d

This is the part that's probably easier for you to remember. The explicit formula is easily derived from this definition. Since a_{n+1}=a_n+d, this means that a_n=a_{n-1}+d, so you plug this into the recursive formula and end up with 

a_{n+1}=(a_{n-1}+d)+d=a_{n-1}+2d

You can continue in this pattern, since every term in the sequence follows this rule:

a_{n+1}=a_{n-1}+2d
a_{n+1}=(a_{n-2}+d)+2d
a_{n+1}=a_{n-2}+3d
a_{n+1}=(a_{n-3}+d)+3d
a_{n+1}=a_{n-3}+4d

and so on. You start to notice a pattern: the subscript of the earlier term in the sequence (on the right side) and the coefficient of the common difference always add up to n+1. You have, for example, (n-2)+3=n+1 in the third equation above.

Continuing this pattern, you can write the formula in terms of a known number in the sequence, typically the first one a_1. In order for the pattern mentioned above to hold, you would end up with

a_{n+1}=a_1+nd

or, shifting the index by one so that the formula gives the nth term explicitly,

a_n=a_1+(n-1)d

Now, geometric sequences behave similarly, but instead of changing additively, the terms of the sequence are scaled or changed multiplicatively. In other words, there is some fixed common ratio r between terms that scales the next term in the sequence relative to the previous one. As a recursive formula,

a_{n+1}=ra_n

Well, since a_n is just the term after a_{n-1} scaled by r, you can write

a_{n+1}=r(ra_{n-1})=r^2a_{n-1}

Doing this again and again, you'll see a similar pattern emerge:

a_{n+1}=r^2a_{n-1}
a_{n+1}=r^2(ra_{n-2})
a_{n+1}=r^3a_{n-2}
a_{n+1}=r^3(ra_{n-3})
a_{n+1}=r^4a_{n-3}

and so on. Notice that the subscript and the exponent of the common ratio both add up to n+1. For instance, in the third equation, 3+(n-2)=n+1. Extrapolating from this, you can write the explicit rule in terms of the first number in the sequence:

a_{n+1}=r^na_1

or, to give the formula for a_n explicitly,

a_n=r^{n-1}a_1
6 0
3 years ago
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