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Advocard [28]
3 years ago
12

polygon P'Q'R'S'T' shown on the grid below is an image of polygon PQRST after dilation with a scale factor of 3, keeping the ori

gin as the center of dilation: ​

Mathematics
1 answer:
ruslelena [56]3 years ago
6 0

Answer:  d) SR and S'R' have the ratio 1:3

<u>Step-by-step explanation:</u>

In order for the polygons to be similar, they must have congruent angles and proportional side lengths.

a) ∠Q and ∠Q' have the ratio 1:3          

  FALSE - The angles must be congruent (not proportional)

b) TS and T'S' have equal lengths

 FALSE - We can see that there is a dilation so they cannot be congruent.

c) RT and R'T' have equal lengths

 FALSE - We can see that there is a dilation so they cannot be congruent.

d) SR and S'R' have a ratio of 1:3

 TRUE! - The sides are proportional so we can use this to prove similarity.

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Give the degree and classify the polynomial by the number of terms.
cestrela7 [59]

Answer:

degree 3, binomial

Step-by-step explanation:

the highest exponent is the degree, it's two terms so bi meaning 2 makes it a binomial

5 0
2 years ago
If the sum of the even integers between 1 and k, inclusive, is equal to 2k, what is the value of k?
Alisiya [41]
If k is odd, then

\displaystyle\sum_{n=1}^{\lfloor k/2\rfloor}2n=2\dfrac{\left\lfloor\frac k2\right\rfloor\left(\left\lfloor\frac k2\right\rfloor+1\right)}2=\left\lfloor\dfrac k2\right\rfloor^2+\left\lfloor\dfrac k2\right\rfloor

while if k is even, then the sum would be

\displaystyle\sum_{n=1}^{k/2}2n=2\dfrac{\frac k2\left(\frac k2+1\right)}2=\dfrac{k^2+2k}4

The latter case is easier to solve:

\dfrac{k^2+2k}4=2k\implies k^2-6k=k(k-6)=0

which means k=6.

In the odd case, instead of considering the above equation we can consider the partial sums. If k is odd, then the sum of the even integers between 1 and k would be

S=2+4+6+\cdots+(k-5)+(k-3)+(k-1)

Now consider the partial sum up to the second-to-last term,

S^*=2+4+6+\cdots+(k-5)+(k-3)

Subtracting this from the previous partial sum, we have

S-S^*=k-1

We're given that the sums must add to 2k, which means

S=2k
S^*=2(k-2)

But taking the differences now yields

S-S^*=2k-2(k-2)=4

and there is only one k for which k-1=4; namely, k=5. However, the sum of the even integers between 1 and 5 is 2+4=6, whereas 2k=10\neq6. So there are no solutions to this over the odd integers.
5 0
3 years ago
the population of a type of local dragonfly can be found using an infinite geometric series where a1= 65 and the common ratio is
LiRa [457]
Sum of  infinite series of a GS  is  a1 / (1-r)

so here it is  65 / (1 - 1/6)  =  65 / 5/6  = 65*6 / 5 = 78  answer
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3 years ago
Jamal is told that it costs $24 plus a certain amount per day to rent a car. He finds that a friend rented a car from this compa
Ad libitum [116K]
40x+24 is the correct answer
3 0
3 years ago
Read 2 more answers
Prime factoring | Example answer: 12=2x2x3 if it is prime, use the factors 1 and itself, example: 13=1x13
Anna007 [38]
32 = 2x2x2x2x2

99 = 11x3x3

100 = 5x5x2x2

400 = 2x2x2x2x5x5

39 = 3x13

35 = 5x7

80 = 2x2x2x2x5

42 = 2x3x7

5 = 1x5

78 = 2x3x13
6 0
3 years ago
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