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4 years ago

6

The triangles are similar.

1 answer:

Natalija [7]4 years ago

4 0

X=23

I say this because you can see that the scale factor is 3 due to just visually scaling both the long and short leg from the original triangle to the dilated triangle. Dividing 39 by 3 is 13, and doing it in reverse. Adding 10 would equal 23, the initial value. So that'd be the missing value.

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Which ordered pair is a solution of the equation? y=8x+3 1/11 or -1/-5 or both

suter [353]

<h3>✽ - - - - - - - - - - - - - - - ~Hello There!~ - - - - - - - - - - - - - - - ✽</h3>

➷ Substitute the values in:

1st option:

y = 8(1) + 3

y = 11

2nd option:

y = 8(-1) + 3

y = -5

The correct answer would be both.

➶ Hope This Helps You!

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4 years ago

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How many different 7-place license plates are possible when 3 of the entries are letters and 4 are digits? Assume that repetitio

timofeeve [1]

Answer:

There are 6,151,600,000 different 7-place license plates are possible when 3 of the entries are letters and 4 are digits,

Step-by-step explanation:

For each of the entries which are letters, there are 26 possible outcomes.

For each of the entries which are digits, there are 10 possible outcomes.

These outcomes can be permutated.

For example, ABC1234 is a different outcome than A1B2C34. This means that we need to use the permutations formula.

The number of permutations of n, divided into two groups of size a and b, is:

$P_{a,b}^{n} = \frac{n!}{a!b!}$ .

In this problem, we have a permutation of 7, divided into a group of 4(digits) and 3(letters).

How many different 7-place license plates are possible when 3 of the entries are letters and 4 are digits?

This is $P_{3,4}^{7}*(26)^{3}*10^{4} = 35*(26)^{3}*10^{4} = 6,151,600,000$

There are 6,151,600,000 different 7-place license plates are possible when 3 of the entries are letters and 4 are digits,

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3 years ago

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Serga [27]

<P = <S
<Q = <T
PQ = ST
both triangles are congruent by ASA

Answer
B. Congruent by ASA

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3 years ago

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Licemer1 [7]

Answer: -1/2

Step-by-step explanation:

Use double-angle identities:

cos(2θ)=cos^2θ−sin2^θ

cos(2θ)=1−2sin^2θ

cos(2θ)=2cos^2θ−1

sin(x)=1/2

cos(2x)=1−2sin^2x=1−2(1/4)=1/2

cos(4x)=2cos2^(2x)−1=2(1/4)−1=-1/2

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2 years ago

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Express 6x5x4 in nPr form

larisa [96]

<span>The number of different permutation of n number of objects can be written as nPr.</span>
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4 years ago

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