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maw [93]
3 years ago
15

Algebra KPL and LJPL are a linear pair, m_ KPL = 2x + 24, and mLJPL = 4x + 36. What are the measures of ZKPL and LJPL? (2x + 24)

(4x + 36) к P
​

Mathematics
1 answer:
denis23 [38]3 years ago
4 0

Answer:

  • 64° = ∠KPL
  • 116° = ∠JPL

Step-by-step explanation:

The sum of the measures of a linear pair of angles is 180°.

  (2x+24) +(4x+36) = 180

  6x = 120 . . . . . . subtract 60, collect terms

  x = 20

  (2(20) +24)° = 64° = ∠KPL

  (4(20) +36)° = 116° = ∠JPL

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pav-90 [236]

Answer: -8x^3y^2  (choice B)

The coefficients multiply to 4 times -2 = -8

the x terms multiply to x^2 times x = x^(2+1) = x^3

The y terms multiply to y times y = y^(1+1) = y^2

So that's how I ended up with -8x^3y^2

7 0
3 years ago
Given that a function, g, has a domain of -1 ≤ x ≤ 4 and a range of 0 ≤ g(x) ≤ 18 and that g(-1) = 2 and g(2) = 8, select the st
Juli2301 [7.4K]

Answer:

Option 4 is correct.

Step-by-step explanation:

Consider a function g, it has a domain of -1 ≤ x ≤ 4 and a range of 0 ≤ g(x) ≤ 18. It is given that g(-1) = 2 and g(2) = 8.

The statement g(5) = 12 is not true because the value of x is 5 which is not in its domain.

The statement g(1) = -2 is not true because the value of function g(x) is -2 which is not in its range.

The statement g(2) = 4 is not true because g is a function and each function has unique output for each input value.

If g(2)=8 and g(2)=4, then the value of g(x) is 8 and 4 at x=2. It means g(x) is not a function, which is contradiction of given statement.

The statement g(3) = 18 is true because the value of x is 3 which is in the domain and the value of function g(x) is 18 which is in its range.

Therefore, the correct option is 4.

8 0
3 years ago
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The ratio of the sides of two similar octagons is 2:5. What is the ratio of the perimeters of these octagons?
Tom [10]
So we see the ratio of the sides of two similar octagons is 2:5 so let's say the first octagon has a side of 2 and the other has a side of 5. 2*8 = 16 and 5*8 = 40. 16:40 = 4:10 = and is again 2:5. So the answer is 2:5.
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\bf \begin{array}{ccll}
x&y\\
\cline{1-2}
1&\stackrel{-1^3}{-1}\\\\
2&\stackrel{-2^3}{-8}\\\\
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x&-x^3
\end{array}~\hspace{5em}\boxed{y=-x^3}~~\checkmark

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erastovalidia [21]
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