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3 years ago

10

In a song competition, a participant has to score a total of at least 20 points in the first four rounds combined to move on to

the fifth and final round. Curtis scored 2 points in the first round. He then went on to score additional points in the second, third, and fourth rounds. In each of those rounds, his score was identical. Which inequality best shows the number of points, p, that Curtis scored in each of the second, third, and fourth rounds if he earned a place in the finals? 2 + 3p ≤ 20 2p + 3 ≥ 20 2 + 3p ≥ 20 2p + 3 ≤ 20

2 answers:

arsen [322]3 years ago

6 0

Answer:

2+3p ≥ 20

Step-by-step explanation:

2 is the amount of points he earned

3 is how many rounds were left before finals

p is the additional points he earned

and ≥ 20 is he had to earn greater than or equal to 20 points total to move on.

Julli [10]3 years ago

6 0

Answer:

2+3p ≥ 20

Step-by-step explanation:

hope i helped

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Answer:

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Step-by-step explanation:

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If both pairs of its opposite sides of a trapezoid are parallel, it becomes a parallelogram

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3 years ago

Prove that sin3a-cos3a/sina+cosa=2sin2a-1

Sloan [31]

Answer:

$\frac{sin(3a)-cos(3a)}{sin(a)+cos(a)} =2sin(2a)-1$

Step-by-step explanation:

we are given

$\frac{sin(3a)-cos(3a)}{sin(a)+cos(a)} =2sin(2a)-1$

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we can use trig identity

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now, we can plug values

$\frac{(3sin(a)-4sin^3(a))-(4cos^3(a)-3cos(a))}{sin(a)+cos(a)}$

now, we can simplify

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$\frac{3(sin(a)+cos(a))-4(sin^3(a)+cos^3(a))}{sin(a)+cos(a)}$

now, we can factor it

$\frac{3(sin(a)+cos(a))-4(sin(a)+cos(a))(sin^2(a)+cos^2(a)-sin(a)cos(a)}{sin(a)+cos(a)}$

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we can use trig identity

$sin^2(a)+cos^2(a)=1$

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we can cancel terms

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3 years ago

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Solve for x, given the equation Square root of x-5+7=11

denis-greek [22]

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Check the answer:

$\sqrt{119-5+7} \\\sqrt{114+7} \\\sqrt{121} \\11$

This answer is correct,

x = 119

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3 years ago

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victus00 [196]

Answer:

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Step-by-step explanation:

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balandron [24]

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