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Komok [63]
3 years ago
14

Which of the following values in the set below will make the equation 2x + 1 = 3 true? (Only input the number.) {0, 1, 2, 3, 4}

Numerical Answers Expected! Answer for Blank 1:
Mathematics
1 answer:
ollegr [7]3 years ago
4 0
The value of x is 1 ( the second choice )
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One solution to the problem below is 4.<br> What is the other solution?<br> y² - 16 = 0
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The 2 answers are 4 and -4

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Which of the following expressions shows the distributive property applied to the expression below?
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C. (z×6)+(z×35)

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PLEASE HELP THIS IS URGENT I AM BEING TIMED!!
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1/30 or 3.333%

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Morgan is working two summer jobs, making $19 per hour lifeguarding and making $6 per hour walking dogs. In a given week, she ca
Alexxx [7]

Answer:

w+d≥14

Step-by-step explanation:

Here is the full question

Morgan is working two summer jobs, washing cars and walking dogs. She must work no less than 14 hours altogether between both jobs in a given week. Write an inequality that would represent the possible values for the number of hours washing cars, w, and the number of hours walking dogs, d, that Morgan can work in a given week.

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So the time she would spend working = w+d≥14

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6 0
3 years ago
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Use the pythagorean theorem to prove that this point lies on the unit circle. Find csc(0) where 0=-120.
exis [7]

By applying Pythagorean theorem, we have proven that the point (-1/2, -√3/2) lies on the unit circle.

<h3>How to prove this point lies on the unit circle?</h3>

In Trigonometry, an angle with a magnitude of -120° is found in the third quarter and as such, both x and y would be negative. Also, we would calculate the reference angle for θ in third quarter as follows:

Reference angle = 180 - θ

Reference angle = 180 - 120

Reference angle = 60°.

For the coordinates, we have:

sin(-120) = -sin(60) = -1/2.

cos(-120) = -cos(60) = -√3/2.

By applying Pythagorean theorem, we have:

z² = x² + y²

z = √((-1/2)² + (-√3/2)²)

z = √(1/4 + 3/4)

z = √1

z = 1.

Read more on unit circle here: brainly.com/question/9797740

#SPJ1

4 0
2 years ago
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