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rosijanka [135]
2 years ago
10

Help me please? tysm!!! Marking the correct answer brainliest

Mathematics
1 answer:
N76 [4]2 years ago
4 0
I think 52 cups
Hope its help
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Pls help i am on a timer
stich3 [128]

Answer:

x = 1 + \sqrt{2} \ \ or \ 1- \sqrt{2}

Step-by-step explanation:

Given;

x² - 2x - 1 = 0

Solve by completing the square method;

⇒ take the constant to the right hand side of the equation.

x² - 2x = 1

⇒ take half of coefficient of x = ¹/₂ x -2 = -1

⇒ square half of coefficient of x and add it to the both sides of the equation

x^2 +  (-1)^2 = 1 + (-1)^2

(x-1)^2 = 1 + 1\\\\(x-1)^2 = 2\\\\

⇒ take the square root of both sides;

x-1 = +/- \ \ \sqrt{2} \\\\x = 1 + \sqrt{2} \ \ or \ 1- \sqrt{2}

Therefore, option B is the right solution.

3 0
2 years ago
Find W to find the measure of the exterior angle
lions [1.4K]

Answer:

da answer iz w=80% coz they are opposite angels

5 0
3 years ago
Read 2 more answers
What is my weighted grade please explain how it's done.
salantis [7]

That's not enough information to calculate your grade from.

-- You've told us the scores you have in things that go together
to make up (35 + 5 + 25) = 65%  of your grade.

-- We don't know your scores in the things that make up the other
35% of your grade.

4 0
3 years ago
a cell phone company charges $42 per month of service the cost of a new cell phone plus 8 months of service is $415.99 how much
salantis [7]
1247.97 so you have to skip count
3 0
3 years ago
The probability that a can of paint contains contamination is 3.23%, and the probability of a mixing error is 2.4%. The probabil
stich3 [128]

Answer:

4.6%.

Step-by-step explanation:

The probability that a can of paint contains contamination(C) is 3.23%

P(C)=3.23%

The probability of a mixing(M) error is 2.4%.

P(M)=2.4%

The probability of both is 1.03%.

P(C \cap M)=1.03\%

We want to determine the probability that a randomly selected can has contamination or a mixing error. i.e. P(C \cup M)

In probability theory:

P(C \cup M) = P(C)+P(M)-P(C \cap M)\\P(C \cup M)=3.23+2.4-1.03\\P(C \cup M)=4.6\%

The probability that a randomly selected can has contamination or a mixing error is 4.6%.

3 0
3 years ago
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