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3 years ago

HELP ME ASAP PLEASE?!?If the pattern below follows the rule "Starting with 10, every consecutive line

Mathematics

1 answer:

Vlad [161]3 years ago

8 0

5 marbles would be in the sixth line

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Complete the square to make a perfect square trinomial. Then, write the result as a binomial squared. c2+3c

Nana76 [90]

Answer:

(c+3/2)^2

Step-by-step explanation:

8 0

3 years ago

7/10 < > = 3/6 3/8 < > = 3/6 7/10 < > = 3/8

IceJOKER [234]

Answer:

a graph would be needed to answer this question

Step-by-step explanation:

3 0

3 years ago

Noah’s recipe for one batch of sparkling orange juice uses 4 liters of orange juice and 5 liters of soda water. If someone uses

navik [9.2K]

Answer:

They would need 500 liters of soda water and this would make 100 batches of the sparkling orange juice.

Step-by-step explanation:

hope this helps. . . <3

good luck! uωu

7 0

2 years ago

Read 2 more answers

Solve for x in the equation x2 - 12x+36 = 90.

cupoosta [38]

Answer:

The correct answer is A. x = 6 + 3 √10

Step-by-step explanation:

Let's solve for x:

x² - 12x + 36 = 90

Factoring this quadratic equation, we have:

(x - 6) (x - 6) = 90

(x - 6)² = 90

x - 6 = √90 (Square root to both sides of the equation)

x = 6 + √90 (Adding 6 to both sides of the equation)

x = 6 + √9 * 10

x = 6 + √3² * 10

x = 6 + 3 √10

The correct answer is A. x = 6 + 3 √10

5 0

3 years ago

Asked what the central limit theorem says, a student replies, As you take larger and larger samples from a population, the histo

Arada [10]

Answer:

A. No, the student is not right. The central limit theorem says nothing about the histogram of the sample values. It deals only with the distribution of the sample means.

Step-by-step explanation:

No, the student is not right. The central limit theorem says nothing about the histogram of the sample values. It deals only with the distribution of the sample means. The central limit theorem says that if we take a large sample (i.e., a sample of size n > 30) of any distribution with finite mean $\mu$ and standard deviation $\sigma$ , then, the sample average is approximately normally distributed with mean $\mu$ and variance $\sigma^2/n$ .

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3 years ago