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3 years ago

Square 24-×whenx=6. Write in radical form

Mathematics

1 answer:

Delicious77 [7]3 years ago

5 0

Answer:

√ 18 = 3 √2

Step-by-step explanation:

So 24-6 is 18 and then the square root of that. You have to divide by a perfect square and the largest perfect square that can go into 18 is nine because 3x3. 9 can go into 18 2 times so it is 2 radical 9

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Which of the expressions below can be factored using the difference of squares method?

zlopas [31]

Answer:

The second choice is correct. It can be factored as:

$9x^2-16y^2=(3x-4y)(3x+4y)$

Step-by-step explanation:

The Difference of Squares Method for Factoring

The expression:

$a^2-b^2=(a-b)(a+b)$

Is a widely used method to factor binomials that are expressed as the subtraction of two perfect squares.

The condition for a binomial to be factored by using this method is that both terms must have an exact square root and they must be subtracted.

The last two choices are not valid because they are not a subtraction but an addition.

The first choice is not valid because none of the terms is a perfect square.

The second choice is correct. It can be factored as:

$9x^2-16y^2=(3x-4y)(3x+4y)$

4 0

3 years ago

Solve this pleaseeeee

Ivan

Answer:

-4 < n ≤ 5

Step-by-step explanation:

________________

7 0

3 years ago

o The day before the exam, Rebecca does a practice run. She attempts 20 glass vases and breaks 3 of them. What is her success ra

Rashid [163]

Answer:

If the goal is to break the vases: 3/20 = 15.0% or 0.15

If the goal is to not break the vases: 17/20 = 85.0% or 0.85

8 0

3 years ago

Complete the equation for the standard form of the line that has an x-intercept of 3 and a y-intercept of 5

SIZIF [17.4K]

Answer:

5x+3y=-15

Step-by-step explanation:

4 0

4 years ago

A couple intends to have two children, and suppose that approximately 52% of births are male and 48% are female.

Pachacha [2.7K]

a) Probability of both being males is 27%

b) Probability of both being females is 23%

c) Probability of having exactly one male and one female is 50%

Step-by-step explanation:

The probability that the birth is a male can be written as

$p(m) = 0.52$ (which corresponds to 52%)

While the probability that the birth is a female can be written as

$p(f) = 0.48$ (which corresponds to 48%)

Here we want to calculate the probability that over 2 births, both are male. Since the two births are two independent events (the probability of the 2nd to be a male does not depend on the fact that the 1st one is a male), then the probability of both being males is given by the product of the individual probabilities:

$p(mm)=p(m)\cdot p(m)$

And substituting, we find

$p(mm)=0.52\cdot 0.52 = 0.27$

So, 27%.

In this case, we want to find the probability that both children are female, so the probability

$p(ff)$

As in the previous case, the probability of the 2nd child to be a female is independent from whether the 1st one is a male or a female: therefore, we can apply the rule for independent events, and this means that the probability that both children are females is the product of the individual probability of a child being a female:

$p(ff)=p(f)\cdot p(f)$