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2 years ago

What is the value of –2|6x – y| when x = –3 and y = 4?

Mathematics

2 answers:

irakobra [83]2 years ago

8 0

Answer:

The correct answer is -44

Step-by-step explanation:

Points to remember

|x| = x

|-x| = x

To find the value of –2|6x – y| 

We have –2|6x – y| When x = -3 and y = 4

Substitute the value of x and y

–2|6x – y| = –2|(6 * -3) – 4|

= -2|-18 - 4|

= -2|-22|

= -2 * 22

= -44

Therefore the value of –2|6x – y| when x = -3 and y = 4

–2|6x – y|

kenny6666 [7]2 years ago

5 0

Answer:

-44

Step-by-step explanation:

We have the expression

$-2|6x - y|$

We need to find the value of the expression when x = -3 and y = 4

Then we must replace the variable with the number -3 and the variable y with the number 4

This is:

$-2|6(-3) - (4)|$

$-2|-18 - 4|$

$-2|-22|$

Remember that the absolute value always results in a positive number.

$-2|-22| = -2(22)=-44$

the answer is -44

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Help with this please lol

NeTakaya

Answer:

x = 107

Step-by-step explanation:

44+29+107=180

a triangle equals 180 with all the degrees added together :)

7 0

3 years ago

Please help me i don’t understand this question at all.

Maru [420]

9514 1404 393

Answer:

Step-by-step explanation:

The wording "when x is an appropriate value" is irrelevant to this question. That phrase should be ignored. (You may want to report this to your teacher.)

When you look at the answer choices, you see that all of them are negative except the last one (D). When you look at the problem fraction, you see that it is positive.

The only reasonable choice is D.

Your calculator can check this for you.

√12/(√3 +3) ≈ 3.4641/(1.7321 +3)

= 3.4641/4.7321 ≈ 0.7321 = -1 +√3

If you want to "rationalize the denominator", then multiply numerator and denominator by the conjugate of the denominator. The conjugate is formed by switching the sign between terms.

$\displaystyle\frac{\sqrt{12}}{\sqrt{3}+3}=\frac{\sqrt{12}}{(\sqrt{3}+3)}\cdot\frac{(\sqrt{3}-3)}{(\sqrt{3}-3)}=\frac{\sqrt{36}-3\sqrt{12}}{3-9}\\\\=\frac{6-3\cdot2\sqrt{3}}{-6}=\boxed{-1+\sqrt{3}}$

_____

Additional comment

We "rationalize the denominator" in this way to take advantage of the relation ...

(a -b)(a +b) = a² -b²

Using this gets rid of the irrational root in the denominator, hence "rationalizes" the denominator.

We could also have multiplied by (3 -√3)/(3 -√3). This would have made the denominator positive, instead of negative. However, I chose to use (√3 -3) so you could see that all we did was change the sign from (√3 +3).

3 0

2 years ago

What’s is 2 1/2 divided 1/3

Marat540 [252]

Answer:

7 $\frac{1}{2}$

Step-by-step explanation:

$2\frac{1}{2} *$ the reciprocal of $\frac{1}{3}$

To find reciprocal, flip the top and the bottom.

So $\frac{1}{3}$ would be $\frac{3}{1}$ .

Multiply like normal.

6 0

3 years ago

United Airlines' flights from Denver to Seattle are on time 50 % of the time. Suppose 9 flights are randomly selected, and the n

Ivanshal [37]

Answer:

a) The probability that exactly 4 flights are on time is equal to 0.0313

b) The probability that at most 3 flights are on time is equal to 0.0293

c) The probability that at least 8 flights are on time is equal to 0.00586

Step-by-step explanation:

The question posted is incomplete. This is the complete question:

United Airlines' flights from Denver to Seattle are on time 50 % of the time. Suppose 9 flights are randomly selected, and the number on-time flights is recorded. Round answers to 3 significant figures.

a) The probability that exactly 4 flights are on time is =

b) The probability that at most 3 flights are on time is =

c)The probability that at least 8 flights are on time is =

<h2>Solution to the problem</h2>

a) Probability that exactly 4 flights are on time

Since there are two possible outcomes, being on time or not being on time, whose probabilities do not change, this is a binomial experiment.

The probability of success (being on time) is p = 0.5.

The probability of fail (note being on time) is q = 1 -p = 1 - 0.5 = 0.5.

You need to find the probability of exactly 4 success on 9 trials: X = 4, n = 9.

The general equation to find the probability of x success in n trials is:

$P(X=x)=_nC_x\cdot p^x\cdot (1-p)^{(n-x)}$

Where $_nC_x$ is the number of different combinations of x success in n trials.

$_nC_x=\frac{x!}{n!(n-x)!}$

Hence,

$P(X=4)=_9C_4\cdot (0.5)^4\cdot (0.5)^{5}$

$_9C_4=\frac{4!}{9!(9-4)!}=126$

$P(X=4)=126\cdot (0.5)^4\cdot (0.5)^{5}=0.03125$

b) Probability that at most 3 flights are on time

The probability that at most 3 flights are on time is equal to the probabiity that exactly 0 or exactly 1 or exactly 2 or exactly 3 are on time:

$P(X\leq 3)=P(X=0)+P(X=1)+P(X=2)+P(X=3)$

$P(X=0)=(0.5)^9=0.00195313$ . . . (the probability that all are not on time)

$P(X=1)=_9C_1(0.5)^1(0.5)^8=9(0.5)^1(0.5)^8=0.00390625$

$P(X=2)=_9C_2(0.5)^2(0.5)^7=36(0.5)^2(0.5)^7=0.0078125$

$P(X=3)= _9C_3(0.5)^3(0.5)^6=84(0.5)^3(0.5)^6=0.015625$

$P(X\leq 3)=0.00195313+0.00390625+0.0078125+0.015625=0.02929688\\\\ P(X\leq 3) \approx 0.0293$

c) Probability that at least 8 flights are on time 

That at least 8 flights are on time is the same that at most 1 is not on time.

That is, 1 or 0 flights are not on time.

Then, it is easier to change the successful event to not being on time, so I will change the name of the variable to Y.

$P(Y=0)=_0C_9(0.5)^0(0.5)^9=0.00195313\\ \\ P(Y=1)=_1C_9(0.5)^1(0.5)^8=0.0039065\\ \\ P(Y=0)+P(Y=1)=0.00585938\approx 0.00586$

6 0

3 years ago

1. Write a two-column proof for the following conjecture. You may not need to use all of the rows of the two-column table provid

marshall27 [118]

Answer:

Given: ABCD is a parallelogram.

Prove: Angle A and B are supplementary and C and D are supplementary.

That is, m∠A + m∠B = 180° and m∠C + m∠D = 180°

Statement Reason

1. ABCD is a parallelogram 1. Given

2.m∠A=m∠C and m∠B=m∠D 2. Definition of parallelogram.

3.m∠A+m∠B+m∠C+m∠D=360° 3. Definition of quadrilateral

4. m∠A+m∠B+m∠A+m∠B=360° 4. By substitution

⇒ 2( m∠A + m∠B ) = 360°

⇒ m∠A + m∠B = 180°

Similarly, m∠C + m∠D = 180°

5.∠A and ∠C are supplementary, 5. Def. of Supplementary

∠ B and ∠D are supplementary

7 0

3 years ago

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