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3 years ago

Find -3a divided by 6 when a=-2

Mathematics

2 answers:

podryga [215]3 years ago

4 0

-3a/6

-3(-2) /6

6/6

1 is the answer

belka [17]3 years ago

3 0

So, your expression would look like this. -3(-2) ÷ 6.

Since a negative times a negative is a positive, -3(-2) is equal to 6. 6 ÷ 6 is equal to 1, which will be your answer.

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What is 200 divided by 150

Zarrin [17]

It is 1.3 because there is a remainder

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3 years ago

Every 3 hours, a machine produces 60 baskets what is the unit rate

uranmaximum [27]

Answer:

20 per hour

Step-by-step explanation:

divide 60 by 3=20

8 0

3 years ago

Read 2 more answers

True or False: A race car driver tested his car for time from 0 to 60 mph, and in 20 tests obtained an average of 4.85 seconds w

vredina [299]

Answer:

False

Step-by-step explanation:

given that a race car driver tested his car for time from 0 to 60 mph, and in 20 tests obtained an average of 4.85 seconds with a standard deviation of 1.47 seconds.

Sample size n = 20

since population standard deviation is not known, we can use t critical value for finding out the confidence interval

df=19

t critical = 2.045

Margin of error = $2.045*\frac{s}{\sqrt{n} } \\=2.045*\frac{1.47}{\sqrt{20} } \\=0.6722$

conidence interval = Mean±Margin of error

= $(4.85-0.6722, 4.85+0.6722)\\= (4.1778, 5.5222)$

The given confidence interval is (4.52, 5.18)

Hence the statement is false.

5 0

3 years ago

Find the Taylor series for f(x) centered at the given value of a. [Assume that f has a power series expansion. Do not show that

kherson [118]

Answer:

The answer is

$f(x) = {\displaystyle 8 + \frac{-28}{1}(x+2)+\frac{-36}{2!}(x+2)^2 + \frac{-18}{3!}(x+2)^2 }$

Step-by-step explanation:

Remember that Taylor says that

$f(x) = {\displaystyle \sum\limits_{k=0}^{\infty} \frac{f^{(k)}(a) }{k!}(x-a)^k }$

For this case

$f^{(0)} (-2) = 8(-2)-3(-2)^3 = 8\\f^{(1)} (-2) = 8-3(3)(-2)^2 = -28\\f^{(2)} (-2) = -3(3)2(-2) = -36\\f^{(2)} (-2) = -3(3)2 = -18$

$f(x) = {\displaystyle 8 + \frac{-28}{1}(x+2)+\frac{-36}{2!}(x+2)^2 + \frac{-18}{3!}(x+2)^2 }$

5 0

3 years ago

1<br>Prove that<br>1: sin/1-cot + cos/1-tan=cos+sin

Anit [1.1K]

Answer:

have:

$\frac{sin}{1-cot}+\frac{cos}{1-tan}\\\\=\frac{sin}{1-\frac{cos}{sin} }+\frac{cos}{1-\frac{sin}{cos} }\\\\=\frac{sin^{2} }{sin-cos}+\frac{cos^{2} }{cos-sin} \\\\=\frac{sin^{2} }{sin-cos}-\frac{cos^{2} }{cos-sin}\\\\=\frac{sin^{2}-cos^{2} }{sin - cos}\\\\=\frac{(sin-cos)(sin+cos)}{sin-cos}\\\\=sin+cos$

Step-by-step explanation:

4 0

3 years ago