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3 years ago

Which of the following numbers is least, or furthest to the left on the number line? {48, 95, 21, 64, 53}

1 answer:

6 0

The number that is the least or furthest is 21 because they are all positive and so that makes the littles number the least, but if they where all negatives it would be the greatest negative.

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Part 9: I need help. please help me

Trava [24]

Answer: A) a² = b² - w² + 2wx

Step-by-step explanation:

b² - (w - x)² = a² - x²

b² - (w² - 2wx + x²) = a² - x²

b² - w² + 2wx - x² = a² - x²

b² - w² + 2wx = a²

3 0

3 years ago

Can someone help me please im very confused and please use step by step please -2/3x+1<7

Brut [27]

Answer:

x>-9

Step-by-step explanation:p

1.) Subtract 1 from both sides

-2/3x+1<7

-1<-1

= -2/3x < 6

2.) Multiple the 3 (denominator) in order to get rid of the fraction

3(-2/3x) < (6)3

= -6/3x < 18 (simplify the 6/3 into 2)

= -2x < 18

3.) Divide by -2 (when you divide with a negative number, your sin is going to switch, meaning it will turn into a '>' sign)

-2x / -2 < 18/-2

= x > -9

*ALSO*

If you want to graph it on a number line, it would look something like the picture above. (the circle is open because it doesn't have the line underneath. You're symbol also indicates which way you shade!! < = left .....> = right.....think of them like arrows :) )

6 0

3 years ago

Please help I need this done by tomorrow Thank you

PtichkaEL [24]

19. C

20. B

21. 200

22. im not sure if im right but i think p+6

23. C

24. Can't see

5 0

3 years ago

Read 2 more answers

Evaluate using integration by parts

PolarNik [594]

Rather than carrying out IBP several times, let's establish a more general result. Let

$I(n)=\displaystyle\int x^ne^x\,\mathrm dx$

One round of IBP, setting

$u=x^n\implies\mathrm du=nx^{n-1}\,\mathrm dx$

$\mathrm dv=e^x\,\mathrm dx\implies v=e^x$

gives

$\displaystyle I(n)=x^ne^x-n\int x^{n-1}e^x\,\mathrm dx$

$I(n)=x^ne^x-nI(n-1)$

This is called a power-reduction formula. We could try solving for $I(n)$ explicitly, but no need. $n=5$ is small enough to just expand $I(5)$ as much as we need to.