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mash [69]
4 years ago
8

Simplify 20-30[4(z+1)-6(z-2)]

Mathematics
1 answer:
algol [13]4 years ago
6 0

Answer:

Step-by-step explanation:

Apply distributive property

20-30(4z+4*1-6(z-2)

multiply 4 by 1

20-30(4z+4-6(z-2))

apply distributive property

20-30(4z+4-6z+12)

multiply -6 by -2

20-30(4z+4-6z+12)

subtract 6z from 4z

20-30(-2z+4+12)

add 4 and 12

20-30(-2z+16)

apply distributive property

20-30(-2z)-30*16

multiply -2 by -30

20+60z-30*16

myltiply 30 by 16

20+60z-480

subtract 480 from 20

60z-460

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A shopkeeper has 438 apples. 12 of them are bad. He packs the rest equally into 6 boxes. How many apples are in each box?
WITCHER [35]

Answer:

71 apples in each box

Step-by-step explanation:

438-12=426

426 divided by 6= 71

6 0
3 years ago
Read 2 more answers
Solve for x. 4=6x/a +5
Semenov [28]
X=-a/6

Because you need to isolate x
8 0
3 years ago
A cube-shaped box has a volume of 27 cubic inches. If the box is packed full of cubes with edge lengths of 1 inch, how many cube
iogann1982 [59]

Answer:

9 along one side.

Step-by-step explanation:

This question may be a bit tricky.

The side should mean the entire side, not just along the bottom edge.

The bottom row will have three 1-inch cubes. Three more in a middle row, and three more on top of that. That is a total of nine on that side.

8 0
3 years ago
(1 point) Each of the following statements is an attempt to show that a given series is convergent or divergent not using the Co
kvv77 [185]

Answer:

I.

CORRECT.

II.

CORRECT.

III.

CORRECT.

IV.

CORRECT.

V.

INCORRECT.

VI.

CORRECT

Step-by-step explanation:

To understand let us restate the comparison test in simple terms.

Comparison test :

Given     \text{Series}_A   and \text{Series}_B such that    \text{Series}_A < \text{Series}_B , then

1.   If    \text{Series}_B converges then \text{Series}_A  converges as well.

2.  If  \text{Series}_A  diverges then \text{Series}_B  diverges as well.

Now to give you a more intuitive idea of what is going on, think about it like this.  When   the series on top converges it is like an "upper bound" for what you have on the bottom, therefore what you have on the bottom has to converge as well.

Similarly if what you have on the bottom explotes, then what you have on top will explote as well.

That's how I like to think about that intuitively.

Now, using those results let us examine the statements.

I.

\frac{1}{n} < \frac{ln(n)}{n}

Since the infinite sum of 1/n  diverges in fact the infinite sum of ln(n)/n does not converge.

Therefore, CORRECT.

II.

\frac{\arctan(n)}{n^3} < \frac{\pi}{2}\frac{1}{n^3}  

Since the infinite sum of    \frac{\pi}{2}\frac{1}{n^3}   is in fact convergent then  \frac{\arctan(n)}{n^3} converges as well using the comparison theorem. Therefore

CORRECT.

III.

\frac{n}{2-n^3} < \frac{1}{n^2}

Once again   1/n^2  does converge so what you have on the bottom converges as well. Therefore

CORRECT.

IV.

\frac{\ln(n)}{n^2} < \frac{1}{n^{1.5}}

Once again   \frac{1}{n^{1.5}}  converges therefore since it is on top what is on the bottom converges as well. Therefore.

CORRECT.

V.

\frac{\ln(n)}{n} < \frac{2}{n}

Now the fact that  \frac{2}{n}    diverges does not necessarily imply that what you have on the bottom diverges. Therefore

INCORRECT.

VI.

That is correct as well since what you have on top converges therefore what you have on the bottom converges as well.

 

6 0
3 years ago
Help me please will mark brainlyist
Leni [432]

Answer:

24.46

Step-by-step explanation:

4 0
4 years ago
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