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ArbitrLikvidat [17]
4 years ago
13

Solve A=1/2bh for b.

Mathematics
1 answer:
Semenov [28]4 years ago
4 0
A=1/2bh
Divide by h and 1/2
2Ah=b
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What is 5.6e12 in standard form
bixtya [17]
-911426.83194642 is the standard form.
3 0
3 years ago
Read 2 more answers
Anyone help pls? No links! :)
Lunna [17]

Answer:

6. D

7. F

8.A

9. B

10 C

13.Question- Write an expression using division and subtraction with a difference of 3.

answer-  (12 divide 3) -1

  Twelve divided by three is four. Four minus 1 is three.

12. Question- Write an expression using multiplication and addition with a sum of 16.

answer- There are many answers for this and this is one of them:

Y=4(3+1)

Y=16

or

Y=2(6+2)

Y=16

11.  Question- Sam bought two CDs for $13 each. Sales tax for both CDs was $3. Write an expression to show how much Sam paid in all.

answer- Expression- (13*2)(3*2)=$32

This is the answer because Sam bought two CD's for 13, 13*2, and the sales tax was $ 3, so 3*2=6

4 0
3 years ago
an administrator is asked to file papers in a box which has the following dimensions:height 15cm, width 300mm and depth 0.2m. ca
nalin [4]

Answer:

900cm^3

Step-by-step explanation:

1 cm= 10 mm

x=300mm (criss cross)

1cm*300mm/10mm= 10mm*x/10mm

30cm=x

30cm=300mm

1m=100cm

0.2m=x

1m*x/1m=0.2m*100cm/1m

x=0.2*100cm

x=20cm

0.2m=20cm

l=15cm

w=30cm

h=20cm

v=l*w*h

v=15cm*30cm*20cm

v=450cm^2*20cm

v=900cm^3

6 0
3 years ago
Read 2 more answers
Determine whether the series is convergent or divergent. 1 2 3 4 1 8 3 16 1 32 3 64 convergent divergent Correct:
Vanyuwa [196]

Answer:

This series diverges.

Step-by-step explanation:

In order for the series to converge, i.e. \lim_{n \to \infty} a_n =A it must hold that for any small \epsilon>0, there must exist n_0\in \mathbb{N} so that starting from that term of the series all of the following terms satisfy that  |a_n-A|n_0 .

It is obvious that this cannot hold in our case because we have three sub-series of this observed series. One of them is a constant series with a_n=1 , the other is constant with a_n=3 , and the third one has terms that are approaching infinity.

Really, we can write this series like this:

a_n=\begin{cases} 1 \ , \ n=4k+1, k\in \mathbb{N}_0\\ 2^{k}\ , \ n=2k, k\in \mathbb{N}_0\\3\ , \ n=4k+3, k\in \mathbb{N}_0\end{cases}

If we  denote the first series as b_n=1, we will have that \lim_{k \to \infty} b_k=1.

The second series is denoted as c_k=2^k and we have that \lim_{k \to \infty} c_k=+\infty.

The third sub-series d_k=3 is a constant series and it holds that \lim_{k \to \infty} d_k=3.

Since those limits of sub-series are different, we can never find such n_0\\ so that every next term of the entire series is close to one number.

To make an example, if we observe the first sub-series if follows that A must be equal to 1. But if we chose \epsilon =1, all those terms associated with the third sub-series will be out of this interval (A-1, A+1)=(0, 2).

Therefore, the observed series diverges.

5 0
4 years ago
I dont get this please help me im smol
muminat

Answer:

what question do you have?

4 0
3 years ago
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