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2 years ago

12

Jane's cat weighs 4.225 kilograms. How does the value represented by the 2 in the hundredths place compare with the value repres

ented by the 2 in the tenths place?

2 answers:

Wittaler [7]2 years ago

5 0

Answer

It is 1/10 of the value

Step-by-step explanation:

denis-greek [22]2 years ago

4 0

The two in the hundredths place is ten times less than the two in the tenths place.

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(–23) – 45 – (–98) = ?

lisabon 2012 [21]

(–23) – 45 – (–98) =–23 – 45 +98=-68+98=30

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2 years ago

Read 2 more answers

Choose the correct translation for the following statement. It is at most ten.

8_murik_8 [283]

Answer:

x ≤ 10

Step-by-step explanation:

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2 years ago

Select all that are true

sertanlavr [38]

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1 year ago

in a AP the first term is 8,nth term is 33 and sum to first n terms is 123.Find n and common difference

allsm [11]

I believe there is no such AP...

Recursively, this sequence is supposed to be given by

$\begin{cases}a_1=8\\a_k=a_{k-1}+d&\text{for }k>1\end{cases}$

so that

$a_k=a_{k-1}+d=a_{k-2}+2d=\cdots=a_1+(k-1)d$

$a_n=a_1+(n-1)d$

$33=8+(n-1)d$

$21=(n-1)d$

$n$ has to be an integer, which means there are 4 possible cases.

Case 1: $n-1=1$ and $d=21$ . But

$\displaystyle\sum_{k=1}^2(8+21(k-1))=37\neq123$

Case 2: $n-1=21$ and $d=1$ . But

$\displaystyle\sum_{k=1}^{22}(8+1(k-1))=407\neq123$

Case 3: $n-1=3$ and $d=7$ . But

$\displaystyle\sum_{k=1}^4(8+7(k-1))=74\neq123$

Case 4: $n-1=7$ and $d=3$ . But

$\displaystyle\sum_{k=1}^8(8+3(k-1))=148\neq123$

8 0

2 years ago

Find the 64th term of the following arithmetic sequence.<br> 8,11,14,17

ruslelena [56]

Answer: 197

Step-by-step explanation:

The nth term is 3n+5

so the 64th term will be

(3 x 64)+5

192+5

= 197

7 0

3 years ago