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statuscvo [17]
3 years ago
6

What is 0.362 in expanded form

Mathematics
2 answers:
olchik [2.2K]3 years ago
8 0
.362 in expanded form is 
.300+.060+.002

Hope this helps!! :D
Alexandra [31]3 years ago
7 0
0.362 in expanded form is 0.3 + 0.06 + 0.002
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Find the volume of the solid below.
9966 [12]

Answer:

3000 cm^3.

Step-by-step explanation:

If the top number is 15 it's

15 * 20 * 10

=  3000 cm^3.

7 0
3 years ago
Read 2 more answers
ILL GIVE BRAINLIEST! PLEASE HELP I DONT UNDERSTAND!
blsea [12.9K]

Answer:

S₁₅ = 645

Step-by-step explanation:

The sum to n terms of an arithmetic sequence is

S_{n} = \frac{n}{2} [ 2a₁ + (n - 1)d ]

where a₁ is the first term and d the common difference

Using the nth term formula a_{n} = 5n + 3 , then

a₁ = 5(1) + 3 = 5 + 3 = 8

a₂ = 5(2) + 3 = 10 + 3 = 13 , then

d = a₂ - a₁ = 13 - 8 = 5

Thus

S₁₅ = \frac{15}{2} [ (2 × 8) + (14 × 5) ]

     = 7.5( 16 + 70)

     = 7.5 × 86

     = 645

7 0
3 years ago
Can you guys answer these questions please i have allot of work and im trying to do multiple things answer will get brainliest
SpyIntel [72]

Answer:

first one - 30%

second - 15%

third - 40%

fourth - 20%

fifth - 75%

sixth - 70%

seventh - 88%

eighth - 25%

sorry but the last question is too blurry, can you type it and I can help you answer it?

Hope this helps.

4 0
3 years ago
HELP PLS LMK IF U CANT READ IT AND I WILL TYPE IT
GenaCL600 [577]

Answer:

cant read it type it pls

Step-by-step explanation:

8 0
3 years ago
What the recursive formula for the sequence
klemol [59]

\bf n^{th}\textit{ term of an arithmetic sequence} \\\\ a_n=a_1+(n-1)d\qquad \begin{cases} n=n^{th}\ term\\ a_1=\textit{first term's value}\\ d=\textit{common difference} \end{cases} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ a_n=2-5(n-1)\implies a_n=\stackrel{\stackrel{a_1}{\downarrow }}{2}+(n-1)(\stackrel{\stackrel{d}{\downarrow }}{-5})


so, we know the first term is 2, whilst the common difference is -5, therefore, that means, to get the next term, we subtract 5, or we "add -5" to the current term.

\bf \begin{cases} a_1=2\\ a_n=a_{n-1}-5 \end{cases}


just a quick note on notation:

\bf \stackrel{\stackrel{\textit{current term}}{\downarrow }}{a_n}\qquad \qquad \stackrel{\stackrel{\textit{the term before it}}{\downarrow }}{a_{n-1}} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{current term}}{a_5}\qquad \quad \stackrel{\textit{term before it}}{a_{5-1}\implies a_4}~\hspace{5em}\stackrel{\textit{current term}}{a_{12}}\qquad \quad \stackrel{\textit{term before it}}{a_{12-1}\implies a_{11}}

8 0
4 years ago
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