What is 98,464 rounded to the nearest ten thousands ?

Mathematics

2 answers:

DENIUS [597]3 years ago

6 0

It is 98,460 because if a unit is 1-4 you round down, 5 to 9 goes up

makkiz [27]3 years ago

4 0

98,464 rounded to the nearest ten thousand<span> is. 8 So the answer will be 100,000</span>

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Adam buys a box of stuff that weighed 6 3/8. if he buys box of fruit that weighed 7 2/5. what is the combined weight.

jolli1 [7]

Answer:

The combined weight is 13 31/40.

Step-by-step explanation:

Given:

Adam buys a box of stuff that weighed 6 3/8. if he buys box of fruit that weighed 7 2/5.

Now, to find the combined weight.

So, we change the mixed fraction to improper first:

Weight of box of stuff = $6\frac{3}{8}=\frac{51}{8}$ .

Weight of box of fruit = $7\frac{2}{5}=\frac{37}{5}$ .

Then, by adding to get the combined weight:

$\frac{51}{8}+\frac{37}{5}$

$=\frac{51\times 5+37\times 8}{40}$

$=\frac{255+296}{40}$

$=\frac{551}{40}$

$=13\frac{31}{40}$

Therefore, the combined weight is 13 31/40.

4 0

3 years ago

Infinite over E n=1 4(0.5)^n-1

yan [13]

Answer:

S = 8

Step-by-step explanation:

An infinite geometric series is defined as limit of partial sum of geometric sequences. Therefore, to find the infinite sum, we have to find the partial sum first then input limit approaches infinity.

However, fortunately, the infinite geometric series has already set up for you. It’s got the formula for itself which is:

$\displaystyle \large{\lim_{n\to \infty} S_n = \dfrac{a_1}{1-r}}$

We can also write in summation notation rather S-term as:

$\displaystyle \large{\sum_{n=1}^\infty a_1r^{n-1} = \dfrac{a_1}{1-r}}$

Keep in mind that these only work for when |r| < 1 or else it will diverge.

Also, how fortunately, the given summation fits the formula pattern so we do not have to do anything but simply apply the formula in.

$\displaystyle \large{\sum_{n=1}^\infty 4(0.5)^{n-1} = \dfrac{4}{1-0.5}}\\\\\displaystyle \large{\sum_{n=1}^\infty 4(0.5)^{n-1} = \dfrac{4}{0.5}}\\\\\displaystyle \large{\sum_{n=1}^\infty 4(0.5)^{n-1} = 8}$