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3 years ago

How to write thirty-four and twelve hundredths in standard form

1 answer:

7 0

To write this value simply write the first part of the value before the decimal point

Which is 34

Then since it says 12 hundredths, make sure to place them after the decimal place.

0.12 = 12 hundredths

34.12 would be the answer.

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How do you use the pythagoreum theorum again? i know the formula is a^2+b^2=c^2 but how do you use it. Use the example 15^2+20^2

kiruha [24]

It would be 225+400=c^2. square both sides to result in 25.

3 0

3 years ago

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What is the inverse of the given coordinates: (1, 2) (3, 8) (-3, 6)?

Vilka [71]

Answer:(2,1) (8,3) (6,-3)

7 0

3 years ago

Please need help on this

Ronch [10]

Ray FM has terminal point F, so that will be the first letter of the name. The second letter of the name can be any of the points to the left of F on the line:

... D, L, M, P, K

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8 0

3 years ago

CORRECT ANSWER GETS BRAINLIEST A frequency table of grades has five classes (A, B, C, D, F) with frequencies of 4,11 ,15 ,

ra1l [238]

Answer:

A = 10.5%

B = 28.9%

C = 39.5%

D = 18.4%

F = 2.6%

Step-by-step explanation:

Add all the frequencies together = 38

4/38 × 100 = 10.5%
11/38 × 100 = 28.9%
15/38 × 100 = 39.5%
7/38 × 100 = 18.4%
1/38 × 100 = 2.6%

10.5 + 28.9 + 39.5 + 18.4 + 2.6 = 100%

5 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.

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

Therefore, the sum will converge to 8.

Please let me know if you have any questions!

6 0

2 years ago