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Sergio [31]
3 years ago
5

What is 1.96 billion in scientific notation?

Mathematics
2 answers:
igomit [66]3 years ago
7 0

Answer:

A.

Step-by-step explanation:

1 billion is equal to 10^9. Then, 1.96 billion in scientific notation is 1.96*10^9.

charle [14.2K]3 years ago
5 0
A billion is 9 zeroes
1.96 is already less than 10 and greater than or equal to 1 so

1.96 times 10^9 is answer

A is answer
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A cube with a side length s can be split into three congruent pyramids which formula represents the volume of a square pyramid
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7 0
3 years ago
At a certain high school, 35 students play only stringed instruments, 10 students play only brass instruments, and 5 students pl
Sophie [7]

Answer:

9/11

Step-by-step explanation:

Well we know there are 55 students in total and only 45 of them can either play a string or brass instrument.

So 45/55 is 9/11

Thus probably is 9/11 of a randomly selected student who plays either a string or brass.

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2 years ago
Use vertical muiltiplication to find the product of x^3+2x+3•x^3-x+1<br> ​​
Alex Ar [27]

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Hope this helps! :) ~Zane

6 0
3 years ago
Please help me, I don't understand this-
Andreas93 [3]

Answer:

i think its J 5,000 pounds

Step-by-step explanation:

5 0
2 years ago
Use the Midpoint Rule with n = 5 to estimate the volume V obtained by rotating about the y-axis the region under the curve y = 1
velikii [3]

Using the shell method, the volume is given exactly by the definite integral,

2\pi\displaystyle\int_0^1x(1+9x^3)\,\mathrm dx

Splitting up the interval [0, 1] into 5 subintervals gives the partition,

[0, 1/5], [1/5, 2/5], [2/5, 3/5], [3/5, 4/5], [4/5, 5]

with left and right endpoints, respectively, for the i-th subinterval

\ell_i=\dfrac{i-1}5

r_i=\dfrac i5

where 1\le i\le5. The midpoint of each subinterval is

m_i=\dfrac{\ell_i+r_i}2=\dfrac{2i-1}{10}

Then the Riemann sum approximating the integral above is

2\pi\displaystyle\sum_{i=1}^5m_i(1+9{m_i}^3)\frac{1-0}5

\dfrac{2\pi}5\displaystyle\sum_{i=1}^5\left(\frac{2i-1}{10}+9\left(\frac{2i-1}{10}\right)^4\right)

\dfrac{2\pi}{5\cdot10^4}\displaystyle\sum_{i=1}^5\left(16i^4-32i^3+24i^2+1992i-999\right)=\frac{112,021\pi}{25,000}\approx\boxed{14.08}

(compare to the actual value of the integral of about 14.45)

3 0
3 years ago
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