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3 years ago

Describe how to write 3,482,000,000 in scientific notation

2 answers:

6 0

∙•❁ The answer to your question "Describe how to write 3,482,000,000 in scientific notation" is:

3.482 x 10⁹

∙•❁<em>I hope this helps!</em>❁•∙

sp2606 [1]3 years ago

4 0

The answer is: 3.482 x 10⁹
I hope that this helped you!!

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What is the percent of 16/25

Lelechka [254]

16/25 = 0.64 = 64%

Another way to do it is to multiply top and bottom by 4
16/25 = (16*4)/(25*4) = 64/100 = 64%

Either way, the answer is 64%

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3 years ago

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How many different arrangements can be made with the letter of the word 'M O R O C C O'?

valentinak56 [21]

Since the order matters, it's a permutation of 7 letters, but mind you in
MOROCCO. you have 3 O's and 2 C's, Hence the arrangements are:

⁷P₇ /(3!2!) = 7!/(3!2!) = 5040/(3!2!) = 420 arrangements

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3 years ago

Solve the equation log^7b>2

ahrayia [7]

Answer:

b > 12.71

Step-by-step explanation:

Here in this question we have to solve the inequality equation given for b (An unknown variable}

Now, the equation is $(\log b)^{7} > 2$

⇒ $\log b > 2^{\frac{1}{7} }$

We are taking the base of the log here is 10.

So, $10^{(\log b)} > 10^{2^{\frac{1}{7} } }$ {Since $10^{\log x} =x$ and $e^{\ln x} = x$ }

⇒ b > 12.71 (Approximate) (Answer)

3 0

3 years ago

calculate gas mileage of car that drives 283 miles on 12.3 gallons of gas andround to nearest tenth of a mpg

Studentka2010 [4]

283 divided by 12.3 = 23.008 and rounded it will just be 23.

5 0

3 years ago

Use the following matrices, A, B, C and D to perform each operation.

Vinvika [58]

Step-by-step explanation:

$A=\left[\begin{array}{ccc}3&1\\5&7\end{array}\right]$

$B=\left[\begin{array}{ccc}4&1\\6&0\end{array}\right]$

$C=\left[\begin{array}{ccc}-2&3&1\\-1&0&4\end{array}\right]$

$D=\left[\begin{array}{ccc}-2&3&4\\0&-2&1\\3&4&-1\end{array}\right]$

$1.\\A+B=\left[\begin{array}{ccc}3&1\\5&7\end{array}\right]+\left[\begin{array}{ccc}4&1\\6&0\end{array}\right]=\left[\begin{array}{ccc}3+4&1+1\\5+6&7+0\end{array}\right]=\left[\begin{array}{ccc}7&2\\11&7\end{array}\right]$

$2.\\B-A=\left[\begin{array}{ccc}4&1\\6&0\end{array}\right]-\left[\begin{array}{ccc}3&1\\5&7\end{array}\right]=\left[\begin{array}{ccc}4-3&1-1\\6-5&0-7\end{array}\right]=\left[\begin{array}{ccc}1&0\\1&-7\end{array}\right]$

$3.\\3C=3\left[\begin{array}{ccc}-2&3&1\\-1&0&4\end{array}\right]=\left[\begin{array}{ccc}(3)(-2)&(3)(3)&(3)(1)\\(3)(-1)&(3)(0)&(3)(4)\end{array}\right]=\left[\begin{array}{ccc}-6&9&3\\-3&0&12\end{array}\right]$

$4.\\C\cdot D=\left[\begin{array}{ccc}-2&3&1\\-1&0&4\end{array}\right]\cdot\left[\begin{array}{ccc}-2&3&4\\0&-2&1\\3&4&-1\end{array}\right]\\\\=\left[\begin{array}{ccc}(-2)(-2)+(3)(0)+(1)(3)&(-2)(3)+(3)(-2)+(1)(4)&(-2)(4)+(3)(1)+(1)(-1)\\(-1)(-2)+(0)(0)+(4)(3)&(-1)(3)+(0)(-2)+(4)(4)&(-1)(4)+(0)(1)+(4)(-1)\end{array}\right]\\=\left[\begin{array}{ccc}7&-8&-6\\14&13&-8\end{array}\right]$

$5.\\2D+3C\\\text{This operation can't be performed because the matrices}\\\text{ are of different dimensions.}$

6 0

3 years ago