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amid [387]
3 years ago
12

Write 0.000073 in scientific notation

Mathematics
2 answers:
Llana [10]3 years ago
7 0

Answer:

7.3x10^-7

Step-by-step explanation:

you move the decimal to the right 7 times.

Hope this helps.

Andrew [12]3 years ago
7 0

<em><u>Answer:</u></em>

10^-^5  *73

<em><u>Step-by-step explanation:</u></em>

So in 0.000073, 7 is the fifth digit in the term so we can write it as 10^-^5 * 73

or 10 to power of negative 5 multiplied by 73.

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Hi, The answer is 2x^2-8x+12, because when you subtract the polynomial with the answer you will get x^2-4
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Please help me identify the terms and like terms
sergey [27]

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4 0
3 years ago
Read 2 more answers
Q12.
SVEN [57.7K]

Answer:

Step-by-step explanation:

1l ........8.5 miles

x l .......6 miles

-----------------------

x=6*1/8.5

x=0.70 l

2*0.7=1.4 l petrol/day ( to work and come back home)

5*1.4=7 l/week ( 5 days works in a week)

7*1.26=8.82 L /week

8.82>8.5

The petrol costs more

So the answer is NO

6 0
3 years ago
lan has a flower pot in the shape of a rectangular prism . The base of the flower pot is 36 square inches . The flower he's plan
JulijaS [17]

Answer:

108

Step-by-step explanation:

  1. volume = area  of the base * height
  2. Area of the base = 36 square inches .
  3. The height = 3 inches
  4. 36 * 3 = 108
4 0
4 years ago
Prove or disprove (from i=0 to n) sum([2i]^4) &lt;= (4n)^4. If true use induction, else give the smallest value of n that it doe
ddd [48]

Answer:

The statement is true for every n between 0 and 77 and it is false for n\geq 78

Step-by-step explanation:

First, observe that, for n=0 and n=1 the statement is true:

For n=0: \sum^{n}_{i=0} (2i)^4=0 \leq 0=(4n)^4

For n=1: \sum^{n}_{i=0} (2i)^4=16 \leq 256=(4n)^4

From this point we will assume that n\geq 2

As we can see, \sum^{n}_{i=0} (2i)^4=\sum^{n}_{i=0} 16i^4=16\sum^{n}_{i=0} i^4 and (4n)^4=256n^4. Then,

\sum^{n}_{i=0} (2i)^4 \leq(4n)^4 \iff \sum^{n}_{i=0} i^4 \leq 16n^4

Now, we will use the formula for the sum of the first 4th powers:

\sum^{n}_{i=0} i^4=\frac{n^5}{5} +\frac{n^4}{2} +\frac{n^3}{3}-\frac{n}{30}=\frac{6n^5+15n^4+10n^3-n}{30}

Therefore:

\sum^{n}_{i=0} i^4 \leq 16n^4 \iff \frac{6n^5+15n^4+10n^3-n}{30} \leq 16n^4 \\\\ \iff 6n^5+10n^3-n \leq 465n^4 \iff 465n^4-6n^5-10n^3+n\geq 0

and, because n \geq 0,

465n^4-6n^5-10n^3+n\geq 0 \iff n(465n^3-6n^4-10n^2+1)\geq 0 \\\iff 465n^3-6n^4-10n^2+1\geq 0 \iff 465n^3-6n^4-10n^2\geq -1\\\iff n^2(465n-6n^2-10)\geq -1

Observe that, because n \geq 2 and is an integer,

n^2(465n-6n^2-10)\geq -1 \iff 465n-6n^2-10 \geq 0 \iff n(465-6n) \geq 10\\\iff 465-6n \geq 0 \iff n \leq \frac{465}{6}=\frac{155}{2}=77.5

In concusion, the statement is true if and only if n is a non negative integer such that n\leq 77

So, 78 is the smallest value of n that does not satisfy the inequality.

Note: If you compute  (4n)^4- \sum^{n}_{i=0} (2i)^4 for 77 and 78 you will obtain:

(4n)^4- \sum^{n}_{i=0} (2i)^4=53810064

(4n)^4- \sum^{n}_{i=0} (2i)^4=-61754992

7 0
4 years ago
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