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irina [24]
3 years ago
10

ANSWER ASAP PLZ

Mathematics
2 answers:
stira [4]3 years ago
5 0
60,000,000

Hope this helps :)
Harrizon [31]3 years ago
5 0
Answer: 60, 000, 000

I hope this helps! :)
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34kurt
12+4x^2+3x^3-2x is the answer.
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2 years ago
How many 7 digit phone numbers are there assuming the first digit can't be a 0 or a 1?
Anettt [7]
The first digit can be any one of the numbers 2-9 That a total of 8 numbers.

The next 6 digits can be any permutation of 6  from the numbers  0 to 9.
(10 numbers)

nPr  =  n! / r!

so here we have  10P6  = 10! / 6!  =  10*9*8*7  =  5040

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7 0
3 years ago
I just need answers to the odd numbered questions.<br> I would appreciate the help<br><br> 35 points
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Answer:

Step-by-step explanation:

Combine real terms and combine complex terms

1) 3 + 2i + 2 - 5i = 3 +2 + 2i - 5i

                          = 5 + (2-5)i

                          = 5 + (-3)i

                          = 5 - 3i

3) 2 - (1 - 2i) + (4 -5i ) - (1 - 3i)  = 2 -1 + 2i + 4 - 5i - 1 + 3i

           {- is distributed to (1 - 2i) & - is distributed to (1- 3i)}

              = 2 - 1  + 4 + 1 + 2i - 5i + 3i

              = 6 +0i = 6

5) 4 - 3i + 4 + 3i = 4 +4 -3i + 3i

                          = 8

7) (3 - 2i)² + (3 +2i) = 3² - 2*3*2i + (2i)² + 3 + 2i     {(a - b)² = a² - 2ab +b²}

= 9 -12i + 4i² + 3 + 2i

= 9 - 12i + 4*(-1) + 3 + 2i       {i²  = -1}

= 9 +3 - 4 - 12i +2i

= 8 - 10i

9) (a +b)^{4} - (a - b)^{4} = 8ab(a^{2} +b^{2})\\\\\\Here, \ a = 1 \ and \ b = i\\\\\\(1+\sqrt{-1})^{4}-(1-\sqrt{-1})^{4}=(1+\sqrt{i^{2}})^{4}-(1-\sqrt{i^{2}})^{4}\\\\=(1+i)^{4}-(1-i)^{4}\\\\=8*1*i(1^{2}+i^{2})\\\\=8i*(1 -1)\\\\=8i*0\\\\=0

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3 years ago
Five times the number of test tubes in a school’s chemistry lab exceeds three times the number of beakers it has by 660. The sum
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B= number of beakers which there are 450 
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38+13      is the equation

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