9.77 to the nearest tenths

first, find the number that is in the tenths place....it is 7....now look at the number directly to the right of it....if that number is 5 or greater, u would round that 7 up to an 8...but if that number is 4 or below, ur 7 would stay the same.
So the number directly to the right of 7 is 7...and since it is greater then 5, u have to round the 7 in the tenths place up to 8.

solution is : 9.8

3 0

3 years ago

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I just need answers to the odd numbered questions.<br> I would appreciate the help<br><br> 35 points

Misha Larkins [42]

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

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

6 0

3 years ago