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Natasha2012 [34]
3 years ago
9

Can someone plz help me solved this problem! I’m giving you 10 points! I need help plz help me! Will mark you as brainiest!

Mathematics
1 answer:
sergeinik [125]3 years ago
4 0

Answer:

a). 8(x + a)

b). 8(h + 2x)

Step-by-step explanation:

a). Given function is, f(x) = 8x²

    For x = a,

    f(a) = 8a²

    Now substitute these values in the expression,

    \frac{f(x)-f(a)}{x-a} = \frac{8x^2-8a^2}{x-a}

                  = \frac{8(x^2-a^2)}{(x-a)}

                  = \frac{8(x-a)(x+a)}{(x-a)}

                  = 8(x + a)

b). \frac{f(x+h)-f(x)}{h} = \frac{8(x+h)^2-8x^2}{h}

                      = \frac{8(x^2+h^2+2xh)-8x^2}{h}

                      = \frac{8x^2+8h^2+16xh-8x^2}{h}

                      = (8h + 16x)

                      = 8(h + 2x)

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8 0
3 years ago
Help Dana find the sum 346 421 152
chubhunter [2.5K]
To add and get the sum 
  346   
  421
+152

We start on the ones place 6 + 1 + 2 = 9, the sum of the digits in the ones place is not up to 10, we will not regroup this ones place.

So, basically the answer to 13a is NO.
Since we are not able to regroup the ones place there is no regrouped tens to add, therefore the answer to 13b is also NO.

Next is the tens place, 4 +2 +5 = 11, the sum of the digits in the ones place is up to 10 so we regroup the tens place.

Thus, the answer to 13c is YES.

Since, we regrouped the ones place, we add the regrouped hundred, thus the answer to 13d is YES.

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