Can someone please tell me the answers to the last two questions please

Mathematics

2 answers:

Sergeu [11.5K]3 years ago

6 0

Are you supposed to estimate or get an exact answer? If you're estimating you can see if the number is close to any perfect squares. In this case 63 is very close to 64, and 48 is very close to 49. The answers can then be estimated to be approximately 8 and 7, respectively.

Veronika [31]3 years ago

3 0

Sqrt(63)=sqrt(9)*sqrt(7)=3sqrt(7)

3 times the square root of 7

sqrt(48)=sqrt(16)*sqrt(3)=4sqrt(3)

4 times the square root of 3

You might be interested in

Mr. Burr and his daughter live in Corpus Christi, Texas. Mr. Burr’s employer pays 100% of his insurance premium and 50% of the p

Setler [38]

2,602 this is probably wrong though so..

5 0

2 years ago

Lim t^4 - 6 / 2t^2 - 3t + 7

Harman [31]

I think you meant to say

$\displaystyle \lim_{t\to2}\frac{t^4-6}{2t^2-3t+7}$

(as opposed to x approaching 2)

Since both the numerator and denominator are continuous at t = 2, the limit of the ratio is equal to a ratio of limits. In other words, the limit operator distributes over the quotient:

$\displaystyle \lim_{t\to2} \frac{t^4 - 6}{2t^2 - 3t + 7} = \frac{\displaystyle \lim_{t\to2}(t^4-6)}{\displaystyle \lim_{t\to2}(2t^2-3t+7)}$

Because these expressions are continuous at t = 2, we can compute the limits by evaluating the limands directly at 2:

$\displaystyle \lim_{t\to2} \frac{t^4 - 6}{2t^2 - 3t + 7} = \frac{\displaystyle \lim_{t\to2}(t^4-6)}{\displaystyle \lim_{t\to2}(2t^2-3t+7)} = \frac{2^4-6}{2\cdot2^2-3\cdot2+7} = \boxed{\frac{10}9}$