Question

(7a−11b)⋅(7a+11b)−49a2, если a=3 и b=0,01

Answer 1

(7 × 3 - 11 × 0,01) × (7 × 3 + 11 × 0,01) - 49 × 3²

(21 - 0,11) × (21 + 0,11) - 49 × 9

20,89 × 21,11 - 441

440,9879 - 441

- 0,0121

Answer 2

$(7a-11b)(7a+11b)-49a^2=49a^2-121b^2-49a^2=-121b^2\\\\-121*(0,01)^2=-121*(\frac{1}{100} )^2=-\frac{121}{10000} =-0,0121$

или

$-121*(0,01)^2=-121*(10^{-2})^2=-121*10^{-4}=-0,0121$

Answer 3

To evaluate the expression $(7a - 11b) \cdot (7a + 11b) - 49a^2$ when $a = 3$ and $b = 0.01$, we'll substitute the given values for $a$ and $b$ and then perform the calculations step by step.

Given $a = 3$ and $b = 0.01$, we can plug these values into the expression:

$$(7 \times 3 - 11 \times 0.01) \times (7 \times 3 + 11 \times 0.01) - 49 \times 3^2$$

Now, we'll perform the calculations step by step:

$$= (21 - 0.11) \times (21 + 0.77) - 49 \times 9 \\ = 20.89 \times 21.77 - 441 \\ \approx 454.7173 - 441 \\ \approx 13.7173$$

So, when $a = 3$ and $b = 0.01$, the value of the expression $(7a - 11b) \cdot (7a + 11b) - 49a^2$ is approximately $13.7173$.