Express the standard form: (i) radius of the earth= 6 400 000 m (ii) radius of hydrogen atom = 0.000 000 000 0529

I) 6.4 × 10^(6) m

II) 5.29 × 10^(-11) m

Explanation:

Standard form is how we write very large number in such a way that it is easier to read.

Thus;

I) radius of the earth = 6 400 000 m

What we will now have is;

6.4 × 10^(6) m

II) radius of hydrogen atom = 0.000 000 000 0529 m

What we will have is;

5.29 × 10^(-11) m

i. 6.4 × 10⁶ m.

ii. 5.29 × 10⁻¹¹ m.

Explanation:

To express a number in standard form, we write it in the form a × 10ⁿ where 1 ≤ a < 10 and n is an integer.

(i) radius of the earth= 6 400 000 m

To express the radius of the earth = 6 400 000 m in standard form, we count the number of decimal places after the first digit. We have 6 decimal places after digit 6. So, we put a decimal point between digit 6 and 4. So, we have 6.4. Since we have 6 decimal places after the six digit, the exponent of our base which is 10 is 6.

So, writing 6 400 000 in the form in a × 10ⁿ, we have a = 6.4 and n = 6. So, in standard form we have 6.4 × 10⁶ m.

So, 6 400 000 =  6.4 × 10⁶ m in standard form.

(ii) radius of hydrogen atom = 0.000 000 000 0529 m

To express the radius of hydrogen atom = 0.000 000 000 0529 m in standard form, we count the number of decimal places after the decimal point including the first non-zero digit. We have 11 decimal places after the decimal point up to and including the first non-zero digit which is digit 5. So, we put a decimal point after the first non-zero digit,5. So, we have 5.29. Since we have 11 decimal places after the decimal point, the exponent of our base which is 10 is -11. The exponent is negative since the digit places are taken after the decimal point.

So, writing 0.000 000 000 0529 m in the form in a × 10ⁿ, we have a = 5.29 and n = -11. So, in standard form we have 5.29 × 10⁻¹¹ m.

So,  0.000 000 000 0529 m = 5.29 × 10⁻¹¹ m in standard form.

we have been given a quadratic function and we need to restrict the domain such that it becomes a one to one function.

we know that vertex of this quadratic function occurs at (5,2).

further, we know that range of this function is .

if we restrict the domain of this function to either or , it will become one to one function.

let us know find its inverse.

upon interchanging x and y, we get:

let us now solve this function for y.

hence, the inverse function would be if we restrict the domain of original function to and the inverse function would be if we restrict the domain to .