If (-c, c) is the set of all values of x for which the expansion is (7 - 5x)-2/3 is valid, then 5c + 7 =
0
12
41
14
We want to find the values of $x$ for which the binomial expansion of $(7 - 5x)^{-2/3}$ is valid.
The binomial expansion is valid when $|5x/7| < 1$.
This means:
$$ \left|\frac{5x}{7}\right| < 1 $$
$$ |5x| < 7 $$
$$ |x| < \frac{7}{5} $$
$$ -\frac{7}{5} < x < \frac{7}{5} $$
The set of all values of $x$ for which the expansion is valid is $(-c, c)$, where $c = \frac{7}{5}$.
Calculating $5c + 7$:
We have $5c = 5 \cdot \frac{7}{5} = 7$.
Therefore, $5c + 7 = 7 + 7 = 14$.
Final Answer:
The final answer is ${14}$.
Solve for \( x \):
\( \log_{10}(x^2) = 2 \).
Let \( K \) be an algebraically closed field containing a finite field \( F \). Let \( L \) be the subfield of \( K \) consisting of elements of \( K \) that are algebraic over \( F \).
Consider the following statements:
S1: \( L \) is algebraically closed.
S2: \( L \) is infinite.
Then, which one of the following is correct?
In a messenger RNA molecule, untranslated regions (UTRs) are present at:
I. 5' end before start codon
II. 3' end after stop codon
III. 3' end before stop codon
IV. 5' end after start codon
Consider z1 and z2 are two complex numbers.
For example, z1 = 3+4i and z2 = 4+3i
Here a=3, b=4, c=4, d=3
∴z1+ z2 = (a+c)+(b+d)i
⇒z1 + z2 = (3+4)+(4+3)i
⇒z1 + z2 = 7+7i
Properties of addition of complex numbers
It is similar to the addition of complex numbers, such that, z1 - z2 = z1 + ( -z2)
For example: (5+3i) - (2+1i) = (5-2) + (-2-1i) = 3 - 3i
Considering the same value of z1 and z2 , the product of the complex numbers are
z1 * z2 = (ac-bd) + (ad+bc) i
For example: (5+6i) (2+3i) = (5×2) + (6×3)i = 10+18i
Properties of Multiplication of complex numbers
Note: The properties of multiplication of complex numbers are similar to the properties we discussed in addition to complex numbers.
Associative law: Considering three complex numbers, (z1 z2) z3 = z1 (z2 z3)
Read More: Complex Numbers and Quadratic Equations
If z1 / z2 of a complex number is asked, simplify it as z1 (1/z2 )
For example: z1 = 4+2i and z2 = 2 - i
z1 / z2 =(4+2i)×1/(2 - i) = (4+i2)(2/(2²+(-1)² ) + i (-1)/(2²+(-1)² ))
=(4+i2) ((2+i)/5) = 1/5 [8+4i + 2(-1)+1] = 1/5 [8-2+1+41] = 1/5 [7+4i]