Using the two functions listed below, insert numbers in place of the letters a, b, c, and d so that f(x) and g(x) are inverses.

Given: 3c + d = f + g, d = f, g = c + 8; Prove: c = 4

agasfer [191]

Answer:

Step-by-step explanation:

7 0

3 years ago

<img src="https://tex.z-dn.net/?f=%28%20%5Cfrac%7B1%7D%7B2%7D%20%20%7Ba%7D%5E%7B2%7D%20%20%2B%20%20%20%5Cfrac%7B1%7D%7B2%7D%20ab

Zigmanuir [339]

Answer:

$\frac{1}{2}$ (a + 2b)(a - b)

Step-by-step explanation:

Assuming you require the expression to be factored

Given

$\frac{1}{2}$ a² + $\frac{1}{2}$ ab - b² ← factor out $\frac{1}{2}$ from each term

= $\frac{1}{2}$ (a² + ab - 2b²) ← factor the quadratic

Consider the factors of the coefficient of the b² term(- 2) which sum to give the coefficient of the ab- term (+ 1)

The factors are + 2 and - 1, since

2 × - 1 = - 2 and 2 - 1 = + 1, thus

a² + ab - 2b² = (a + 2b)(a - b) and

$\frac{1}{2}$ a² + $\frac{1}{2}$ ab - b² = $\frac{1}{2}$ (a + 2b)(a - b)

5 0

3 years ago

at a particular high school, 42% of the students participate in sports and 25% of the students participate in drama. if 53% of t

baherus [9]

The probability that a student participates in both sports and drama is $0.14$ .

<h3>What is the formula for P(AUB), where A and B are any two events?</h3>

If $A$ and $B$ are any two events, then the probability of the joint event $(A\cup B)$ is given by the following formula: $P(A\cup B)=P(A)+P(B)-P(A\cap B)$

Given that 42% of the students participate in sports and 25% of the students participate in drama and 53% of the students participate in either sports or drama.

Suppose $S$ denotes that "a student participates in sports" and $D$ denotes that "a student participates in drama".
So, we have $P(S)=\frac{42}{100}=0.42$ , $P(D)=\frac{25}{100}=0.25$ , $P(S\cup D)=\frac{53}{100}=0.53$ .

We want to find the probability that a student participates in both sports and drama i.e., we want to find $P(S\cap D)$ .

By the above formula, we obtain:

$P(S\cup D)=P(S)+P(D)-P(S\cap D)\\\Longrightarrow P(S\cap D)=P(S)+P(D)-P(S\cup D)\\\Longrightarrow P(S\cap D)=0.42+0.25-0.53\\\therefore P(S\cap D)=0.14=\frac{14}{100}$