When you have a problem like this you just have to thing on a reason to multiply by a fraction and a withdrawal from the total, if you are multiplying by a fraction, that means you are dividing money, so thats why we put julie and her sisters as subjects, and her grandma givin them money is to withdraw from the total amount those $3.20

4 0

3 years ago

If P(x) = P<img src="https://tex.z-dn.net/?f=_%7Bn%7D" id="TexFormula1" title="_{n}" alt="_{n}" align="absmiddle" class="latex-f

seropon [69]

If $P(x) = p_nx^n + p_{n-1}x^{n-1}+\ldots+p_0$ is divided by $(x-a)$ , then $P(x) = (x-a) \cdot Q(x) + R(x)$ for some polynomials $Q,R$ . Moreover, $\deg R < 1$ (because $\deg (x-a) = 1$ ), so there exists $\alpha \in \mathbb{R}$ such that $R(x) = \alpha$ for all $x \in \mathbb{R}$ . But if we calculate $P(a)$ , it turns out that $P(a) = (a-a)\cdot Q(a) + \alpha$ , so $R(x) = \alpha = P(a)$ . $\blacksquare$

3 0

3 years ago

Please help!! I need Thai answer soon, will mark brainless

olga2289 [7]

Answer:

d. 127.41

hope it helps....

4 0

3 years ago