Simplify 3{[3(9 – 2)] + [7(8 – 1)]} A 210 B 250 C 235 D 240

Mathematics

2 answers:

marta [7]3 years ago

5 0

Answer:

Step-by-step explanation:

I have had this before.

hope this helps

Sauron [17]3 years ago

3 0

Answer:

Step-by-step explanation:

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State the number of complex zeros, the possible number of real and imaginary zeros, and the possible rational zeros for each fun

ArbitrLikvidat [17]

Since the degree of this polynomial is 5, there will be 5 possible zeros. To find the possible rational 0s, use the rational root theorem (p/q). P is the last, non x value, which here it is the four on the end. The q is the leading coefficient, which is also q. Next, find all of the factors of q and p, which since they are both 4, are ±1, ±2, and ±4. Next do all possible values of p/q, which are ±1, ±2, ±4, ±1/2, and ±1/4. These are all your possible rational zeros. complex 0s only come in pairs, so the maximum there can be is 4 complex zeros, meaning there is at least one rational, real 0. (i graphed it it is -1/2, so all others must be rational or imaginary)

4 0

4 years ago

<img src="https://tex.z-dn.net/?f=y%20%3D%20%20%5Cfrac%7B%28190%29%5E%7B3%7D%20%7B68%5E%7B2%7D%20%7D%20%20%5Ctimes%20%28%20%5Cfr

mrs_skeptik [129]

Answer:

$\dfrac{\sqrt[3]{95^2}}{17\cdot95^4}=\dfrac{\sqrt[3]{9\,025}}{1\,384\,660\,625}$

Step-by-step explanation:

The applicable rules of exponents are ...

(ab)^c = (a^c)(b^c)

(a^b)/(a^c) = a^(b-c)

$\dfrac{190^3}{68^2}\times\dfrac{34}{95^{\frac{19}{3}}}=\dfrac{(2\cdot 95)^3}{(2\cdot 34)^2}\cdot\dfrac{34}{95^6\cdot 95^{\frac{1}{3}}}=2^{3-2}95^{3-6-\frac{1}{3}}34^{1-2}\\\\=2\cdot 95^{-3\frac{1}{3}}\cdot 34^{-1}=2\cdot 95^{-4+\frac{2}{3}}\cdot 34^{-1}\\\\=\dfrac{2\sqrt[3]{95^2}}{95^4\cdot 34}=\dfrac{\sqrt[3]{95^2}}{17\cdot95^4}\\\\=\dfrac{\sqrt[3]{9\,025}}{1\,384\,660\,625}$

6 0

3 years ago

What is the lowest common mutiple of 6 and 9

Lilit [14]

Answer:

Step-by-step explanation:

$9 = 3 \times 3 \\ 6 = 2 \times 3 \\ common \: factor = 3 \\ uncommon \: factors = 2, \: 3 \\ lcm = common \times uncommon \: factors \\ = 3 \times 2 \times 3 = 18$