All categories

3 years ago

Which statement correctly describes the expression (-60)(-5)?

Mathematics

2 answers:

Elina [12.6K]3 years ago

8 0

It would be b. If two numbers are negative and your multiplying them, it would be positive!

Ludmilka [50]3 years ago

5 0

Answer:

If you're multiplying then the answer would be B.

Step-by-step explanation:

If you multiply two negative numbers, your product would be positive.

If you multiply 1 positive and 1 negative then your product will be negative.

You might be interested in

Solve 152x = 36. Round to the nearest ten-thousandth. 1.7509 2.6466 0.6616 1.9091

Vlad [161]

The answer is 0.6616
Hope it helped :)

6 0

3 years ago

Read 2 more answers

Review the proof of de Moivre’s theorem.

RideAnS [48]

Answer:

Step-by-step explanation:

Just took it.

5 0

3 years ago

Which is the value of this expression when p=-2 and q=-1?

saul85 [17]

Answer:

D. 4

Step-by-step explanation:

$[(p^2) (q^{-3}) ]^{-2}.[(p)^{-3}(q)^5] ^{-2}\\\\=[(p^2) (q^{-3}) \times(p)^{-3}(q)^5 ]^{-2}\\\\=[(p^{2}) \times(p)^{-3} \times (q^{-3}) \times(q)^5 ]^{-2}\\\\=[(p^{2-3}) \times (q^{5-3}) ]^{-2}\\\\=[(p^{-1}) \times (q^{2}) ]^{-2}\\\\=(p^{-1\times (-2)}) \times (q^{2\times (-2) }) \\\\=p^{2}\times q^{-4} \\\\= \frac{p^2}{q^4}\\\\= \frac{(-2)^2}{(-1)^4}\\\\= \frac{4}{1}\\\\= 4$

8 0

2 years ago

Write out the first few terms of the series Summation from n equals 0 to infinity (StartFraction 2 Over 3 Superscript n EndFract

anyanavicka [17]

Answer:

$\sum\limits_{n=0}^{\infty} \frac{2}{3^n}\frac{(-1)^n}{5^n} = 15/8$

Step-by-step explanation:

The sum you are trying to understand is this.

$\sum\limits_{n=0}^{\infty} \frac{2}{3^n}\frac{(-1)^n}{5^n}$

Remember that in general when you have a geometric series

$\sum\limits_{n = 0}^{\infty} a*r^n$ you have that

$\sum\limits_{n = 0}^{\infty} a*r^n = \frac{a}{1-r}$ and that equality is true as long as $|r| < 1$ .

Therefore here we have

$\sum\limits_{n=0}^{\infty} \frac{2}{3^n}\frac{(-1)^n}{5^n} = \sum\limits_{n=0}^{\infty} 2* \big(\frac{-1}{3*5} \big)^n = \sum\limits_{n=0}^{\infty} 2* \big(\frac{-1}{15} \big)^n$ and $\big|\frac{-1}{15} \big| = \frac{1}{15} < 1$

Therefore we can use the formula and

$\sum\limits_{n=0}^{\infty} 2* \big(\frac{-1}{15} \big)^n = \frac{2}{1-(-1/15)} = \frac{2}{1+1/15} = 30/16 = 15/8$

5 0

2 years ago

How are factors related to prime numbers?

melomori [17]

They have some of the same numbers

7 0

3 years ago