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3 years ago

I need to know this help please

Mathematics

2 answers:

Vikentia [17]3 years ago

6 0

= 2x^10y^12

answer is B 2x^10y^12

wlad13 [49]3 years ago

6 0

2 x^10 y^12

1. 60 divided by 30 = 2
2. x^20 - x^10 = x^10 (just minus exponents)
3. y^24 - y^12 = y^12

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A tour group with 44 people in it is getting rooms in a motel to stay in for the night. no more than 5 people can stay in each r

Anna11 [10]

Hello!

Since we want to find out how many rooms will be needed for 44 people, we must use division.

44 ÷ 5 = 8.8

Check by multiplying:

5 × 8 = 40

Because 40 is less than the amount of people (44), we'll need to multiply by the next number in line (9):

5 × 9 = 45 (1 extra person, no biggie!)

The tour group will need 9 rooms.

5 0

3 years ago

Pls help me with my practice quiz

hjlf

Answer:

its C

Step-by-step explanation:

8 0

3 years ago

Determine which of the sets of vectors is linearly independent. A: The set where p1(t) = 1, p2(t) = t2, p3(t) = 3 + 3t B: The se

defon

Answer:

The set of vectors A and C are linearly independent.

Step-by-step explanation:

A set of vector is linearly independent if and only if the linear combination of these vector can only be equalised to zero only if all coefficients are zeroes. Let is evaluate each set algraically:

$p_{1}(t) = 1$ , $p_{2}(t)= t^{2}$ and $p_{3}(t) = 3 + 3\cdot t$ :

$\alpha_{1}\cdot p_{1}(t) + \alpha_{2}\cdot p_{2}(t) + \alpha_{3}\cdot p_{3}(t) = 0$

$\alpha_{1}\cdot 1 + \alpha_{2}\cdot t^{2} + \alpha_{3}\cdot (3 +3\cdot t) = 0$

$(\alpha_{1}+3\cdot \alpha_{3})\cdot 1 + \alpha_{2}\cdot t^{2} + \alpha_{3}\cdot t = 0$

The following system of linear equations is obtained:

$\alpha_{1} + 3\cdot \alpha_{3} = 0$

$\alpha_{2} = 0$

$\alpha_{3} = 0$

Whose solution is $\alpha_{1} = \alpha_{2} = \alpha_{3} = 0$ , which means that the set of vectors is linearly independent.

$p_{1}(t) = t$ , $p_{2}(t) = t^{2}$ and $p_{3}(t) = 2\cdot t + 3\cdot t^{2}$

$\alpha_{1}\cdot p_{1}(t) + \alpha_{2}\cdot p_{2}(t) + \alpha_{3}\cdot p_{3}(t) = 0$

$\alpha_{1}\cdot t + \alpha_{2}\cdot t^{2} + \alpha_{3}\cdot (2\cdot t + 3\cdot t^{2})=0$

$(\alpha_{1}+2\cdot \alpha_{3})\cdot t + (\alpha_{2}+3\cdot \alpha_{3})\cdot t^{2} = 0$

The following system of linear equations is obtained:

$\alpha_{1}+2\cdot \alpha_{3} = 0$

$\alpha_{2}+3\cdot \alpha_{3} = 0$

Since the number of variables is greater than the number of equations, let suppose that $\alpha_{3} = k$ , where $k\in\mathbb{R}$ . Then, the following relationships are consequently found: