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kirza4 [7]
3 years ago
15

Multiply. Write your answer as a fraction in simplest form. 10/7 x 3/5

Mathematics
1 answer:
ziro4ka [17]3 years ago
6 0
10/7*3/5: 30/35 or 6/7
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Write an exxpression to represent nine minus the quoitient of two and a number x
Bond [772]

Answer:

9-(2/x)

Step-by-step explanation:

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3 years ago
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For every 13 birds born at the zoo there are two animals born. There were 44 if you were animals born last month then birds so h
Daniel [21]
The ratio birds : animals = 13 : 2
so for every 1 animal, 6.5 birds were born

last month: animals = birds - 44
birds - animals = 44
use trial and improvement to find:
if there were 52 birds, there must be 8 animals
CHECK: 52 - 8 = 44
4 0
3 years ago
Heellllpppp plzzzzzzzzzzz
ser-zykov [4K]

Answer:

3/32

Step-by-step explanation:

the reason it’s 3/32 is because you have to divide 3/4 divide by 8. So that means 3/4x1/8 and that gives you 3/32. I hope that helped you!

4 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:

\alpha_{1} = -2\cdot \alpha_{3}

\alpha_{1} = -2\cdot k

\alpha_{2}= -2\cdot \alpha_{3}

\alpha_{2} = -3\cdot k

It is evident that \alpha_{1} and \alpha_{2} are multiples of \alpha_{3}, which means that the set of vector are linearly dependent.

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

\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+t^{2}) = 0

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

The following system of linear equations is obtained:

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

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

3\cdot \alpha_{3} = 0

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

The set of vectors A and C are linearly independent.

4 0
3 years ago
What is the solution to the equation 1 over 2 multiplied by n equals 8 ? (5 points) n = 10 n = 6 n = 4 n = 16
Leno4ka [110]

Answer: n=16

Step-by-step explanation:

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3 years ago
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