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3 years ago

Help please! asap!

Mathematics

1 answer:

aliya0001 [1]3 years ago

8 0

Answer: 50% or 1/2

Step-by-step explanation:

number of possible combos:

1+1, 1+2, 1+3, 1+4, 1+5, 1+6

2+1, 2+2, 2+3, 2+4, 2+5, 2+6

3+1, 3+2, 3+3, 3+4, 3+5, 3+6

4+1, 4+2, 4+3, 4+4, 4+5, 4+6

5+1, 5+2, 5+3, 5+4, 5+5, 5+6

6+1, 6+2, 6+3, 6+4, 6+5, 6+6

number of even sums within combos:

1+1, 1+3, 1+5

2+2, 2+4, 2+6

3+1, 3+3, 3+5

4+2, 4+4, 4+6

5+1, 5+3, 5+5

6+2, 6+4, 6+6

<u>TOTAL combos:</u>

<u>TOTAL even sums:</u>

so that means 18/36 or 1/2 of the possible combinations has a sum of an even number.

sorry it’s not a table but I literally worked it out while answering so that’s all my work

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Firdavs [7]

Answer:

Step-by-step explanation:

-4/2 = -2

-2 + 5 = 3

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3 years ago

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I dont understand this at all. I really need help!

Westkost [7]

Hello!

$\large\boxed{x^{4}}$

Recall that:

$\sqrt[z]{x^{y} }$ is equal to $x^{\frac{y}{z} }$ . Therefore:

$\sqrt[3]{x^{2} } = x^{\frac{2}{3} }$

There is also an exponent of '6' outside. According to exponential properties, when an exponent is within an exponent, you multiply them together. Therefore:

$(x^{\frac{2}{3} })^{6} = x^{\frac{2}{3}* 6 } = x^{\frac{12}{3} } = x^{4}$

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3 years ago

Determine whether The given segments have the same length. Justify your answer

Semenov [28]

We know the distance formula is

$\sqrt{ (x_{2}-x_{1})^2+ (y_{2}-y_{1})^2 }$

Here A( -4,2) and B(1,4)

So length of AB

= $\sqrt{(1-(-4))^2+(4-2)^2} =\sqrt{5^2+2^2} =\sqrt{29}$

Also C(2,1)

Length of BC

= $\sqrt{(2-1)^2+(-1-4)^2} =\sqrt{1^2+(-5)^2} =\sqrt{26}$

So we can see that length of AB is not equal to length of BC

11.

Now AB = $\sqrt{29}$

Also C(2,-1) & D(4,4)

Length of CD

= $\sqrt{(4-2)^2+(4-(-1)) ^2} =\sqrt{2^2+5^2} =\sqrt{29}$

Yes AB = CD

5 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: