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3 years ago

What is the slope of this line

Mathematics

1 answer:

kvasek [131]3 years ago

4 0

The slope is zero because the line is not moving

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Suppose you construct lines L, M, and N so that L is perpendicular to M and L is parallel to N. which of the following is true.

Blizzard [7]

In the question we are given three lines L,M and N such that:

Line L is perpendicular to M

and line L is parallel to line N

so the option is (C) M is perpendicular N

Step-by-step explanation

We are given three lines L,M and N such that:

Line L is perpendicular to M
Line L is parallel to line N

<u>We know that when one line parallel to the second line and is perpendicular to the third line then the second line is also perpendicular to the third line.</u>

<u></u>

So,the answer is (C) M is perpendicular N

3 0

3 years ago

What is the quotient?

Goryan [66]

Answer:

1/9

Step-by-step explanation:

this is (-3)^(0-2) = (-3)^(-2) = 1 / 9.

6 0

3 years ago

Read 2 more answers

Let C(n, k) = the number of k-membered subsets of an n-membered set. Find (a) C(6, k) for k = 0,1,2,...,6 (b) C(7, k) for k = 0,

vladimir1956 [14]

Answer:

(a) $C(6,0) = 1$ , $C(6,1) = 6$ , $C(6,2) = 15$ , $C(6,3) = 20$ , $C(6,4) = 15$ , $C(6,5) = 6$ , $C(6,6) = 1$ .

(b) $C(7,0) = 1$ , $C(7,1) = 7$ , $C(7,2) = 21$ , $C(7,3) = 35$ , $C(7,4) = 35$ , $C(7,5) = 21$ , $C(7,6) = 7$ , $C(7,7)=1$ .

Step-by-step explanation:

In this exercise we only need to recall the formula for C(n,k):

$C(n,k) = \frac{n!}{k!(n-k)!}$

where the symbol $n!$ is the factorial and means

$n! = 1\cdot 2\cdot 3\cdot 4\cdtos (n-1)\cdot n$ .

By convention 0!=1. The most important property of the factorial is $n!=(n-1)!\cdot n$ , for example 3!=1*2*3=6.

(a) The explanations to the solutions is just the calculations.

$C(6,0) = \frac{6!}{0!(6-0)!} = \frac{6!}{6!} = 1$

$C(6,1) = \frac{6!}{1!(6-1)!} = \frac{6!}{5!} = \frac{5!\cdot 6}{5!} = 6$

$C(6,2) = \frac{6!}{2!(6-2)!} = \frac{6!}{2\cdot 4!} = \frac{5!\cdot 6}{2\cdot 4!} = \frac{4!\cdot 5\cdot 6}{2\cdot 4!} = \frac{5\cdot 6}{2} = 15$

$C(6,3) = \frac{6!}{3!(6-3)!} = \frac{6!}{3!\cdot 3!} = \frac{5!\cdot 6}{6\cdot 6} = \frac{5!}{6} = \frac{120}{6} = 20$

$C(6,4) = \frac{6!}{4!(6-4)!} = \frac{6!}{4!\cdot 2!} = frac{5!\cdot 6}{2\cdot 4!} = \frac{4!\cdot 5\cdot 6}{2\cdot 4!} = \frac{5\cdot 6}{2} = 15$

$C(6,5) = \frac{6!}{5!(6-5)!} = \frac{6!}{5!} = \frac{5!\cdot 6}{5!} = 6$

$C(6,6) = \frac{6!}{6!(6-6)!} = \frac{6!}{6!} = 1$ .

(b) The explanations to the solutions is just the calculations.

$C(7,0) = \frac{7!}{0!(7-0)!} = \frac{7!}{7!} = 1$

$C(7,1) = \frac{7!}{1!(7-1)!} = \frac{7!}{6!} = \frac{6!\cdot 7}{6!} = 7$

$C(7,2) = \frac{7!}{2!(7-2)!} = \frac{7!}{2\cdot 5!} = \frac{6!\cdot 7}{2\cdot 5!} = \frac{5!\cdot 6\cdot 7}{2\cdot 5!} = \frac{6\cdot 7}{2} = 21$

$C(7,3) = \frac{7!}{3!(7-3)!} = \frac{7!}{3!\cdot 4!} = \frac{6!\cdot 7}{6\cdot 4!} = \frac{5!\cdot 6\cdot 7}{6\cdot 4!} = \frac{120\cdot 7}{24} = 35$

$C(7,4) = \frac{7!}{4!(7-4)!} = \frac{6!\cdot 7}{4!\cdot 3!} = frac{5!\cdot 6\cdot 7}{4!\cdot 6} = \frac{120\cdot 7}{24} = 35$

$C(7,5) = \frac{7!}{5!(7-2)!} = \frac{7!}{5!\cdot 2!} = 21$

$C(7,6) = \frac{7!}{6!(7-6)!} = \frac{7!}{6!} = \frac{6!\cdot 7}{6!} = 7$

$C(7,7) = \frac{7!}{7!(7-7)!} = \frac{7!}{7!} = 1$

For all the calculations just recall that 4! =24 and 5!=120.

6 0

3 years ago

What values of a and b would make the equation shown have infinitely many solutions? 3x=ax+b

Wittaler [7]

Infinitely many solutions means that you have the same thing on both sides of the equation no matter what value of x you plug in, right?

We just need both sides to be 3x then, correct?

If a were equal to 3 and b were equal to 0, we'd have

3x = (3)x + 0

Which is essentially 3x = 3x

So that means a = 3 and b = 0 must work!
Let's say x = 5

3(5) = 3(5) + 0

15 = 15 + 0

15 = 15

That means that a = 3 and b = 0 is your final answer :)

7 0

3 years ago

. (6.03) Which of the following best describes the expression 6(y + 3)? (4 points) The product of two constant factors six and t

kolezko [41]

C. The product of a constant factor of six and a factor with the sum of two terms. In this case, 6 is constant, and the sum of two terms is: y+3. Since they are products, you multiple them. Therefore C would be correct.

8 0

3 years ago