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3 years ago

Which expression is equivalent to (3x² + 4x - 7)(x - 2)? (Apex)

Mathematics

1 answer:

katovenus [111]3 years ago

3 0

Answer:

A i think

Step-by-step explanation:

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X is 13 less than 17. WRITE AN EQUATION TO REPRESENT THE statement.

vovikov84 [41]

Answer:

X=4

Step-by-step explanation:

13 less than 17 would be 4 becayse if you do 17-13=4

6 0

3 years ago

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

Determine the solutuon for x^2-3x-28>0

natita [175]

Answer:

x < -4 or x > 7.

Step-by-step explanation:

We first determine the critical points by solving x^2 - 3x - 28 = 0:

x^2 - 3x - 28 = 0

(x - 7)(x + 4) = 0

x = 7, - 4

so the critical points are -4 and 7.

Create a Table (pos = positive and neg = negative):

Value of x< - 4 -4 < x < 7 x > 7

---------------------|----------- |--------------------- |---------------------

x + 4 NEG POS POS

x - 7 NEG NEG POS

(x + 4)(x - 7) POS NEG POS

So the function is positive (>0) for x < -4 or x > 7.

You can also do this by drawing the graph of the function.

6 0

3 years ago

There are 22 animals in the barn. Some are geese and some are goats. There are 82 legs in all. How many of each animal are there

Free_Kalibri [48]

Let the number of geese be x.

Then number of goats is 22-x.

A goose has 2 legs, and a goat has 4.

There are a total of 82 legs.

So,

2x + 4(22-x) = 82

2x + 88 - 4x = 82

-2x = -6

x = 3

There are 3 geese and (22-3=) 19 goats.

Please mark Brainliest if this helps and feel free to ask doubts!

7 0

3 years ago

What is the most difficult math problem in the world?

solmaris [256]

xn+yn=zn where n>2 and n is an integer

<h2>bolded letters are exponents!!!!</h2>

4 0

3 years ago

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