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4 years ago

Which binomials are a difference of squares?

Mathematics

1 answer:

kow [346]4 years ago

3 0

Answer:

Options B and C.

Step-by-step explanation:

Difference of squares is defined as

$a^2-b^2$

where, a and b both area real numbers.

$y^4+9\Rightarrow (y^2)^2+3^2$

It is sum of squares.

$9x^2-16\Rightarrow (3x)^2-4^2$

It is difference of squares.

$m^6-25\Rightarrow (m^3)^2-5^2$

It is difference of squares.

$m^7-1\Rightarrow m^7-1^2$

It is not a difference of squares.

Therefore, the correct options are B and C.

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Use the discriminant to determine how many and what kind of solutions the quadratic equation 3x^2 + 4x = - 5 has.

leonid [27]

Answer:

Step-by-step explanation:

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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.

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3 years ago

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3 years ago

5. The table below represents fees for a parkirls lot. Graph the

Vladimir [108]

Answer:

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3 years ago

Which expression is the factored form of x2 – 7x 10? (x 3)(x 4) (x – 3)(x – 4) (x – 2)(x – 5) (x 2)(x 5)

alexandr402 [8]

The expression x^2 -7x + 10 can be written in factored form as given by Option C: (x-2)(x-5)

<h3>How to find the factors of a quadratic expression?</h3>

If the given quadratic expression is of the form $ax^2 + bx + c$

then its factored form is obtained by two numbers alpha( α ) and beta( β) such that:

$b = \alpha + \beta \\ ac =\alpha \times \beta$

Then writing b in terms of alpha and beta would help us getting common factors out.

Sometimes, it is not possible to find factors easily, so using the quadratic equation formula can help out without any trial and error.

For this case, we're provided the expression:

$x^2 -7x + 10$

We can write 10 as multiple of -5 and -2 and

we can write -7 as sum of -5 and -2. Thus, we get:

$x^2 -7x + 10 = x^2 -5x - 2x + 10 = x(x-5+ -2(x-5) = (x-2)(x-5)$

Thus, the expression x^2 -7x + 10 can be written in factored form as given by Option C: (x-2)(x-5)

Learn more about factorization of quadratic expression here:

brainly.com/question/26675692

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2 years ago