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4 years ago

If aſc and a +b = C, prove that a|b.

Mathematics

1 answer:

mars1129 [50]4 years ago

7 0

Answer with Step-by-step explanation:

Since we have given that

a + b = c

and a|c

i.e. a divides c.

We need to prove that a|b.

⇒ a = mb for some integer m

Since a|c,

So, mathematically, it is expressed as

c= ka

Now, we put the above value in a + b = c.

So, it becomes,

$a+b=c\\\\a+b=ka\\\\b=ka-a\\\\b=a(k-1)\\\\\implies a|b$

a=mb, here, m = k-1

Hence, proved.

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Q is a degree 5 polynomial that passes through the origin, has zeros i and 7 - i, and q(-1)= -520. find the equation for q

Lunna [17]

Using the factor theorem, the equation for q is:

$q(x) = 4(x^5 - 14x^4 + 51x^3 - 14x^2 + 50x)$

The <em>Factor Theorem</em> states that a polynomial function with roots $x_1, x_2, ..., x_n$ is given by:

$f(x) = a(x - x_1)(x - x_2)...(x - x_n)$

In which a is the leading coefficient.

In this problem:

Passes through the origin, so x = 0 is a zero, that is $x_1 = 0$ .
x = i is a zero, and thus, it's conjugate also is, that is, x = -i, hence $x_2 = i, x_3 = -i$

x = 7 - i is a zero, and thus, it's conjugate also is, that is, x = 7 + i, hence $x_4 = 7 - i, x_3 = 7 + i$

Then, the equation is:

$q(x) = a(x - 0)(x - i)(x + i)(x - 7 + i)(x - 7 - i)$

$q(x) = ax(x^2 - i^2)((x-7)^2 - i^2)$

Considering that $i^2 = -1$

$q(x) = ax(x^2 + 1)(x^2 - 14x + 50)$

$q(x) = ax(x^4 - 14x^3 + 51x^2 - 14x + 50)$

$q(x) = a(x^5 - 14x^4 + 51x^3 - 14x^2 + 50x)$

q(-1) = -520 means that when $x = -1, q = -520$ , and this is used to find a.

$q(x) = a(x^5 - 14x^4 + 51x^3 - 14x^2 + 50x)$

$-520 = a(-1 - 14 - 51 - 14 - 50)$

$130a = 520$

$a = \frac{520}{130}$

$a = 4$

Hence, the equation is:

$q(x) = 4(x^5 - 14x^4 + 51x^3 - 14x^2 + 50x)$

A similar problem is given at brainly.com/question/24380382

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3 years ago

Read 2 more answers

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mafiozo [28]

9514 1404 393

Answer:

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Step-by-step explanation:

The point of inflection of the parent cubic function f(x) = x³ is at the origin, the same place it is shown on the graph. So, there is no translation involved in creating the graph.

We know the parent function gives f(1) = 1, but here, we have (x, y) = (1, -5). This suggests a vertical scale factor of -5. So, the transformed function is ...

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