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3 years ago

What are the zeros of f(x) = x3 + 6x2 + 3x – 10 ?

1 answer:

3 0

One of the zeroes would be : x = 1

f (1) = 1^3 + 6(1) ^2 + 3 (1) - 10

f(1) = 1 + 6 + 3 - 10

f (1) = 10 - 10

f (1) = 0

Hope this helps

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2 years ago

Let f(x)= x^2-4x-c. Find a nonzero value of c such that f(c)=c.

Dimas [21]

Answer:

The nonzero value of c will be:

$c = 6$

Step-by-step explanation:

Given the function

$f\left(x\right)=\:x^2-4x-c$

$f\left(c\right)=\:c^2-4c-c$

$f(c) = c$

$c=\:c^2-4c-c$

switching the sides

$c^2-4c-c=c$

subtract c from both sides

$c^2-4c-c-c=c-c$

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$c\left(c-6\right)=0$

Using the zero factor principle

$\:If}\:ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0\:\left(\mathrm{or\:both}\:a=0\:\mathrm{and}\:b=0\right)$

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$c=0,\:c=6$

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2 years ago

Please help!! I need to go to the next question!

tino4ka555 [31]

Given:

The inequality is:

$3t+1\leq 5$

To find:

The graph of the given inequality.

Solution:

We have,

$3t+1\leq 5$

Subtract both sides by 1.

$3t\leq 5-1$

$3t\leq 4$

Divide both sides by 3.

$t\leq \dfrac{4}{3}$

The value of t is less than or equal to $\dfrac{4}{3}$ .

Since $t\leq \dfrac{4}{3}$ , it means $\dfrac{4}{3}$ is included in the solution, therefore there is a closed circle at $t=\dfrac{4}{3}$ and an arrow approaches to left from $t=\dfrac{4}{3}$ .

Therefore, the correct option is A.

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