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2 years ago

What is the vertex of the absolute value function defined by ƒ(x) = |x + 5| + 7?

2 answers:

7 0

hello :
اthe vertex of the absolute value function defined by ƒ(x) = |x + 5| + 7 is :
(-5,7) because when : x= - 5 ...f(-5) = |- 5 + 5| + 7 = 0+ 7 = 7

Fofino [41]2 years ago

6 0

The graph of f(x) = |x + 5| will be shaped like a V with the vertex at (-5,0)

the + 7 moves it up 7 units so your answer is (-5,7)

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Which is the equation of an asymptote of the hyperbola whose equation is <img src="https://tex.z-dn.net/?f=%5Cfrac%7B%28x-2%29x%

tatuchka [14]

The equation of the asymptote is y = 3x - 5. The correct answer is option C

<h3>What is Asymptote of an Hyperbola ?</h3>

The distance from a point and the distance to a line in hyperbola is known as asymptote. The general equation is $x^{2}/ a^{2} - y^{2}/b^{2} = 1$

From the given equation of hyperbola, which is

$\frac{(x - 2)^{2}}{4}$ - $\frac{(y - 1)^{2} }{36}$ = 1

The center (h , k) of the hyperbola = C(2, 1)

a = 2

b = 6

Where C = $\sqrt{a^{2} + b^{2} }$

C = $\sqrt{4 + 36}$

C = $\sqrt{40}$

C = $2\sqrt{10}$

The equation of the asymptote will be y - K = +/-(b/a)(x - h)

That is,

y - 1 = +/-(6/2)(x - 2)

y - 1 = +/-3(x - 2)

y - 1 = +/-3x - 6

y = +/-3x - 6 + 1

y = +/- 3x - 5

Therefore, the equation of the asymptote is y = 3x - 5.

Learn more about Asymptote here: brainly.com/question/4138300

#SPJ1

3 0

1 year ago

Find a polynomial $f(x)$ of degree $5$ such that both of these properties hold: $\bullet$ $f(x)$ is divisible by $x^3$. $\bullet

frosja888 [35]

There seems to be one character missing. But I gather that f(x) needs to satisfy

• $x^3$ divides $f(x)$

• $(x-1)^3$ divides $f(x)^2$

I'll also assume f(x) is monic, meaning the coefficient of the leading term is 1, or

$f(x) = x^5 + \cdots$

Since $x^3$ divides $f(x)$ , and

$f(x) = x^3 p(x)$

where $p(x)$ is degree-2, and we can write it as

$f(x)=x^3 (x^2+ax+b)$

Now, we have

$f(x)^2 = \left(x^3p(x)\right)^2 = x^6 p(x)^2$

so if $(x - 1)^3$ divides $f(x)^2$ , then $p(x)$ is degree-2, so $p(x)^2$ is degree-4, and we can write

$p(x)^2 = (x-1)^3 q(x)$

where $q(x)$ is degree-1.

Expanding the left side gives

$p(x)^2 = x^4 + 2ax^3 + (a^2+2b)x^2 + 2abx + b^2$

and dividing by $(x-1)^3$ leaves no remainder. If we actually compute the quotient, we wind up with

$\dfrac{p(x)^2}{(x-1)^3} = \underbrace{x + 2a + 3}_{q(x)} + \dfrac{(a^2+6a+2b+6)x^2 + (2ab-6a-8)x +2a+b^2+3}{(x-1)^3}$

If the remainder is supposed to be zero, then

$\begin{cases}a^2+6a+2b+6 = 0 \\ 2ab-6a-8 = 0 \\ 2a+b^2+3 = 0\end{cases}$

Adding these equations together and grouping terms, we get

$(a^2+2ab+b^2) + (2a+2b) + (6-8+3) = 0 \\\\ (a+b)^2 + 2(a+b) + 1 = 0 \\\\ (a+b+1)^2 = 0 \implies a+b = -1$

Then $b=-1-a$ , and you can solve for a and b by substituting this into any of the three equations above. For instance,

$2a+(-1-a)^2 + 3 = 0 \\\\ a^2 + 4a + 4 = 0 \\\\ (a+2)^2 = 0 \implies a=-2 \implies b=1$

So, we end up with

$p(x) = x^2 - 2x + 1 \\\\ \implies f(x) = x^3 (x^2 - 2x + 1) = \boxed{x^5-2x^4+x^3}$

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2 years ago

How do you write 4/ 9 as a decimal?

olasank [31]

Answer:

.4 repeating

Step-by-step explanation: divide the top number by the bottom number so 4 divided by 9

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2 years ago

What is the value of x in 5x−3(x−11)=7(2x−5)+2

riadik2000 [5.3K]

The value of x is 5.5. I got this answer by first distributing -3 amongst (x-11), and distributing 7 amongst (2x-5). Next I combined like terms and subtracted 3x from 5x, and added -35 and 2. Next I subtracted 33 from both sides. Next I subtracted 14x from both sides. And finally I divided both sides by -12 to get the final answer of 5.5.

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3 years ago

What is this question?

Ira Lisetskai [31]

Albert because 1/2 or .5 is grater than .4

7 0

3 years ago