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3 years ago

What is the range of the function on the graph?

Mathematics

1 answer:

saw5 [17]3 years ago

6 0

Answer:

IT AMOUNG SUS DEZZ balls

Step-by-step explanation:

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Ummmmm please help me

ivanzaharov [21]

Just look up how to solve volume

4 0

3 years ago

A rock outcrop was found to have 95.82% of its parent U- 238 isotope remaining. Approximate the age of the outcrop. The half-lif

mixer [17]

Answer:

A) 277 million years

Step-by-step explanation:

Here, decay of U-238 is first order reaction. (since most of the nuclear decays are first order).

so, the equation t = $\frac{1}{k}$ × ln( $\frac{A(0)}{A}$ )

where, t = time from start of reaction

k = reaction constant

A(0) = initial concentration; A = present concentration.

⇒ 4.5 billion = 4500 million years = $\frac{1}{k}$ × ln2.

Now, given 95.82% of parent U-238 isotope is left.

⇒ t = $\frac{1}{k}$ × ln( $\frac{A(0)}{A}$ )

t = $\frac{4500}{ln2}$ × ln( $\frac{100}{95.82}$ )

⇒ t ≅ 277 million years.

5 0

3 years 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

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

6 0

3 years ago

How do I evaluate a+b for a= 47 and b=31

melisa1 [442]

A + b
47 + 31
78

a= 47
b = 31

6 0

3 years ago

Work about the mid-points of these line segments. GH: G(-3,-4) and H (0,2)

Svet_ta [14]

The x-coordinate of the midpoint is the average of the x-coordinates
of the ends.
The average of the ends is 1/2(-3+0) = -1.5 .

The y-coordinate of the midpoint is the average of the y-coordinates
of the ends.
The average of the ends is 1/2(-4+2) = -1

The midpoint of the segment is (-1.5, -1).

3 0

4 years ago