Ask question

Ask question

All categories

3 years ago

15

Complete the equation of the line through (3,-8) and (6,-4). Use exact numbers

1 answer:

MakcuM [25]3 years ago

7 0

Answer: y= 4/3x -8

Step-by-step explanation:

-4 - -8 over 6 - 3

-4 + 8 over 3

4 over 3

Y intercept is y1 which is -8

You might be interested in

What is the image of the point (-8, 3) after a rotation of 270°

Leya [2.2K]

Answer:

(3,8)

I hope it helps.

6 0

3 years ago

First Identify the Angle relationship, then solve for x.

Tanya [424]

Answer:

x=2

Step-by-step explanation:

46x-2=45x

+2 +2

46x=45x+2

-45x -45x

x=2

5 0

3 years ago

Read 2 more answers

Bob's tire outlet sold a record of tires last month one sales ma sold 165 tires which was 60 percent of the record what was the

Arisa [49]

The record number was 275 tires.

165/x=60/100

165 times 100=16500
16500/60=275
x=275

3 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 <em>f(x)</em> needs to satisfy

• $x^3$ divides $f(x)$

• $(x-1)^3$ divides $f(x)^2$

I'll also assume <em>f(x)</em> 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 <em>a</em> and <em>b</em> 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

Larissa argued that since log2(2) = 1 and log2(4) = 2, then it must be true that log2(3) = 1.5. Is she correct?

mestny [16]

Answer:

$2^{1.5} = 2.8284$ . This means that Larissa is not correct.

Step-by-step explanation:

To solve logarithms, we have to consider the following:

$\log_{a}(b)$ = c means that $a^{c} = b$ .

So, for example:

$log_{2}(2) = 1$ because $2^{1} = 2$ .

$log_{2}(4) = 2$ because $2^{2} = 4$ .

If $2^{1.5} = 3$ , then Larissa is correct. However, $2^{1.5} = 2.8284$ . This means that Larissa is not correct.

6 0

3 years ago