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2 years ago

Pls help i will mark brainliest

Mathematics

2 answers:

s344n2d4d5 [400]2 years ago

6 0

Answer:

(4,6)

Step-by-step explanation:

hope this helps you

adell [148]2 years ago

3 0

I was gonna comment but my answer is the same as the person above

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Need help with this one question

loris [4]

Answer:

71.6599

Step-by-step explanation:

I think if I’m wrong please tell me

wich ik I’m probably wrong

8 0

2 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

2 years ago

What is (f - 5)(-2) + 11f

n200080 [17]

Answer:

-2f+10+11f

= 9f+10

Step-by-step explanation:

just expand it

7 0

2 years ago

Read 2 more answers

would someone be able to explain to me how to solve these problems? I don't need the solutions, I just need an explanation. I ne

Leni [432]

The first 3 are examples of the difference of 2 squares so you use the identity
a^2 - b^2 = (a + b)(a - b)
x^2 - 49 = 0
so (x + 7)(x - 7) = 0
so either x + 7 = 0 or x - 7 = 0
giving x = -7 and 7.
Number 7 reduces to 3x^2 =12, x^2 = 4 so x = +/- 2
Number 8 take out GCf (d) to give
d(d - 2) = 0 so d = 0 , 2
9 and 10 are more difficult to factor
you use the 'ac' method Google it to get more details
2x^2 - 5x + 2
multiply first coefficient by the constant at the end
that is 2 * 2 = 4
Now we want 2 numbers which when multiplied give + 4 and when added give - 5:- -1 and -4 seem promising so we write the equation as:-

2x^2 - 4x - x + 2 = 0

now factor by grouping
2x(x - 2) - 1(x - 2) = 0

(x - 2) is common so

(2x - 1)(x - 2) = 0
and 2x - 1 = 0 or x - 2 = 0 and now you can find x.

The last example is solved in the same way.

8 0

3 years ago

Three forces of 300 N in the direction of N30E, 400N in the direction of N60E and

andreyandreev [35.5K]

Split up each force into horizontal and vertical components.

• 300 N at N30°E :

(300 N) (cos(30°) i + sin(30°) j)

• 400 N at N60°E :

(400 N) (cos(60°) i + sin(60°) j)

• 500 N at N80°E :

(500 N) (cos(80°) i + sin(80°) j)

The resultant force is the sum of these forces,

∑ F = (300 cos(30°) + 400 cos(60°) + 500 cos(80°)) i

… … … + (300 sin(30°) + 400 sin(60°) + 500 sin(80°)) j N

∑ F ≈ (546.632 i + 988.814 j) N

so ∑ F has a magnitude of approximately 1129.85 N and points in the direction of approximately N61.0655°E.

4 0

2 years ago