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andrey2020 [161]
3 years ago
7

What is the value of x?

Mathematics
1 answer:
Lerok [7]3 years ago
7 0

Answer:

  3√2

Step-by-step explanation:

The hypotenuse of an isosceles right triangle is √2 times the side length.

  x = 3√2

_____

All these triangles are similar, so the ratio of hypotenuse to leg is the same for all.

  hypotenuse/leg = x/3 = (3+3)/x

  x² = 3·6 = 3²·2 . . . . cross multiply (identify square factors)

  x = 3√2 . . . . . take the square root

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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
2 years ago
We need a minimum of 30 book sales by today. If 10 books were sold yesterday, what are the possible numbers for sales today
amm1812

I mean it would be A and D because I don't know why they are both the same answer. If 10 books were sold yesterday, at LEAST 20 are needed to make the goal.

6 0
3 years ago
What is the solution to the equation below? Round your answer to two decimal places. 4•e^x=17
bixtya [17]

Answer:

\large\boxed{x\approx1.45}

Step-by-step explanation:

4e^x=17\qquad\text{divide both sides by 4}\\\\e^x=\dfrac{17}{4}\qquad \ln\ \text{of both sides}\\\\\ln e^x=\ln\dfrac{17}{4}\qquad\text{use}\ \log_aa^n=n\\\\x=\ln\dfrac{17}{4}\\\\x\approx1.45

7 0
3 years ago
Simplify: sec(θ) sin(θ) cot(θ)
krek1111 [17]

We have to simplify

sec(θ) sin(θ) cot(θ)

Now first of all let's simplify these separately , using reciprocal identities.

Sec(θ) = 1/cos(θ)

Sin(θ) is already simplified

Cot(θ)= cos(θ) / sin(θ) ,

Let's plug these values in the expression

sec(θ) sin(θ) cot(θ)

= ( 1/cos(θ) ) * ( sin(θ) ) * ( cos(θ) / sin(θ) )

= ( sin(θ) /cos(θ) ) * ( cos(θ) /sin(θ) )

sin cancels out with sin and cos cancels out with cos

So , answer comes out to be

=( sin(θ) /cos(θ) ) * ( cos(θ) /sin(θ) )

= 1

4 0
3 years ago
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Is my answers correct just want to be sure that my answers is right. Thanks
sergey [27]
Yes 4/4= 1 whole hope this helps
8 0
3 years ago
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