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3 years ago

Without graphing, determine the x-intercepts and vertex of the quadratic function f(x)=x²+11x+18. Show all of your work.

1 answer:

5 0

Answer: https://www.khanacademy.org/math/algebra/x2f8bb11595b61c86:quadratic-functions-equations/x2f8bb11595b61c86:standard-form-quadratic/v/ex3-completing-the-square

Step-by-step explanation: This will help you a lot

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Trey started to run on a treadmill after setting its timer for 56 minutes. The display says that he has finished 62% of his run.

Ilia_Sergeevich [38]

Answer:

34.7 minutes

Step-by-step explanation:

56 minutes times 62% (0.62) = 34.72

5 0

2 years ago

How to solve part ii and iii

iragen [17]

(i) Given that

$\tan^{-1}(x) + \tan^{-1}(y) + \tan^{-1}(xy) = \dfrac{7\pi}{12}$

when $x=1$ this reduces to

$\tan^{-1}(1) + 2 \tan^{-1}(y) = \dfrac{7\pi}{12}$

$\dfrac\pi4 + 2 \tan^{-1}(y) = \dfrac{7\pi}{12}$

$2 \tan^{-1}(y) = \dfrac\pi3$

$\tan^{-1}(y) = \dfrac\pi6$

$\tan\left(\tan^{-1}(y)\right) = \tan\left(\dfrac\pi6\right)$

$\implies \boxed{y = \dfrac1{\sqrt3}}$

(ii) Differentiate $\tan^{-1}(xy)$ implicitly with respect to $x$ . By the chain and product rules,

$\dfrac d{dx} \tan^{-1}(xy) = \dfrac1{1+(xy)^2} \times \dfrac d{dx}xy = \boxed{\dfrac{y + x\frac{dy}{dx}}{1 + x^2y^2}}$

(iii) Differentiating both sides of the given equation leads to

$\dfrac1{1+x^2} + \dfrac1{1+y^2} \dfrac{dy}{dx} + \dfrac{y + x\frac{dy}{dx}}{1+x^2y^2} = 0$

where we use the result from (ii) for the derivative of $\tan^{-1}(xy)$ .

Solve for $\frac{dy}{dx}$ :

$\dfrac1{1+x^2} + \left(\dfrac1{1+y^2} + \dfrac x{1+x^2y^2}\right) \dfrac{dy}{dx} + \dfrac y{1+x^2y^2} = 0$

$\left(\dfrac1{1+y^2} + \dfrac x{1+x^2y^2}\right) \dfrac{dy}{dx} = -\left(\dfrac1{1+x^2} + \dfrac y{1+x^2y^2}\right)$

$\dfrac{1+x^2y^2 + x(1+y^2)}{(1+y^2)(1+x^2y^2)} \dfrac{dy}{dx} = - \dfrac{1+x^2y^2 + y(1+x^2)}{(1+x^2)(1+x^2y^2)}$

$\implies \dfrac{dy}{dx} = - \dfrac{(1 + x^2y^2 + y + x^2y) (1 + y^2) (1 + x^2y^2)}{(1 + x^2y^2 + x + xy^2) (1+x^2) (1+x^2y^2)}$

$\implies \dfrac{dy}{dx} = -\dfrac{(1 + x^2y^2 + y + x^2y) (1 + y^2)}{(1 + x^2y^2 + x + xy^2) (1+x^2)}$

From part (i), we have $x=1$ and $y=\frac1{\sqrt3}$ , and substituting these leads to

$\dfrac{dy}{dx} = -\dfrac{\left(1 + \frac13 + \frac1{\sqrt3} + \frac1{\sqrt3}\right) \left(1 + \frac13\right)}{\left(1 + \frac13 + 1 + \frac13\right) \left(1 + 1\right)}$

$\dfrac{dy}{dx} = -\dfrac{\left(\frac43 + \frac2{\sqrt3}\right) \times \frac43}{\frac83 \times 2}$

$\dfrac{dy}{dx} = -\dfrac13 - \dfrac1{2\sqrt3}$

as required.

3 0

2 years ago

To own and operate a poster printer, it costs $250 for the printer and an additional $0.20 per poster for ink. To print out post

NARA [144]

Answer:

312.5

Step-by-step explanation:

Since it is $250 plus 20 cents per poster, the equation is 0.20p+250. It is $1 per poster at store and you want to know when the cost is going to be the same so you set them equal to each other.

0.20p+250=p

250=0.80p

312.5=p

6 0

3 years ago

Solve each inequality. <br> 1. 5x + 2 <17

butalik [34]

Answer:

x<3

Step-by-step explanation:

5x+2<17

Subtract 2 from both sides.

5x<17−2

Subtract 2 from 17 to get 15.

5x<15

Divide both sides by 5. Since 5 is positive, the inequality direction remains the same.

x=15/5

Divide 15 by 5 to get 3.

x<3

8 0

3 years ago

⚠️NEED HELP ASPA⚠️<br> Determine whether each pair of triangles is similar. Justify your answer.

bekas [8.4K]

So the answer is A. The way to check if triangles are similar is find the ratio and make sure they are equal.
For example if you divide 42 by 27 you get 1.55
And if you do that to all the sides you also get 1.55 so therefore they are similar

7 0

3 years ago