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Find the points on the given curve where the tangent line is horizontal or vertical. (Assume 0 ≤ θ < π. Enter your answers as a comma-separated list of ordered pairs.)r = 6 cos(θ)

Answers

Answer 1

Answer:

Answer:

point of horizontal tangent is $(0^(o),6)$ and point of vertical tangent is $(-15.17^(o),5.79)$

Step-by-step explanation:

For a horizontal tangent it's slope should be zero thus

$r=6cos(\theta )\n\n(dr)/(d\theta)=-6sin(\theta )$ $\therefore -6sin(\theta)=0\n\n \Rightarrow \theta =0,\pi$

Thus the ordered pair of $(\theta ,r)$ becomes (0,6) at this point tangent is horizontal

For a vertical tangent it's slope should be $(\pi )/(2)$

Again differentiating the given curve we get

$r=6cos(\theta )\n\n (dr)/(d\theta)=-6sin(\theta )$ $\therefore -6sin(\theta)=(\pi )/(2)\n\n\Rightarrow \theta =sin^(-1)(-\pi )/(12)$

$\therefore \theta =-15.17^(o)$

Thus the ordered pair of vertical tangent becomes ( $\theta =-15.17^(o),5.79)$

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N<2^2. mathematical induction

Answers

Answer:

Inequality Form:
n < 4

Interval Notation:
(-, 4)

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Answers

The answer is 115 degrees too

Need answer asap The manager of a video game store found that 35 of the 140 people who preordered the latest baseball game canceled their orders the day before the game was released. He used that data to create a simulation to predict the probability that future customers will cancel their preorders.

According to the manager’s model, what is the probability that two customers who preorder the newest golf game will both cancel their orders the day before the game is released?
StartFraction 1 over 16 EndFraction
StartFraction 1 over 8 EndFraction
One-fourth
One-half

Answers

Answer:

1/4 will cancel

Step-by-step explanation:

Simplify the fraction

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140

Divide the top and bottom by 7

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divide the top and bottom by 5

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I think it from 4 how I know this cause I tried

H= -16t^2+16t+192 When will the ball hit the ground?

Answers

Answer:

t = 4 s

Step-by-step explanation

We can infer the origin from the t⁰ term of the quadratic as being 192 feet above the ground. Initial velocity is 16 ft/s from the t¹ term, gravity is -32 ft/s² from the t² term

0 = -16t² + 16t + 192

0 = -t² + t + 12

0 = t² - t - 12

0 = (t - 4)(t + 3)

so either

t - 4 = 0

t = 4 s

t + 3 = 0

t = -3 s which we ignore as it occurs before the ball is released.

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Answers

Answer:

Option B

118

Step-by-step explanation:

$4{x}^(2) + 3x + 3$

By putting the value of x = 5

$= 4 {(5)}^(2) + 3 * 5 + 3$

= 4 × 25 + 15 + 3

= 100 + 18

= 118 (Ans)

Answer:

Step-by-step explanation:

4 × 5 to the power of 2 + 3 × 5 + 3

For the characteristic polynomialp(s) =s5+ 2s4+ 24s3+ 48s2−25s−50(a) Use the Routh-Hurwitz Criterion to determine the number of roots ofp(s) in the right-half plane, in the left-half plane, and on thejω-axis.(b) Use Matlab to determine the roots ofp(s), and verify your results in part 2a.

Answers

Answer:

1 root in the right half-plane
1 conjugate pair on the imaginary axis
2 roots in the left half-plane

Step-by-step explanation:

Without using the Routh-Hurwitz criterion at all, you know there is one positive real root. Descartes' rule of signs tells you the number of positive real roots is equal to the number of sign changes in the coefficients (perhaps less a multiple of 2). There is one sign change in + + + + - - , so there is one positive real root.

_____

(a) The Routh array starts as two rows of the polynomial's coefficients, alternate coefficients on each row. For this odd-degree polynomial, the number of coefficients is even, so no zero-padding is necessary at the right end of the second row. That is, we start with ...

$\begin{array}{cccc}s^5&1&24&-25\ns^4&2&48&-50\end{array}$

The next row is formed from combinations of coefficients in the two rows above. The computation is similar to that of a determinant. By matching the numbers to those in the array, you can see the pattern of the computation.

The next row values are ...

$\begin{array}{ccc}s^3&((2)(24)-(1)(48))/(2)&((2)(-25)-(1)(-50))/(2)\end{array}$

Simplifying, we find this row to be ...

$\begin{array}{ccc}s^3&0&0\end{array}$

The zero row is a special case that requires we proceed as follows. The row above (identified with s⁴) represents an "auxiliary polynomial":

$2s^4 +48s^2 -50$

To continue the process, we replace the zero row by the coefficients of the derivative of this auxiliary polynomial. Proceeding as before, the array now becomes ...

$\begin{array}{cccc}s^5&1&24&-25\ns^4&2&48&-50\ns^3&8&96\ns^2&24&-50\ns^1&112(2)/(3)&0\ns^0&-50\end{array}$

The number of sign changes in the first column (1) tells the number of roots in the right half-plane. The auxiliary polynomial will give us the remaining two pairs of roots:

$2s^4+48s^2-50=0\n\n2(s^2+25)(s^2-1)=0\n\ns=\pm 5i,\ s=\pm 1$

So, we have determined there to be ...

1 root in the right half-plane
2 roots on the jω axis
2 roots in the left half-plane

(b) The original polynomial can be factored as ...

p(s) = (s +2)(s² +25)(s +1)(s -1)

p(s) = (s +2)(s +1)(s -5i)(s +5i)(s -1)

This verifies our result from part (a).

_____

Additional comments

Any row can be multiplied by a convenient factor to simplify the arithmetic. Here, it would be convenient to divide the second row by 2 and the third row by 8.

A zero element (not row) in the first column is replaced by "epsilon" (a small positive number) and the rest of the arithmetic is continued as normal. That row is not counted (it is ignored) when counting sign changes in the first column.

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