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3 years ago

Help! Algebra!!!

1 answer:

4 0

Thet contradict each other, that's why both of them are incorrect.
Suppose that a polynomial has four roots: s, t, u, and v. If the polynomial were evaluated at any of these values, it would have to be zero. Therefore, the polynomial can be written in this form.

p(x)(x - s)(x - t)(x - u)(x - v), where p(x) is some non-zero polynomial

This polynomial has a degree of at least 4. It therefore cannot be cubic.

Now prove Kelsey correct. We have already proved that there can be no more than three roots. To prove that a cubic polynomial with three roots is possible, all we have to do is offer a single example of that. This one will do.

(x - 1)(x - 2)(x - 3)

This is a cubic polynomial with three roots, and four or more roots are not possible for a cubic polynomial. Kelsey is correct.

Incidentally, if this is a roller coaster we are discussing, then a cubic polynomial is not such a good idea, either for a vertical curve or a horizontal curve. I hope this helps

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Karl's team score is 39 points. this was one point less than twice Todd's score. find Todd's score

EleoNora [17]

Answer: i'm pretty sure todd's score would be 19

39=2x-1

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3 years ago

Read 2 more answers

If 13cos theta -5=0 find sin theta +cos theta / sin theta -cos theta

Ivahew [28]

Step-by-step explanation:

We have to find the value of (sinθ + cosθ)/(sinθ - cosθ), when 13 cosθ - 5 = 0.

$\red{\frak{Given}} \begin{cases} & \sf {13\ cos \theta\ -\ 5\ =\ 0\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \big\lgroup Can\ also\ be\ written\ as \big\rgroup} \\ & \sf {cos \theta\ =\ {\footnotesize{\dfrac{5}{13}}}} \end{cases}$

Here, we're asked to find out the value of (sinθ + cosθ)/(sinθ - cosθ), when 13 cosθ - 5 = 0. In order to find the solution we're gonna use trigonometric ratios to find the value of sinθ and cosθ. Let us consider, a right angled triangle, say PQR.

Where,

PQ = Opposite side
QR = Adjacent side
RP = Hypotenuse
∠Q = 90°
∠C = θ

As we know that, 13 cosθ - 5 = 0 which is stated in the question. So, it can also be written as cosθ = 5/13. As per the cosine ratio, we know that,

$\rightarrow {\underline{\boxed{\red{\sf{cos \theta\ =\ \dfrac{Adjacent\ side}{Hypotenuse}}}}}}$

Since, we know that,

cosθ = 5/13
QR (Adjacent side) = 5
RP (Hypotenuse) = 13

So, we will find the PQ (Opposite side) in order to estimate the value of sinθ. So, by using the Pythagoras Theorem, we will find the PQ.

Therefore,

$\red \bigstar {\underline{\underline{\pmb{\sf{According\ to\ Question:-}}}}}$

$\rule{200}{3}$

$\sf \dashrightarrow {(PQ)^2\ +\ (QR)^2\ =\ (RP)^2} \\ \\ \\ \sf \dashrightarrow {(PQ)^2\ +\ (5)^2\ =\ (13)^2} \\ \\ \\ \sf \dashrightarrow {(PQ)^2\ +\ 25\ =\ 169} \\ \\ \\ \sf \dashrightarrow {(PQ)^2\ =\ 169\ -\ 25} \\ \\ \\ \sf \dashrightarrow {(PQ)^2\ =\ 144} \\ \\ \\ \sf \dashrightarrow {PQ\ =\ \sqrt{144}} \\ \\ \\ \dashrightarrow {\underbrace{\boxed{\pink{\frak{PQ\ (Opposite\ side)\ =\ 12}}}}_{\sf \blue{\tiny{Required\ value}}}}$

∴ Hence, the value of PQ (Opposite side) is 12. Now, in order to determine it's value, we will use the sine ratio.

$\rightarrow {\underline{\boxed{\red{\sf{sin \theta\ =\ \dfrac{Opposite\ side}{Hypotenuse}}}}}}$

Where,

Opposite side = 12
Hypotenuse = 13

Therefore,

$\sf \rightarrow {sin \theta\ =\ \dfrac{12}{13}}$

Now, we have the values of sinθ and cosθ, that are 12/13 and 5/13 respectively. Now, finally we will find out the value of the following.

$\rightarrow {\underline{\boxed{\red{\sf{\dfrac{sin \theta\ +\ cos \theta}{sin \theta\ -\ cos \theta}}}}}}$

By substituting the values, we get,

$\rule{200}{3}$

$\sf \dashrightarrow {\dfrac{sin \theta\ +\ cos \theta}{sin \theta\ -\ cos \theta}\ =\ {\footnotesize{\dfrac{\Big( \dfrac{12}{13}\ +\ \dfrac{5}{13} \Big)}{\Big( \dfrac{12}{13}\ -\ \dfrac{5}{13} \Big)}}}} \\ \\ \\ \sf \dashrightarrow {\dfrac{sin \theta\ +\ cos \theta}{sin \theta\ -\ cos \theta}\ =\ {\footnotesize{\dfrac{\dfrac{17}{13}}{\dfrac{7}{13}}}}} \\ \\ \\ \sf \dashrightarrow {\dfrac{sin \theta\ +\ cos \theta}{sin \theta\ -\ cos \theta}\ =\ \dfrac{17}{13} \times \dfrac{13}{7}} \\ \\ \\ \sf \dashrightarrow {\dfrac{sin \theta\ +\ cos \theta}{sin \theta\ -\ cos \theta}\ =\ \dfrac{17}{\cancel{13}} \times \dfrac{\cancel{13}}{7}} \\ \\ \\ \dashrightarrow {\underbrace{\boxed{\pink{\frak{\dfrac{sin \theta\ +\ cos \theta}{sin \theta\ -\ cos \theta}\ =\ \dfrac{17}{7}}}}}_{\sf \blue{\tiny{Required\ value}}}}$

∴ Hence, the required answer is 17/7.

6 0

2 years ago

Three cards are chosen at random from a standard 52-card deck. What is the probability that they are not all the same color?

rewona [7]

Three cards are selected from a standard deck of 52 cards. Disregarding the order in which they are drawn, the possible outcomes are (52/3). Out of these, how many include all cards of the same color (say red)? There are (13/3) ways in which you can get all 13 red cards.

8 0

3 years ago

THE QUESTION IS SOLVE FOR X

Damm [24]

Answer:

Step-by-step explanation:

i honestly don't care i came on here to cheat but it's saying answer so here you go :)

3 0

3 years ago

Thomas took three 100-point quizzes, and the mean score for the three quizzes was 87 points. If the highest score is taken away,

Olegator [25]

Answer:

The highest score is 94.

Step-by-step explanation:

Let the highest score be denoted by, A, the lowest score be, B and the middle score be X.

Then,

A + X + B = 87 × 3 = 261 ...(i)

X + B = 83.5 × 2 = 167 ...(ii)

⇒ B = 167 - X ...(iii)

A + X = 89 × 2 = 178 ...(iv)

⇒ A = 178 - X ...(v)

Substitute (iii) and (v) in (i) and solve for X as follows:

A + X + B = 261

178 - X + X + 167 - X = 261

345 - X = 261

X = 84

Substitute the value of X in (iii) and solve for B as follows:

B = 167 - X

= 167 - 84

= 83

Substitute the value of X in (v) and solve for A as follows:

A = 178 - X

= 178 - 84

= 94

Thus, the highest score is 94.

5 0

3 years ago