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bonufazy [111]
3 years ago
12

There are values of t so that sin t=.35 and cos t=.6

Mathematics
1 answer:
NISA [10]3 years ago
7 0

Recall the fundamental rule of trig:

\sin^2(x)+\cos^2(x)=1 \quad\forall x \in \mathbb{R}

So, there exists an angle t such that

(0.6,0.35)=(\sin(t),\cos(t))

if and only if

\sin^2(t)+\cos^2(t)=0.6^2+0.35^2=1

Working out the numbers, we get

0.6^2+0.35^2=0.36+0.1225=0.4825\neq 1

So, there doesn't exist a number t such that

(0.6,0.35)=(\sin(t),\cos(t))

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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
3 years ago
Someone plz help! Photo Above
TiliK225 [7]

Answer:

  • -704

Step-by-step explanation:

<u>Given the sequence</u>

  • A = -4 - 6i

<u>To find </u>

  • Sum of terms from i=5 to i=15
<h3>Solution</h3>

We see the sequence is AP

<u>The required sum is</u>

  • S₅₋₁₅ = S₁₅ - S₄

<u>Using sum of AP formula</u>

  • Sₙ = 1/2n(i₁ + iₙ)

<u>Finding the required terms</u>

  • i₁ = - 4- 6 = -10
  • i₄ = -4 -6*4 = - 28
  • i₁₅ = -4 -6*15 = -94

<u>Getting the sum</u>

  • S₄ = 1/2*4*(-10 - 28) = -76
  • S₁₅ = 1/2*15*(-10 - 94) = -780
  • S₅₋₁₅ = S₁₅ - S₄ = -780 - (-76) = - 704
4 0
3 years ago
What is another way to name 1/4
Sauron [17]
Another way to name 1/4 would be .25
4 0
3 years ago
Read 2 more answers
Luke’s basketball team went to an amusement park at the end of the season. The cost of the admission for 5 coaches and 12 player
Finger [1]

Answer:

The admission cost for each player was <u>22.50</u>.

Step-by-step explanation:

Given:

Luke’s basketball team went to an amusement park at the end of the season.

The cost of the admission for 5 coaches and 12 players was 407.50.

The admission cost for each coach was 27.50.

Now, to find the admission cost for each player.

Total admission cost = 407.50.

Number of coaches = 5.

Admission cost for each = 27.50.

So, the cost of admission of all coaches:

27.50\times 5

=137.50.

Then, we deduct the cost of admission of all coaches from the total admission cost:

407.50-137.50

=270.

<em>Remaining cost = 270.</em>

Number of player = 12.

Now, to get the admission cost for each player we divide the remaining cost  by number of players:

270\div 12

=22.50.

Therefore, the admission cost for each player was 22.50.

7 0
3 years ago
What is the average rate of change
algol [13]
The average rate of change between two input values is the total change of the function values (output values) divided by the change in the input values.
6 0
3 years ago
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