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3 years ago

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 f(x) needs to satisfy

• $x^3$ divides $f(x)$

• $(x-1)^3$ divides $f(x)^2$

I'll also assume f(x) 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 a and b 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:

Given the sequence

A = -4 - 6i

To find

Sum of terms from i=5 to i=15

<h3>Solution</h3>

We see the sequence is AP

The required sum is

S₅₋₁₅ = S₁₅ - S₄

Using sum of AP formula

Sₙ = 1/2n(i₁ + iₙ)

Finding the required terms

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

Getting the sum

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 22.50.

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:

$407.50-137.50$

$=270.$

Remaining cost = 270.

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.

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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.

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3 years ago