All categories

2 years ago

Solve for x. Do not type degree symbol or any spaces in your answer. 85 40° X

Mathematics

1 answer:

Komok [63]2 years ago

6 0

145 cause 40+85+90 equals 215 n then subtract that from 360

You might be interested in

Find the second derivative for y=e^10x+y

vesna_86 [32]

Answer:

10 e ^10 x

Step-by-step explanation:

Find the derivative using the chain and power rules.

3 0

3 years ago

I do not know this answer

Ivenika [448]

Answer:

The answer is -19

Step-by-step explanation:

Given information -19y^2 , y=-1

-19 * y^2

-19 * (-1)^2

-19 * 1 = -19

7 0

2 years ago

an airplaine descends 1.5 miles to an elevation of 5.25 miles. find the elevation of the plane before its descent

Maksim231197 [3]

We can then write an equation representing this problem as:

e−1.5mi=5.25mi

Now, add 1.5mi to each side of the equation to solve for e while keeping the equation balanced:

e−1.5mi+1.5mi=5.25mi+1.5mi

e−0=6.75mi

e=6.75mi

The plane's starting elevation was 6.75 miles

Hope this helps!

8 0

3 years ago

PLEASE HELP ASAP! The quotient of two rational numbers is positive. What can you conclude about the signs of the dividend and th

Anon25 [30]

Answer:

They either both have to be positive or negative.

Step-by-step explanation:

1 / 1 = 1

-1 / -1 = 1

This gets you positive, when both dividend and divisors are positive or negative.

-1 / 1 = -1

1 / -1 = -1

This gets you negative, when both dividend and divisors are different signs.

3 0

3 years ago

Read 2 more answers

y′′ −y = 0, x0 = 0 Seek power series solutions of the given differential equation about the given point x 0; find the recurrence

sukhopar [10]

Let

$\displaystyle y(x) = \sum_{n=0}^\infty a_nx^n = a_0 + a_1x + a_2x^2 + \cdots$

Differentiating twice gives

$\displaystyle y'(x) = \sum_{n=1}^\infty na_nx^{n-1} = \sum_{n=0}^\infty (n+1) a_{n+1} x^n = a_1 + 2a_2x + 3a_3x^2 + \cdots$

$\displaystyle y''(x) = \sum_{n=2}^\infty n (n-1) a_nx^{n-2} = \sum_{n=0}^\infty (n+2) (n+1) a_{n+2} x^n$

When x = 0, we observe that y(0) = a₀ and y'(0) = a₁ can act as initial conditions.

Substitute these into the given differential equation:

$\displaystyle \sum_{n=0}^\infty (n+2)(n+1) a_{n+2} x^n - \sum_{n=0}^\infty a_nx^n = 0$

$\displaystyle \sum_{n=0}^\infty \bigg((n+2)(n+1) a_{n+2} - a_n\bigg) x^n = 0$

Then the coefficients in the power series solution are governed by the recurrence relation,

$\begin{cases}a_0 = y(0) \\ a_1 = y'(0) \\\\ a_{n+2} = \dfrac{a_n}{(n+2)(n+1)} & \text{for }n\ge0\end{cases}$

Since the n-th coefficient depends on the (n - 2)-th coefficient, we split n into two cases.

• If n is even, then n = 2k for some integer k ≥ 0. Then

$k=0 \implies n=0 \implies a_0 = a_0$

$k=1 \implies n=2 \implies a_2 = \dfrac{a_0}{2\cdot1}$

$k=2 \implies n=4 \implies a_4 = \dfrac{a_2}{4\cdot3} = \dfrac{a_0}{4\cdot3\cdot2\cdot1}$

$k=3 \implies n=6 \implies a_6 = \dfrac{a_4}{6\cdot5} = \dfrac{a_0}{6\cdot5\cdot4\cdot3\cdot2\cdot1}$

It should be easy enough to see that

$a_{n=2k} = \dfrac{a_0}{(2k)!}$

• If n is odd, then n = 2k + 1 for some k ≥ 0. Then

$k = 0 \implies n=1 \implies a_1 = a_1$

$k = 1 \implies n=3 \implies a_3 = \dfrac{a_1}{3\cdot2}$

$k = 2 \implies n=5 \implies a_5 = \dfrac{a_3}{5\cdot4} = \dfrac{a_1}{5\cdot4\cdot3\cdot2}$

$k=3 \implies n=7 \implies a_7=\dfrac{a_5}{7\cdot6} = \dfrac{a_1}{7\cdot6\cdot5\cdot4\cdot3\cdot2}$

so that

$a_{n=2k+1} = \dfrac{a_1}{(2k+1)!}$

So, the overall series solution is

$\displaystyle y(x) = \sum_{n=0}^\infty a_nx^n = \sum_{k=0}^\infty \left(a_{2k}x^{2k} + a_{2k+1}x^{2k+1}\right)$

$\boxed{\displaystyle y(x) = a_0 \sum_{k=0}^\infty \frac{x^{2k}}{(2k)!} + a_1 \sum_{k=0}^\infty \frac{x^{2k+1}}{(2k+1)!}}$

4 0

3 years ago