All categories

2 years ago

Use the quadratic formula to solve for the zeros: 2x^2 + 39 = -18

Mathematics

1 answer:

mars1129 [50]2 years ago

3 0

Answer:

Step-by-step explanation:

−

2x^{2}+39=-18

2x2+39=−18

−

(

−

)

You might be interested in

Please help will mark brainliest

Advocard [28]

Answer:

y=-1/2x+8

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

How can you find the surface area of a composite solid made up of prisms ? (Help)

sammy [17]

Oh, this is easy. For any prism, you find the surface area by finding the are of all the sides. I will post all surface area formulas here, for reference
Cube:

Surface area = 6 × a2

Right circular cylinder:

Surface area = 2 × pi × r2 + 2 × pi × r × h

pi = 3.14
h is the height
r is the radius

Rectangular prism:

Surface area = 2 × l × w + 2 × l × h + 2 × w × h

l is the length
w is the width
h is the height

Sphere:

Surface area = 4 × pi × r2

pi = 3.14
r is the radius

Right circular cone:

Surface area = pi × r2 + pi × r ×( √(h2 + r2))

pi = 3.14
r is the radius
h is the height
l is the slant height

Right square pyramid:

Surface area = s2 + 2 × s × l

s is the length of the base
h is the height
l is the slant height

8 0

3 years ago

Find the 10th term of the sequence defined by the rule, f (n) = 4n - 3.

DiKsa [7]

Answer:

Step-by-step explanation:

f(n) = 4n - 3

f(10) = 4×10 - 3 = 40 - 3 = 37

6 0

3 years ago

Find the slope of the line below.

creativ13 [48]

A.
$- \frac{4}{5}$

6 0

3 years ago

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

2 years ago