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3 years ago

Solve the equation for x, where x is a real number (5 points): -13x^2 + 17x + 39 = 44

1 answer:

7 0

We first subtract 44 from both sides so that the equation becomes -13x^2 + 17x - 5 = 0.

To find the solutions to this equation, we can apply the quadratic formula. This quadratic formula solves equations of the form ax^2 + bx + c = 0
                     x = [ -b ± √(b^2 - 4ac) ] / (2a)
                     x = [ -17 ± √(17^2 - 4(-13)(-5)) ] / ( 2(-13) )
                     x = [ -17 ± √(289 - (260) ) ] / ( -26 )
                     x = [ -17 ± √(29) ] / ( -26)
                     x = [ -17 ± sqrt(29) ] / ( -26 )
                     x = 17/26 ± -sqrt(29)/26
The answers are 17/26 + sqrt(29)/26 and 17/26 - sqrt(29)/26.

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Find the equation of the axis of symmetry of the following parabola algebraically.

mr Goodwill [35]

Answer:

the equation of the axis of symmetry is $x=8$

Step-by-step explanation:

Recall that the equation of the axis of symmetry for a parabola with vertical branches like this one, is an equation of a vertical line that passes through the very vertex of the parabola and divides it into its two symmetric branches. Such vertical line would have therefore an expression of the form: $x=constant$ , being that constant the very x-coordinate of the vertex.

So we use for that the fact that the x position of the vertex of a parabola of the general form: $y=ax^2+bx+c$ , is given by:

$x_{vertex}=\frac{-b}{2\,a}$

which in our case becomes:

$x_{vertex}=\frac{-b}{2\,a} \\x_{vertex}=\frac{48}{2\,(3)} \\x_{vertex}=\frac{48}{6} \\x_{vertex}=8$

Then, the equation of the axis of symmetry for this parabola is:

$x=8$

4 0

4 years ago

Consider the function graphed below. What is the average rate of change of the function over the interval [1, 3]?

EleoNora [17]

The correct answer is B. 36.

8 0

3 years ago

Manchu has drawn a

timama [110]

Answer:

Height of each triangle: 14cm

Area of each triangle: 168cm²

Step-by-step explanation:

So we know that the base of the area is 24. Lets find the height.

336 ÷ 24 = 14

If he makes two triangles out of the rectangle that just means he cuts it in half.

To find the area of one triangle lets do:

336 ÷ 2 = 168

Let's double check our answer:

(14 × 24) ÷ 2 = 168

Seems great!

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

3 years ago

Use the Distributive Property to multiply 8 and 45 in your head.

Andre45 [30]

Answer:

320+40=360

Step-by-step explanation:

8(40+5)

320+40=360

5 0

3 years ago

Read 2 more answers