What is the value of x in the triangles to the right? A. 9 B. 12 C. 15 D. 17.5

Ivan

Trying to factor by splitting the middle term

Factoring b2-4b+4
The first term is, b2 its coefficient is 1 .
The middle term is, -4b its coefficient is -4 .
The last term, "the constant", is +4 

Step-1 : Multiply the coefficient of the first term by the constant 1 • 4 = 4

Step-2 : Find two factors of 4 whose sum equals the coefficient of the middle term, which is -4 .

-4 + -1 = -5 -2 + -2 = -4 That's it

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, -2 and -2
 b2 - 2b - 2b - 4

Step-4 : Add up the first 2 terms, pulling out like factors :
 b • (b-2)
 Add up the last 2 terms, pulling out common factors :
 2 • (b-2)
Step-5 : Add up the four terms of step 4 :
 (b-2) • (b-2)
 Which is the desired factorization

8 0

3 years ago

Read 2 more answers

Round 23.782 to the nearest tenth

LekaFEV [45]

Answer:

23.800

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers

What is the slope of the line through the points (4, 2) and (-16, -6)?

Harrizon [31]

Gradient = change in y /change in x so -6-2 =-8 / -16-4 = -20 so -8/-20 = 0.4 so gradient = 0.4

3 0

2 years ago

A parabola has zeros at (5,0) and (-3,0) and passes through point (6,18) determine the axis of symmetry

worty [1.4K]

Answer:

The axis of symmetry is $x=1$

Step-by-step explanation:

we know that

In a vertical parabola, the axis of symmetry is equal to the x-coordinate of the vertex

In this problem we have a vertical parabola open upward

The x-coordinate of the vertex is equal to the midpoint between the zeros of the parabola

so

$x=\frac{5-3}{2}=1$

therefore

The axis of symmetry is $x=1$

5 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