Ask question

Ask question

All categories

3 years ago

11

please help me i don't know how to do this and i need help i been struggling for 2 hours please help me i really need help help

is really appreciated please

2 answers:

Rasek [7]3 years ago

4 0

Vertex: (4,0)

Axis of symmetry: x = 4

Klio2033 [76]3 years ago

4 0

Answer:

Vertex: (4, 0)

Axis of symmetry: x = 4

Step-by-step explanation:

This quadratic is in vertex form. Vertex form is

y = a(x - h)² + k where the point (h, k) is the vertex.

For this equation we have

y = -(1/8)(x - 4)² + 0, so our vertex is at the point (4, 0)

*It's positive 4 because the equation is (x - h), so we're subtracting positive 4

The axis of symmetry is the equation of the vertical line that goes through the x coordinate of the vertex (quadratics are symmetric from the vertex)

So the axis of symmetry is x = 4

You might be interested in

What is 1/8-10(3/4-3/8x)+5/8x

cluponka [151]

Answer:

x = 59/35

Step-by-step explanation:

<u>Step 1: Distribute</u>

1/8 - 10(3/4 - 3/8x) + 5/8x

1/8 - 30/4 + 30/8x + 5/8x

<u>Step 2: Combine like terms</u>

1/8 - 30*2/4*2 - 30/8x + 5/8x

1/8 - 60/8 + 35/8x

-59/8 + 35/8x

<u>Step 3: Solve for x</u>

-59/8 + 35/8x + 59/8 = 0 + 59/8

35/8x * 8/35 = 59/8 * 8/35

x = 59/35

Answer: x = 59/35

4 0

3 years ago

Consider the following. (A computer algebra system is recommended.) y'' + 3y' = 2t4 + t2e−3t + sin 3t (a) Determine a suitable f

drek231 [11]

First look for the fundamental solutions by solving the homogeneous version of the ODE:

$y''+3y'=0$

The characteristic equation is

$r^2+3r=r(r+3)=0$

with roots $r=0$ and $r=-3$ , giving the two solutions $C_1$ and $C_2e^{-3t}$ .

For the non-homogeneous version, you can exploit the superposition principle and consider one term from the right side at a time.

$y''+3y'=2t^4$

Assume the ansatz solution,

${y_p}=at^5+bt^4+ct^3+dt^2+et$

$\implies {y_p}'=5at^4+4bt^3+3ct^2+2dt+e$

$\implies {y_p}''=20at^3+12bt^2+6ct+2d$

(You could include a constant term <em>f</em> here, but it would get absorbed by the first solution $C_1$ anyway.)

Substitute these into the ODE:

$(20at^3+12bt^2+6ct+2d)+3(5at^4+4bt^3+3ct^2+2dt+e)=2t^4$

$15at^4+(20a+12b)t^3+(12b+9c)t^2+(6c+6d)t+(2d+e)=2t^4$

$\implies\begin{cases}15a=2\\20a+12b=0\\12b+9c=0\\6c+6d=0\\2d+e=0\end{cases}\implies a=\dfrac2{15},b=-\dfrac29,c=\dfrac8{27},d=-\dfrac8{27},e=\dfrac{16}{81}$

$y''+3y'=t^2e^{-3t}$

$e^{-3t}$ is already accounted for, so assume an ansatz of the form

$y_p=(at^3+bt^2+ct)e^{-3t}$

$\implies {y_p}'=(-3at^3+(3a-3b)t^2+(2b-3c)t+c)e^{-3t}$

$\implies {y_p}''=(9at^3+(9b-18a)t^2+(9c-12b+6a)t+2b-6c)e^{-3t}$

Substitute into the ODE:

$(9at^3+(9b-18a)t^2+(9c-12b+6a)t+2b-6c)e^{-3t}+3(-3at^3+(3a-3b)t^2+(2b-3c)t+c)e^{-3t}=t^2e^{-3t}$

$9at^3+(9b-18a)t^2+(9c-12b+6a)t+2b-6c-9at^3+(9a-9b)t^2+(6b-9c)t+3c=t^2$

$-9at^2+(6a-6b)t+2b-3c=t^2$

$\implies\begin{cases}-9a=1\\6a-6b=0\\2b-3c=0\end{cases}\implies a=-\dfrac19,b=-\dfrac19,c=-\dfrac2{27}$

$y''+3y'=\sin(3t)$

Assume an ansatz solution

$y_p=a\sin(3t)+b\cos(3t)$

$\implies {y_p}'=3a\cos(3t)-3b\sin(3t)$

$\implies {y_p}''=-9a\sin(3t)-9b\cos(3t)$

Substitute into the ODE:

$(-9a\sin(3t)-9b\cos(3t))+3(3a\cos(3t)-3b\sin(3t))=\sin(3t)$

$(-9a-9b)\sin(3t)+(9a-9b)\cos(3t)=\sin(3t)$

$\implies\begin{cases}-9a-9b=1\\9a-9b=0\end{cases}\implies a=-\dfrac1{18},b=-\dfrac1{18}$

So, the general solution of the original ODE is

$y(t)=\dfrac{54t^5 - 90t^4 + 120t^3 - 120t^2 + 80t}{405}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-\dfrac{3t^3+3t^2+2t}{27}e^{-3t}-\dfrac{\sin(3t)+\cos(3t)}{18}$

3 0

3 years ago

Every integer is an irrational number.<br><br> True<br><br> False

Westkost [7]

Answer: true

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

I need help on this problem

Tatiana [17]

Answer:

D

Step-by-step explanation:

8 0

2 years ago

Value of the expression

Luda [366]

$\bf \cfrac{1}{4}\div \cfrac{4}{10}~~+~~\cfrac{5}{7}\div \cfrac{5}{7}\implies \cfrac{1}{4}\cdot \cfrac{10}{4}~~+~~\cfrac{5}{7}\cdot \cfrac{7}{5}\implies \cfrac{1\cdot 10}{4\cdot 4}+\cfrac{5\cdot 7}{7\cdot 5}
\\\\\\
\cfrac{10}{16}+\cfrac{35}{35}\implies \cfrac{10}{16}+1\implies \cfrac{(1\cdot 10)~+~16}{16}\implies \cfrac{26}{16}\implies \cfrac{13}{8}\implies 1\frac{5}{8}$

4 0

3 years ago

Read 2 more answers