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2 years ago

Define the axis of symmetry, vertex, focus and directrix for

Mathematics

1 answer:

Papessa [141]2 years ago

8 0

Answer:

see attached

Step-by-step explanation:

The equation is in the form ...

4p(y -k) = (x -h)^2 . . . . . (h, k) is the vertex; p is the focus-vertex distance

Comparing this to your equation, we see ...

p = 4, (h, k) = (3, 4)

p > 0, so the parabola opens upward. The vertex is on the axis of symmetry. That axis has the equation x=x-coordinate of vertex. This tells you ...

vertex: (3, 4)

axis of symmetry: x = 3

focus: (3, 8) . . . . . 4 units up from vertex

directrix: y = 0 . . . horizontal line 4 units down from vertex

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Which two properties of equality could Zoe use to finish solving for x

7nadin3 [17]

3x-4= - 10
add both side 4 so its a addition property
3x=- 6
now divide both side 3 so it is a <span>division property.
so x=-2</span>

4 0

3 years ago

-1/2m=-9 what does m=

lilavasa [31]

Answer:

m = 18

Step-by-step explanation:

Step 1: Write out equation

-1/2m = -9

Step 2: Multiply both sides by -2

m = 18

5 0

3 years ago

Three co-workers recorded the amount of water in a fish tank being replaced at regular intervals. A) Maggie measured ½ gallon of

IRINA_888 [86]

Answer:

A and C

Step-by-step explanation:

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4 0

3 years ago

NEED HELP ASAP. A garden in the shape of an equilateral triangle measures 6 feet on each side. An irrigation hose runs from one

Step2247 [10]

Answer:

Find the length of the hose = 5.2 feet

Step-by-step explanation:

The garden is divided into two similar triangle.

The hose divided one side of the triangle to two equal parts.

Side a = 6ft

Side b = 3ft

Side c = ? (length of hose)

a² = b² + c²

6² = 3² + c²

36 = 9 + c²

27 = c²

√27 = c

5.196 = c

8 0

3 years ago

can someone show me how to find the general solution of the differential equations? really need to know how to do it for the upc

mariarad [96]

The first equation is linear:

$x\dfrac{\mathrm dy}{\mathrm dx}-y=x^2\sin x$

Divide through by $x^2$ to get

$\dfrac1x\dfrac{\mathrm dy}{\mathrm dx}-\dfrac1{x^2}y=\sin x$

and notice that the left hand side can be consolidated as a derivative of a product. After doing so, you can integrate both sides and solve for $y$ .

$\dfrac{\mathrm d}{\mathrm dx}\left[\dfrac1xy\right]=\sin x$
$\implies\dfrac1xy=\displaystyle\int\sin x\,\mathrm dx=-\cos x+C$
$\implies y=-x\cos x+Cx$

- - -

The second equation is also linear:

$x^2y'+x(x+2)y=e^x$

Multiply both sides by $e^x$ to get

$x^2e^xy'+x(x+2)e^xy=e^{2x}$

and recall that $(x^2e^x)'=2xe^x+x^2e^x=x(x+2)e^x$ , so we can write

$(x^2e^xy)'=e^{2x}$
$\implies x^2e^xy=\displaystyle\int e^{2x}\,\mathrm dx=\frac12e^{2x}+C$
$\implies y=\dfrac{e^x}{2x^2}+\dfrac C{x^2e^x}$

- - -

Yet another linear ODE:

$\cos x\dfrac{\mathrm dy}{\mathrm dx}+\sin x\,y=1$

Divide through by $\cos^2x$ , giving

$\dfrac1{\cos x}\dfrac{\mathrm dy}{\mathrm dx}+\dfrac{\sin x}{\cos^2x}y=\dfrac1{\cos^2x}$
$\sec x\dfrac{\mathrm dy}{\mathrm dx}+\sec x\tan x\,y=\sec^2x$
$\dfrac{\mathrm d}{\mathrm dx}[\sec x\,y]=\sec^2x$
$\implies\sec x\,y=\displaystyle\int\sec^2x\,\mathrm dx=\tan x+C$
$\implies y=\cos x\tan x+C\cos x$
$y=\sin x+C\cos x$

- - -

In case the steps where we multiply or divide through by a certain factor weren't clear enough, those steps follow from the procedure for finding an integrating factor. We start with the linear equation

$a(x)y'(x)+b(x)y(x)=c(x)$

then rewrite it as

$y'(x)=\dfrac{b(x)}{a(x)}y(x)=\dfrac{c(x)}{a(x)}\iff y'(x)+P(x)y(x)=Q(x)$

The integrating factor is a function $\mu(x)$ such that

$\mu(x)y'(x)+\mu(x)P(x)y(x)=(\mu(x)y(x))'$

which requires that

$\mu(x)P(x)=\mu'(x)$

This is a separable ODE, so solving for $\mu$ we have

$\mu(x)P(x)=\dfrac{\mathrm d\mu(x)}{\mathrm dx}\iff\dfrac{\mathrm d\mu(x)}{\mu(x)}=P(x)\,\mathrm dx$
$\implies\ln|\mu(x)|=\displaystyle\int P(x)\,\mathrm dx$
$\implies\mu(x)=\exp\left(\displaystyle\int P(x)\,\mathrm dx\right)$

and so on.

6 0

3 years ago