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3 years ago

If the vertex of a parabola is (-8,-6), what is the axis of symmetry

Mathematics

1 answer:

umka21 [38]3 years ago

3 0

Answer:

(8,6)

Step-by-step explanation:

I used the formula -b/2a

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1. One skater paid $108.00 for

aliya0001 [1]

Answer:

each jewel was 3 $

Step-by-step explanation:

divide 108.00/36= 3

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2 years ago

Read 2 more answers

Which values for x, y, and z satisfy the equation 7xy/z=28

nirvana33 [79]

Answer:

try x= 4z/y

Step-by-step explanation:

7 0

3 years ago

The mean of 67004, 67004, 67005, 67005, 67005, 67005, 67006, 67006

romanna [79]

Answer:

the answer is 59467

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2 years ago

Cassie has test grades of 62, 85 and 89 on the first three tests in herpre-algebra class. What are the possible scores she can m

Vsevolod [243]

Explanation:

The test grade for 1st three tests = 62, 85, 89

Let the fourth test score = x

The result we hoping to get is grade B.

Let's find the average of the four test score

$\text{Average = }\frac{Su\text{ m of all test score}}{nu\text{mber of test score}}$ $\begin{gathered} s\text{ um of test scores = 62 + 85 + 89 + x} \\ nu\text{mber of test score = 4} \\ \frac{\text{62 + 85 + 89 + x}}{4}=\text{ Average} \end{gathered}$ $\begin{gathered} A\text{ grade B represents atleast 80} \\ \text{Atleast represents }\ge \\ \frac{\text{62 + 85 + 89 + x}}{4}\text{ }\ge\text{ 80} \\ \frac{\text{236 + x}}{4}\text{ }\ge\text{ 80} \end{gathered}$ $\begin{gathered} \text{236 + x }\ge\text{ }80(4) \\ \text{236 + x }\ge\text{ }320 \\ \text{subtract }236\text{ from both sides:} \\ 236\text{ - 236 + x }\ge\text{ }320\text{ - 23}6 \\ x\text{ }\ge\text{ 84} \end{gathered}$

This means Cassie needs to score 84 or more to get a grade of B

In ineq

3 0

1 year 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