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3 years ago

How to write a cubic function that passes through the points?

Mathematics

1 answer:

Bad White [126]3 years ago

8 0

Please be more clear with what you want for your answer.

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Lena [83]

Answer:

Step-by-step explanation:

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2 years ago

Write the slope intercept form of the equation of each line

Thepotemich [5.8K]

The slope intercept it (0,4)

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2 years ago

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What is the measure of arc CD?

CaHeK987 [17]

Option C:

The measure of arc CD is 40°.

Solution:

Given data:

m∠X = 11° and m(arc AB) = 18°

To find the measure of arc CD:

We know that,

Angle formed by two intersecting secants outside the circle is equal to half of the difference between the intercepted arcs.

$$m \angle X =\frac{1}{2}({arc \ CD}-{arc} \ AB)$

$$ 11^\circ =\frac{1}{2}({arc} \ CD-{18^\circ})$

Multiply by 2 on both sides.

22° = arc CD - 18°

Add 18° from both sides.

40° = arc CD

Switch the sides.

arc CD = 40°

Hence the measure of arc CD is 40°.

Option B is the correct answer.

7 0

2 years ago

(b) dy/dx = (x - y+ 1)^2

Elanso [62]

Substitute $v(x)=x-y(x)+1$ , so that

$\dfrac{\mathrm dv}{\mathrm dx}=1-\dfrac{\mathrm dy}{\mathrm dx}$

Then the resulting ODE in $v(x)$ is separable, with

$1-\dfrac{\mathrm dv}{\mathrm dx}=v^2\implies\dfrac{\mathrm dv}{1-v^2}=\mathrm dx$

On the left, we can split into partial fractions:

$\dfrac12\left(\dfrac1{1-v}+\dfrac1{1+v}\right)\mathrm dv=\mathrm dx$

Integrating both sides gives

$\dfrac{\ln|1-v|+\ln|1+v|}2=x+C$

$\dfrac12\ln|1-v^2|=x+C$

$1-v^2=e^{2x+C}$

$v=\pm\sqrt{1-Ce^{2x}}$

Now solve for $y(x)$ :

$x-y+1=\pm\sqrt{1-Ce^{2x}}$

$\boxed{y=x+1\pm\sqrt{1-Ce^{2x}}}$

3 0

3 years ago

Convert the following percent to a fraction and decimal. (THIS IS NOT MULTIPLE CHOICE!!)

Mkey [24]

Answer:

90%=9/10=0.9

0.9=90%=9/10

900%=900/1000=0.9

9/10%=0.9=90%

Step-by-step explanation:

6 0

2 years ago

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