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MA_775_DIABLO [31]
2 years ago
13

Write the number in expanded form 68,020

Mathematics
1 answer:
allochka39001 [22]2 years ago
4 0

Standard Form:

68,020

Expanded Notation Form:

 68,020 =

 60,000  

+ 8,000  

+ 0  

+ 20  

+ 0  

Expanded Factors Form:

   68,020 =

 6 × 10,000  

+ 8 × 1,000  

+ 0 × 100  

+ 2 × 10  

+ 0 × 1  

Expanded Exponential Form:

 68,020 =

       6 × 10^4

+ 8 × 10^3

+ 0 × 10^2

+ 2 × 10^1

+ 0 × 10^0

Word Form:

68,020 =

sixty-eight thousand twenty

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Which of the values shown are potential roots of f(x) = 3x3 – 13x2 – 3x + 45? Select all that apply.
galina1969 [7]

Answer:

All potential roots are 3,3 and -\frac{5}{3}.

Step-by-step explanation:

Potential roots of the polynomial is all possible roots of f(x).

f(x)=3x^3-13x^2-3x+45

Using rational root theorem test. We will find all the possible or potential roots of the polynomial.

p=All the positive/negative factors of 45

q=All the positive/negative factors of 3

p=\pm 1,\pm 3,\pm 5\pm \pm 9,\pm 15\pm 45

q=\pm 1,\pm 3

All possible roots

\frac{p}{q}=\pm 1,\pm 3,\pm 5\pm \pm 9,\pm 15\pm 45,\pm \frac{1}{3},\pm \frac{5}{3}

Now we check each rational root and see which are possible roots for given function.

f(1)= 3\times 1^3-13\times 1^2-3\times 1+45\Rightarrow 32\neq 0

f(-1)= 3\times (-1)^3-13\times (-1)^2-3\times (-1)+45\Rightarrow \neq 32

f(-3)= 3\times (-3)^3-13\times (-3)^2-3\times (-3)+45\Rightarrow \neq -144

f(3)= 3\times (3)^3-13\times (3)^2-3\times (3)+45\Rightarrow =0\\\\ \therefore x=3\text{ Potential roots of function}

Similarly, we will check for all value of p/q and we get

f(-5/3)=0

Thus, All potential roots are 3,3 and -\frac{5}{3}.


5 0
3 years ago
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Y varies directly as x squared. y is 18 when x is 3. Find y if x is 4. Any help? thanks ​
stealth61 [152]

Answer:

Step-by-step explanation:

y ∝ x^2

Introducing the proportionality constant, we have  

y = kx^2

Given : y= 18 when x = 3

substitute the given values in order to get the constant

i.e 18 = k x 3^2

     18 = 9k

      k = 2

therefore the formula connecting  x and y

⇒ y = 2x^2

To find y if x is 4, just substitute x = 4 into the formula connecting x and y

i.e y = 2 x 4^2

       = 2 x 16

       = 32

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3 years ago
Find the surface area of x^2+y^2+z^2=9 that lies above the cone z= sqrt(x^@+y^2)
Mashcka [7]
The cone equation gives

z=\sqrt{x^2+y^2}\implies z^2=x^2+y^2

which means that the intersection of the cone and sphere occurs at

x^2+y^2+(x^2+y^2)=9\implies x^2+y^2=\dfrac92

i.e. along the vertical cylinder of radius \dfrac3{\sqrt2} when z=\dfrac3{\sqrt2}.

We can parameterize the spherical cap in spherical coordinates by

\mathbf r(\theta,\varphi)=\langle3\cos\theta\sin\varphi,3\sin\theta\sin\varphi,3\cos\varphi\right\rangle

where 0\le\theta\le2\pi and 0\le\varphi\le\dfrac\pi4, which follows from the fact that the radius of the sphere is 3 and the height at which the sphere and cone intersect is \dfrac3{\sqrt2}. So the angle between the vertical line through the origin and any line through the origin normal to the sphere along the cone's surface is

\varphi=\cos^{-1}\left(\dfrac{\frac3{\sqrt2}}3\right)=\cos^{-1}\left(\dfrac1{\sqrt2}\right)=\dfrac\pi4

Now the surface area of the cap is given by the surface integral,

\displaystyle\iint_{\text{cap}}\mathrm dS=\int_{\theta=0}^{\theta=2\pi}\int_{\varphi=0}^{\varphi=\pi/4}\|\mathbf r_u\times\mathbf r_v\|\,\mathrm dv\,\mathrm du
=\displaystyle\int_{u=0}^{u=2\pi}\int_{\varphi=0}^{\varphi=\pi/4}9\sin v\,\mathrm dv\,\mathrm du
=-18\pi\cos v\bigg|_{v=0}^{v=\pi/4}
=18\pi\left(1-\dfrac1{\sqrt2}\right)
=9(2-\sqrt2)\pi
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Answer:

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Step-by-step explanation:

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