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3 years ago

Is this correct? helppp

Mathematics

2 answers:

Marta_Voda [28]3 years ago

8 0

Answer:

Yes its correct the unhighlighted is complement

Scilla [17]3 years ago

3 0

Answer:

maybe?

Step-by-step explanation:

i think its correct let me know if is not correct :)))

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Shirley purchased a plot of land for $19,500. The land appreciates about 3.9% each year. What is the value of the land after fiv

Leona [35]

Answer:

3802,5

Step-by-step explanation:

19500*3,9=760,5

760,5*5=3802,5

7 0

3 years ago

Solve the inequality 3/7(35x-14)less then or equal to 21x/2+3

Bingel [31]

15x-6 $\leq$ 21x/2+3. Multiply everything by 2: 30x-12=21x+6. Combine like terms: 9x=18. Get the unknown by itself: 9x/9=X. 18/9=2. X is less than or equal to 2 :)

5 0

4 years ago

Xsquared+11x+28+xsquared+13x+40

UNO [17]

X² + 11x + 28 + x² + 13x + 40

combine like terms

x² + x² = 2x²
11x + 13x = 24x
28 + 40 = 68

2x² + 24x + 68 is your answer

if you want it simplified:

2(x² + 12x + 34)

hope this helps

5 0

4 years ago

CAN SOMEONE HELP ME IN THIS INTEGRAL QUESTION PLS

finlep [7]

Due to the symmetry of the paraboloid about the <em>z</em>-axis, you can treat this is a surface of revolution. Consider the curve $y=x^2$ , with $1\le x\le2$ , and revolve it about the <em>y</em>-axis. The area of the resulting surface is then

$\displaystyle2\pi\int_1^2x\sqrt{1+(y')^2}\,\mathrm dx=2\pi\int_1^2x\sqrt{1+4x^2}\,\mathrm dx=\frac{(17^{3/2}-5^{3/2})\pi}6$

But perhaps you'd like the surface integral treatment. Parameterize the surface by

$\vec s(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec\jmath+u^2\,\vec k$

with $1\le u\le2$ and $0\le v\le2\pi$ , where the third component follows from

$z=x^2+y^2=(u\cos v)^2+(u\sin v)^2=u^2$

Take the normal vector to the surface to be

$\dfrac{\partial\vec s}{\partial u}\times\dfrac{\partial\vec s}{\partial u}=-2u^2\cos v\,\vec\imath-2u^2\sin v\,\vec\jmath+u\,\vec k$

The precise order of the partial derivatives doesn't matter, because we're ultimately interested in the magnitude of the cross product:

$\left\|\dfrac{\partial\vec s}{\partial u}\times\dfrac{\partial\vec s}{\partial v}\right\|=u\sqrt{1+4u^2}$

Then the area of the surface is

$\displaystyle\int_0^{2\pi}\int_1^2\left\|\dfrac{\partial\vec s}{\partial u}\times\dfrac{\partial\vec s}{\partial v}\right\|\,\mathrm du\,\mathrm dv=\int_0^{2\pi}\int_1^2u\sqrt{1+4u^2}\,\mathrm du\,\mathrm dv$

which reduces to the integral used in the surface-of-revolution setup.

7 0

3 years ago

Round 1888.9275 to three significant figures

lorasvet [3.4K]

Answer:

1890

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Is this correct? helppp​

Is this correct? helppp