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2 years ago

Is 6 a solution to the equation? Show your work.

Mathematics

2 answers:

meriva2 years ago

7 0

Answer:

yes I think 6 is a solution to the equation

weqwewe [10]2 years ago

4 0

Can't see the equation sorry... Don't think you can change your question you typed either

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Find the surface area of the part of the circular paraboloid z=x^2+y^2 that lies inside the cylinder X^2+y^2=1

hichkok12 [17]

Answer:

$\mathbf{\dfrac{\pi}{6}[5 \sqrt{5}-1]}$

Step-by-step explanation:

Given that:

The surface area (S.A) $z = x^2 +y^2$

Hence the S.A is of form z = f(x,y)

Then the S.A can be represented with the equation

$A(S) = \iint _D \sqrt{1+ (\dfrac{\partial z}{\partial x})^2+ (\dfrac{\partial z}{\partial y})^2} \ dA$

where :

D = cylinder $x^2 +y^2 =1$

In polar co-ordinates:

D = {(r, θ): 0≤ r ≤ 1, 0 ≤ θ ≤ 2π)

Similarly, $\dfrac{\partial z}{\partial x} = 2x$ and $\dfrac{\partial z}{\partial y} = 2y$

Therefore;

$S.A = \iint_D \sqrt{1+4x^2+4y^2} \ dA$

$= \iint_D \sqrt{1+4(x^2+y^2)} \ dA$

$= \int^{2 \pi}_{0} \int^{1}_{0} \sqrt{1+4r^2} \ r \ dr \d \theta$

$= [\theta]^{2 \pi}_{0} \dfrac{1}{8}\times \dfrac{2}{3}\begin {bmatrix} (1+4r^2)^{\dfrac{3}{2}}\end {bmatrix}^1_0$

$= 2 \pi \times \dfrac{1}{12}[5^{\dfrac{3}{2}} - 1]$

$\mathbf{=\dfrac{\pi}{6}[5 \sqrt{5}-1]}$

6 0

2 years ago

The ratio of men to women working for a company is 5 to 4. If there are 261 employees total, how many men work for the company?

fiasKO [112]

Answer:

There would be 145 men employees.

Step-by-step explanation:

5/9=x/261

9x=1305

x=145

6 0

2 years ago

janelle wishes to finance a car for $33,000. the bank's annual interest rate is 3.5%, and she can choose between durations of fi

Harman [31]

Answer:

For a duration of 5 years, Monthly Payment =$600.42

For a duration of 6 years, Monthly Payment =$508.83

Step-by-step explanation:

$P=\dfrac{ar (1+r)^n}{(1+r)^n-1} \\$

where a= Amount to Finance=$33,000

Annual interest rate = 3.5%=0.035

r=Monthly Interest Rate= 0.035 ÷ 12 = $\frac{7}{2400}$

n=number of months to pay

For a duration of 5 years

n=5X12=60 months

$P=\dfrac{ar (1+r)^n}{(1+r)^n-1} \\\\P=\dfrac{33000 X\frac{7}{2400} (1+\frac{7}{2400} )^{60}}{(1+\frac{7}{2400})^{60}-1} \\=\dfrac{96.25 (1.1909)}{1.1909-1}\\=\dfrac{96.25 (1.1909)}{0.1909}\\=\dfrac{114.62}{0.1909}=\$600.42$

For a duration of 6 years

n=6X12=72 months

$P=\dfrac{ar (1+r)^n}{(1+r)^n-1} \\\\P=\dfrac{33000 X\frac{7}{2400} (1+\frac{7}{2400} )^{72}}{(1+\frac{7}{2400})^{72}-1} \\=\dfrac{96.25 (1.2333)}{1.2333-1}\\=\dfrac{96.25 (1.2333)}{0.2333}\\=\dfrac{118.71}{0.2333}=\$508.83$