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2 years ago

7 *the difference of a number and 7

Mathematics

1 answer:

mash [69]2 years ago

5 0

Nothing is different but that both are odds

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Evaluate the triple integral ∭EzdV where E is the solid bounded by the cylinder y2+z2=81 and the planes x=0,y=9x and z=0 in the

dem82 [27]

Answer:

I = 91.125

Step-by-step explanation:

Given that:

$I = \int \int_E \int zdV$ where E is bounded by the cylinder $y^2 + z^2 = 81$ and the planes x = 0 , y = 9x and z = 0 in the first octant.

The initial activity to carry out is to determine the limits of the region

since curve z = 0 and $y^2 + z^2 = 81$

∴ $z^2 = 81 - y^2$

$z = \sqrt{81 - y^2}$

Thus, z lies between 0 to $\sqrt{81 - y^2}$

GIven curve x = 0 and y = 9x

$x =\dfrac{y}{9}$

As such,x lies between 0 to $\dfrac{y}{9}$

Given curve x = 0 , $x =\dfrac{y}{9}$ and z = 0, $y^2 + z^2 = 81$

y = 0 and

$y^2 = 81 \\ \\ y = \sqrt{81} \\ \\ y = 9$

∴ y lies between 0 and 9

Then $I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \int^{\sqrt{81-y^2}}_{z=0} \ zdzdxdy$

$I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \begin {bmatrix} \dfrac{z^2}{2} \end {bmatrix} ^ {\sqrt {{81-y^2}}}_{0} \ dxdy$

$I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \begin {bmatrix} \dfrac{(\sqrt{81 -y^2})^2 }{2}-0 \end {bmatrix} \ dxdy$

$I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \begin {bmatrix} \dfrac{{81 -y^2} }{2} \end {bmatrix} \ dxdy$

$I = \int^9_{y=0} \begin {bmatrix} \dfrac{{81x -xy^2} }{2} \end {bmatrix} ^{\dfrac{y}{9}}_{0} \ dy$

$I = \int^9_{y=0} \begin {bmatrix} \dfrac{{81(\dfrac{y}{9}) -(\dfrac{y}{9})y^2} }{2}-0 \end {bmatrix} \ dy$

$I = \int^9_{y=0} \begin {bmatrix} \dfrac{{81 \ y -y^3} }{18} \end {bmatrix} \ dy$

$I = \dfrac{1}{18} \int^9_{y=0} \begin {bmatrix} {81 \ y -y^3} \end {bmatrix} \ dy$

$I = \dfrac{1}{18} \begin {bmatrix} {81 \ \dfrac{y^2}{2} - \dfrac{y^4}{4}} \end {bmatrix}^9_0$

$I = \dfrac{1}{18} \begin {bmatrix} {40.5 \ (9^2) - \dfrac{9^4}{4}} \end {bmatrix}$

$I = \dfrac{1}{18} \begin {bmatrix} 3280.5 - 1640.25 \end {bmatrix}$

$I = \dfrac{1}{18} \begin {bmatrix} 1640.25 \end {bmatrix}$

I = 91.125

4 0

3 years ago

Helpp!!<br> Simplify : [( 4 1/6 + 2 1/3 ) ÷ 4 1/3 ] - 1 1/2

Nadusha1986 [10]

Answer:

$\boxed{ \bold{ \huge{ \boxed{ \sf{0}}}}}$

Step-by-step explanation:

$\sf{ [ (4 \frac{1}{6} + 2 \frac{1}{3} ) \div 4 \frac{1}{3}] - 1\frac{1}{2} }$

Convert the mixed numbers into improper fraction

$\longrightarrow{ \sf{ [ ( \frac{25}{6} + \frac{7}{3} ) \div \frac{13}{3}] - \frac{3}{2}}}$

Add the fractions : 25 / 6 and 7 / 3

While performing addition or subtraction of unlike fractions, you have to express the given fractions into equivalent fractions of common denominator and add or subtract as we do with like fraction.

To do so, first take the L.C.M of 6 and 3 which results to 6

$\longrightarrow\sf{ [( \frac{25 + 7 \times 2}{6} ) \div \frac{13}{3} ] - \frac{3}{2}}$

$\longrightarrow{ \sf{ [( \frac{25 + 14}{6} ) \div \frac{13}{3} ] - \frac{3}{2} }}$

$\longrightarrow{ \sf{ [ \frac{39}{6} \div \frac{13}{3}] - \frac{3}{2} }}$

Multiply the dividend by the reciprocal of the divisor.

Reciprocal of any number or fraction can be obtained by interchanging the position of numerator and denominator

$\longrightarrow{ \sf{ [ \frac{39}{6} \times \frac{3}{13} ] - \frac{3}{2}}}$

To multiply one fraction by another, multiply the numerators for the numerator and multiply the denominators for its denominator and reduce the fraction obtained after multiplication into lowest term

$\longrightarrow{ \sf{ [ \frac{39 \times 3}{6 \times 13} ] - \frac{3}{2}}}$

$\longrightarrow{ \sf{ [ \frac{117}{78} ] - \frac{3}{2} }}$

$\longrightarrow{ \sf{ \frac{3}{2} - \frac{3}{2} }}$

While performing the addition or subtraction of like fractions , you just have to add or subtract the numerator respectively in which the denominator is retained same

$\longrightarrow{ \sf{ \frac{3 - 3}{2} }}$

Subtract 3 from 3

$\longrightarrow{ \sf{ \frac{0}{2}}}$

Divide 0 by 2

$\longrightarrow{ \sf{0}}$

Hope I helped!

Best regards!

3 0

3 years ago

PLS ANSWER ASAP 30 POINTS!!! CHECK PHOTO! WILL MARK BRAINLIEST TO WHO ANSWERS

Sveta_85 [38]

I'll do Problem 8 to get you started

a = 4 and c = 7 are the two given sides

Use these values in the pythagorean theorem to find side b

$a^2 + b^2 = c^2\\\\4^2 + b^2 = 7^2\\\\16 + b^2 = 49\\\\b^2 = 49 - 16\\\\b^2 = 33\\\\b = \sqrt{33}\\\\$

With respect to reference angle A, we have:

opposite side = a = 4
adjacent side = b = $\sqrt{33}$
hypotenuse = c = 7

Now let's compute the 6 trig ratios for the angle A.

We'll start with the sine ratio which is opposite over hypotenuse.

$\sin(\text{angle}) = \frac{\text{opposite}}{\text{hypotenuse}}\\\\\sin(A) = \frac{a}{c}\\\\\sin(A) = \frac{4}{7}\\\\$

Then cosine which is adjacent over hypotenuse

$\cos(\text{angle}) = \frac{\text{adjacent}}{\text{hypotenuse}}\\\\\cos(A) = \frac{b}{c}\\\\\cos(A) = \frac{\sqrt{33}}{7}\\\\$

Tangent is the ratio of opposite over adjacent

$\tan(\text{angle}) = \frac{\text{opposite}}{\text{adjacent}}\\\\\tan(A) = \frac{a}{b}\\\\\tan(A) = \frac{4}{\sqrt{33}}\\\\\tan(A) = \frac{4\sqrt{33}}{\sqrt{33}*\sqrt{33}}\\\\\tan(A) = \frac{4\sqrt{33}}{(\sqrt{33})^2}\\\\\tan(A) = \frac{4\sqrt{33}}{33}\\\\$

Rationalizing the denominator may be optional, so I would ask your teacher for clarification.

So far we've taken care of 3 trig functions. The remaining 3 are reciprocals of the ones mentioned so far.

cosecant, abbreviated as csc, is the reciprocal of sine
secant, abbreviated as sec, is the reciprocal of cosine
cotangent, abbreviated as cot, is the reciprocal of tangent

So we'll flip the fraction of each like so:

$\csc(\text{angle}) = \frac{\text{hypotenuse}}{\text{opposite}} \ \text{ ... reciprocal of sine}\\\\\csc(A) = \frac{c}{a}\\\\\csc(A) = \frac{7}{4}\\\\\sec(\text{angle}) = \frac{\text{hypotenuse}}{\text{adjacent}} \ \text{ ... reciprocal of cosine}\\\\\sec(A) = \frac{c}{b}\\\\\sec(A) = \frac{7}{\sqrt{33}} = \frac{7\sqrt{33}}{33}\\\\\cot(\text{angle}) = \frac{\text{adjacent}}{\text{opposite}} \ \text{ ... reciprocal of tangent}\\\\\cot(A) = \frac{b}{a}\\\\\cot(A) = \frac{\sqrt{33}}{4}\\\\$

------------------------------------------------------

Summary:

The missing side is $b = \sqrt{33}$

The 6 trig functions have these results

$\sin(A) = \frac{4}{7}\\\\\cos(A) = \frac{\sqrt{33}}{7}\\\\\tan(A) = \frac{4}{\sqrt{33}} = \frac{4\sqrt{33}}{33}\\\\\csc(A) = \frac{7}{4}\\\\\sec(A) = \frac{7}{\sqrt{33}} = \frac{7\sqrt{33}}{33}\\\\\cot(A) = \frac{\sqrt{33}}{4}\\\\$

Rationalizing the denominator may be optional, but I would ask your teacher to be sure.

7 0

1 year ago

If a = √3-√11 and b = 1 /a, then find a² - b²

Serga [27]

If $b=\frac1a$ , then by rationalizing the denominator we can rewrite

$b = \dfrac1{\sqrt3-\sqrt{11}} \times \dfrac{\sqrt3+\sqrt{11}}{\sqrt3+\sqrt{11}} = \dfrac{\sqrt3+\sqrt{11}}{\left(\sqrt3\right)^2-\left(\sqrt{11}\right)^2} = -\dfrac{\sqrt3+\sqrt{11}}8$

Now,

$a^2 - b^2 = (a-b) (a+b)$

and

$a - b = \sqrt3 - \sqrt{11} + \dfrac{\sqrt3 + \sqrt{11}}8 = \dfrac{9\sqrt3 - 7\sqrt{11}}8$

$a + b = \sqrt3 - \sqrt{11} - \dfrac{\sqrt3 + \sqrt{11}}8 = \dfrac{7\sqrt3 - 9\sqrt{11}}8$

$\implies a^2 - b^2 = \dfrac{\left(9\sqrt3 - 7\sqrt{11}\right) \left(7\sqrt3 - 9\sqrt{11}\right)}{64} = \boxed{\dfrac{441 - 65\sqrt{33}}{32}}$

5 0

1 year ago

Simplify into one fraction -2/x+3 - 4/x-5

sergiy2304 [10]

Multiply and divide by the product of the denominators.
$\dfrac{-2}{x+3}+\dfrac{-4}{x-5}=\dfrac{-2(x-5)-4(x+3)}{(x+3)(x-5)}\\\\=\dfrac{-6x-2}{x^{2}-2x-15}$

6 0

3 years ago

Read 2 more answers

7 *the difference of a number and 7​

7 *the difference of a number and 7