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3 years ago

9

PTINO Can you unscramble this... It's the answer to this riddle; Decimals always win in arguments with fractions. What do decima

ls have that fractions don't?

2 answers:

mr_godi [17]3 years ago

6 0

Could it be a point?

Simora [160]3 years ago

4 0

Decimal Point 1.25 fraction 1 1/4

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What is 4log1/2^w(2log1/2^u-3log1/2^v written as a single logarithm?

IrinaK [193]

Given:

4log1/2^w (2log1/2^u-3log1/2^v)

Req'd:

Single logarithm = ?

Sol'n:

First remove the parenthesis,

4 log 1/2 (w) + 2 log 1/2 (u) - 3 log 1/2 (v)

Simplify each term,

Simplify the 4 log 1/2 (w) by moving the constant 4 inside the logarithm;
Simplify the 2 log 1/2 (u) by moving the constant 2 inside the logarithm;
Simplify the -3 log 1/2 (v) by moving the constant -3 inside the logarithm:

log 1/2 (w^4) + 2 log 1/2 (u) - 3 log 1/2 (v)
log 1/2 (w^4) + log 1/2 (u^2) - log 1/2 (v^3)

We have to use the product property of logarithms which is log of b (x) + log of b (y) = log of b (xy):

Thus,

Log of 1/2 (w^4 u^2) - log of 1/2 (v^3)

then use the quotient property of logarithms which is log of b (x) - log of b (y) = log of b (x/y)

Therefore,

log of 1/2 (w^4 u^2 / v^3)

and for the final step and answer, reorder or rearrange w^4 and u^2:

log of 1/2 (u^2 w^4 / v^3)

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3 years ago

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13.) The leading scorer in your high school's hockey league finished the season with an average of 4 goals per game for 15 games

Novosadov [1.4K]

Answer:

5 goals

Step-by-step explanation:

4 goals per game for 15 games means a total of 60 goals in that season. You have 4 goals per game for 14 games meaning you have a total of 56 goals. To <u>overtake</u> the leading team you must score a minimum of 5 goals to become the leading scorer. (Double-check to make sure I am correct)

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2 years ago

Please help me as soon as possible

garik1379 [7]

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3 years ago

A rectangle has a diagonal of 277 centimeters and a length of 252cm. How long is the width?

Crank

Answer:

width = 115 cm

Step-by-step explanation:

width² = 277² - 252²

width² = 76729 - 63504 = 13225

width = 115 cm

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2 years ago

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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