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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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If Sunil earns $3672 per month, what is his annual salary?

snow_tiger [21]

Answer:

$44064

Step-by-step explanation:

Sunil earns $3672 every month. To find out his annual salary, we will need to multiply this monthly salary by the number of months in a year; 12.

3672 * 12 = $44064

7 0

3 years ago

Find the length of the curve. R(t) = cos(8t) i + sin(8t) j + 8 ln cos t k, 0 ≤ t ≤ π/4

arsen [322]

we are given

$R(t)=cos(8t)i+sin(8t)j+8ln(cos(t))k$

now, we can find x , y and z components

$x=cos(8t),y=sin(8t),z=8ln(cos(t))$

Arc length calculation:

we can use formula

$L=\int\limits^a_b {\sqrt{(x')^2+(y')^2+(z')^2} } \, dt$

$x'=-8sin(8t),y=8cos(8t),z=-8tan(t)$

now, we can plug these values

$L=\int _0^{\frac{\pi }{4}}\sqrt{(-8sin(8t))^2+(8cos(8t))^2+(-8tan(t))^2} dt$

now, we can simplify it

$L=\int _0^{\frac{\pi }{4}}\sqrt{64+64tan^2(t)} dt$

$L=\int _0^{\frac{\pi }{4}}8\sqrt{1+tan^2(t)} dt$

$L=\int _0^{\frac{\pi }{4}}8\sqrt{sec^2(t)} dt$

$L=\int _0^{\frac{\pi }{4}}8sec(t) dt$

now, we can solve integral

$\int \:8\sec \left(t\right)dt$

$=8\ln \left|\tan \left(t\right)+\sec \left(t\right)\right|$

now, we can plug bounds

and we get

$=8\ln \left(\sqrt{2}+1\right)-0$

so,

$L=8\ln \left(1+\sqrt{2}\right)$ ..............Answer

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2 years ago

Helpppo for an exammmm

Mademuasel [1]

Answer:

AAS

Step-by-step explanation:

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2 years ago

According to the distributive property, 6(a + b) =

kkurt [141]

With the distributive property, you multiply whatever is one the outside of the parenthesis, by each term on the inside.

6a + 6b

Also, whats with the image? Lol.

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3 years ago

Read 2 more answers

The expression 16-3/4 is equivalent to the expression

Natali5045456 [20]

64/4 or 15.25 decimal

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3 years ago