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Question Number 192536 Answers: 1 Comments: 0

R=((3sin 5°+4cos 5°−5cos 58°+35(√2) cos 13°)/(cos 5°))=?

$$\:\:\mathrm{R}=\frac{\mathrm{3sin}\:\mathrm{5}°+\mathrm{4cos}\:\mathrm{5}°−\mathrm{5cos}\:\mathrm{58}°+\mathrm{35}\sqrt{\mathrm{2}}\:\mathrm{cos}\:\mathrm{13}°}{\mathrm{cos}\:\mathrm{5}°}=? \\ $$

Question Number 192545 Answers: 3 Comments: 0

Question Number 192514 Answers: 1 Comments: 2

Question Number 192508 Answers: 1 Comments: 1

Question Number 192503 Answers: 0 Comments: 0

Question Number 192497 Answers: 1 Comments: 0

tan 66°+tan 12°=(√3) +tan x x=?

$$\:\:\mathrm{tan}\:\mathrm{66}°+\mathrm{tan}\:\mathrm{12}°=\sqrt{\mathrm{3}}\:+\mathrm{tan}\:\mathrm{x} \\ $$$$\:\:\mathrm{x}=? \\ $$

Question Number 192492 Answers: 0 Comments: 0

Question Number 192481 Answers: 1 Comments: 0

Question Number 192477 Answers: 2 Comments: 0

Find: ((((25)/(42)) − (5/(16)) + ((10)/9) − (2/3))/((3/8) + (4/5) − (5/7) − (4/3))) = ?

$$\mathrm{Find}:\:\:\:\:\:\frac{\frac{\mathrm{25}}{\mathrm{42}}\:−\:\frac{\mathrm{5}}{\mathrm{16}}\:+\:\frac{\mathrm{10}}{\mathrm{9}}\:−\:\frac{\mathrm{2}}{\mathrm{3}}}{\frac{\mathrm{3}}{\mathrm{8}}\:+\:\frac{\mathrm{4}}{\mathrm{5}}\:−\:\frac{\mathrm{5}}{\mathrm{7}}\:−\:\frac{\mathrm{4}}{\mathrm{3}}}\:=\:? \\ $$

Question Number 192472 Answers: 1 Comments: 0

A heat transfer of 9.0×10^5 J is required to convert a block of ice at 20°C to water at 15°C, what was the mass of the block of ice ?

$$\mathrm{A}\:\mathrm{heat}\:\mathrm{transfer}\:\mathrm{of}\:\mathrm{9}.\mathrm{0}×\mathrm{10}^{\mathrm{5}} \mathrm{J}\:\mathrm{is}\:\mathrm{required} \\ $$$$\mathrm{to}\:\mathrm{convert}\:\mathrm{a}\:\mathrm{block}\:\mathrm{of}\:\mathrm{ice}\:\mathrm{at}\:\mathrm{20}°\mathrm{C}\:\mathrm{to} \\ $$$$\mathrm{water}\:\mathrm{at}\:\mathrm{15}°\mathrm{C},\:\mathrm{what}\:\mathrm{was}\:\mathrm{the}\:\mathrm{mass}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{block}\:\mathrm{of}\:\mathrm{ice}\:? \\ $$

Question Number 192470 Answers: 1 Comments: 0

Question Number 192469 Answers: 2 Comments: 0

Question Number 192466 Answers: 2 Comments: 1

Question Number 192464 Answers: 0 Comments: 1

Question Number 192463 Answers: 3 Comments: 0

Question Number 192458 Answers: 1 Comments: 0

Question Number 192454 Answers: 0 Comments: 0

Evaluate the following improper intergrals ∫_(π/4) ^(π/2) sec xdx if it convergent

$${Evaluate}\:{the}\:{following}\:{improper}\:{intergrals} \\ $$$$\int_{\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{sec}\:{xdx}\:\:{if}\:{it}\:{convergent} \\ $$

Question Number 192453 Answers: 2 Comments: 0

Find lim_(n→∞) ((1/n^2 )+(2/n^2 )+(3/n^2 )+...(n/n^2 ))

$${Find}\:\:\:\:\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\left(\frac{\mathrm{1}}{{n}^{\mathrm{2}} }+\frac{\mathrm{2}}{{n}^{\mathrm{2}} }+\frac{\mathrm{3}}{{n}^{\mathrm{2}} }+...\frac{{n}}{{n}^{\mathrm{2}} }\right) \\ $$

Question Number 192452 Answers: 2 Comments: 0

Evaluate ∫_0 ^1 2^(x ) dx

$${Evaluate}\:\:\:\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{2}^{{x}\:} {dx}\:\:\: \\ $$$$ \\ $$

Question Number 192451 Answers: 0 Comments: 0

Question Number 192450 Answers: 1 Comments: 0

The first term of an arithmetic series is 7 and the last is 70. and the sum is 385. find the number of terms in the series and its common difference.

$$\mathrm{The}\:\mathrm{first}\:\mathrm{term}\:\mathrm{of}\:\mathrm{an}\:\mathrm{arithmetic}\:\mathrm{series} \\ $$$$\mathrm{is}\:\mathrm{7}\:\mathrm{and}\:\mathrm{the}\:\mathrm{last}\:\mathrm{is}\:\mathrm{70}.\:\mathrm{and}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{is} \\ $$$$\mathrm{385}.\:\mathrm{find}\:\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{terms}\:\mathrm{in}\:\mathrm{the} \\ $$$$\mathrm{series}\:\mathrm{and}\:\mathrm{its}\:\mathrm{common}\:\mathrm{difference}. \\ $$

Question Number 192444 Answers: 1 Comments: 4

The sum of three numbers in arith- metic progression is 18 and sum of square is 206. find the numbers

$$\mathrm{The}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{three}\:\mathrm{numbers}\:\mathrm{in}\:\mathrm{arith}- \\ $$$$\mathrm{metic}\:\mathrm{progression}\:\mathrm{is}\:\mathrm{18}\:\mathrm{and}\:\mathrm{sum}\:\mathrm{of} \\ $$$$\mathrm{square}\:\mathrm{is}\:\mathrm{206}.\:\mathrm{find}\:\mathrm{the}\:\mathrm{numbers} \\ $$

Question Number 192443 Answers: 1 Comments: 0

Find the sum of the first 16th term of the series 3(1/2) + 4(3/4) + 6 + 7(1/4) ...

$$\mathrm{Find}\:\mathrm{the}\:\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{first}\:\mathrm{16th}\:\mathrm{term} \\ $$$$\:\mathrm{of}\:\mathrm{the}\:\mathrm{series}\:\mathrm{3}\frac{\mathrm{1}}{\mathrm{2}}\:+\:\mathrm{4}\frac{\mathrm{3}}{\mathrm{4}}\:+\:\mathrm{6}\:+\:\mathrm{7}\frac{\mathrm{1}}{\mathrm{4}}\:... \\ $$

Question Number 192440 Answers: 1 Comments: 0

Given { ((A=(((p^2 +q^2 +r^2 )^2 )/((pq)^2 +(pr)^2 +(qr)^2 )))),((B=((q^2 −pr)/(p^2 +q^2 +r^2 )) )) :} If p+q+r=0 then A^2 −4B=?

$$\:\:\:\:\mathrm{Given}\:\begin{cases}{\mathrm{A}=\frac{\left(\mathrm{p}^{\mathrm{2}} +\mathrm{q}^{\mathrm{2}} +\mathrm{r}^{\mathrm{2}} \right)^{\mathrm{2}} }{\left(\mathrm{pq}\right)^{\mathrm{2}} +\left(\mathrm{pr}\right)^{\mathrm{2}} +\left(\mathrm{qr}\right)^{\mathrm{2}} }}\\{\mathrm{B}=\frac{\mathrm{q}^{\mathrm{2}} −\mathrm{pr}}{\mathrm{p}^{\mathrm{2}} +\mathrm{q}^{\mathrm{2}} +\mathrm{r}^{\mathrm{2}} }\:}\end{cases}\:\:\:\:\:\: \\ $$$$\:\mathrm{If}\:\mathrm{p}+\mathrm{q}+\mathrm{r}=\mathrm{0}\:\mathrm{then}\:\mathrm{A}^{\mathrm{2}} −\mathrm{4B}=? \\ $$$$ \\ $$

Question Number 192437 Answers: 1 Comments: 0

(1/a) + (1/b) + (1/c) = (1/(a + b + c)) . Prove that (1/a^5 ) + (1/b^5 ) + (1/c^5 ) = (1/(a^5 + b^5 + c^5 )) = (1/((a + b + c)^5 ))

$$\frac{\mathrm{1}}{{a}}\:+\:\frac{\mathrm{1}}{{b}}\:+\:\frac{\mathrm{1}}{{c}}\:=\:\frac{\mathrm{1}}{{a}\:+\:{b}\:+\:{c}}\:.\:\mathrm{Prove}\:\mathrm{that} \\ $$$$\frac{\mathrm{1}}{{a}^{\mathrm{5}} }\:+\:\frac{\mathrm{1}}{{b}^{\mathrm{5}} }\:+\:\frac{\mathrm{1}}{{c}^{\mathrm{5}} }\:=\:\frac{\mathrm{1}}{{a}^{\mathrm{5}} \:+\:{b}^{\mathrm{5}} \:+\:{c}^{\mathrm{5}} }\:=\:\frac{\mathrm{1}}{\left({a}\:+\:{b}\:+\:{c}\right)^{\mathrm{5}} } \\ $$

Question Number 192433 Answers: 1 Comments: 0

lim_(x→0) ((1−cos (ln (3x+1)))/(3x^6 −tan (3x^2 )))=?

$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−\mathrm{cos}\:\left(\mathrm{ln}\:\left(\mathrm{3x}+\mathrm{1}\right)\right)}{\mathrm{3x}^{\mathrm{6}} −\mathrm{tan}\:\left(\mathrm{3x}^{\mathrm{2}} \right)}=? \\ $$

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