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Question Number 64853    Answers: 1   Comments: 0

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Question Number 64851    Answers: 0   Comments: 6

Find all solutions of x real numbers f^(−1) (x) = f(x)

$${Find}\:\:{all}\:\:{solutions}\:\:{of}\:\:{x}\:\:{real}\:\:{numbers}\: \\ $$$$\:\:\:\:\:\:\:\:{f}\:^{−\mathrm{1}} \left({x}\right)\:\:=\:\:{f}\left({x}\right) \\ $$

Question Number 64850    Answers: 0   Comments: 4

let A_λ =∫_0 ^π (dx/(λ +cosx +sinx)) (λ ∈ R) 1) find a explicit form of A_λ 2) find also B_λ =∫_0 ^π (dx/((λ +cosx +sinx)^2 )) 3) calculate ∫_0 ^π (dx/(2+cosx +sinx)) and ∫_0 ^π (dx/((3+cosx +sinx)^2 ))

$${let}\:{A}_{\lambda} \:=\int_{\mathrm{0}} ^{\pi} \:\:\:\frac{{dx}}{\lambda\:\:+{cosx}\:+{sinx}}\:\:\:\:\left(\lambda\:\in\:{R}\right) \\ $$$$\left.\mathrm{1}\right)\:{find}\:{a}\:{explicit}\:{form}\:{of}\:{A}_{\lambda} \\ $$$$\left.\mathrm{2}\right)\:\:{find}\:{also}\:{B}_{\lambda} \:=\int_{\mathrm{0}} ^{\pi} \:\:\frac{{dx}}{\left(\lambda\:+{cosx}\:+{sinx}\right)^{\mathrm{2}} } \\ $$$$\left.\mathrm{3}\right)\:{calculate}\:\int_{\mathrm{0}} ^{\pi} \:\:\:\frac{{dx}}{\mathrm{2}+{cosx}\:+{sinx}}\:\:{and}\:\int_{\mathrm{0}} ^{\pi} \:\:\frac{{dx}}{\left(\mathrm{3}+{cosx}\:+{sinx}\right)^{\mathrm{2}} } \\ $$

Question Number 64829    Answers: 2   Comments: 4

Question Number 64837    Answers: 1   Comments: 0

{ ((x^(√y) + y^(√x) = ((49)/(48)))),(((√x) + (√y) = (7/2))) :} find x and y

$$\begin{cases}{{x}^{\sqrt{{y}}} \:+\:{y}^{\sqrt{{x}}} \:=\:\frac{\mathrm{49}}{\mathrm{48}}}\\{\sqrt{{x}}\:+\:\sqrt{{y}}\:=\:\frac{\mathrm{7}}{\mathrm{2}}}\end{cases} \\ $$$$ \\ $$$${find}\:{x}\:{and}\:{y} \\ $$$$ \\ $$

Question Number 64824    Answers: 0   Comments: 0

Question Number 64823    Answers: 0   Comments: 5

Question Number 64820    Answers: 0   Comments: 1

calculate Σ_(n=1) ^∞ (((−1)^n )/(n^2 (n+1)(n+2)))

$${calculate}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)} \\ $$

Question Number 64819    Answers: 0   Comments: 2

calculate Σ_(n=1) ^∞ (((2n+1)(−1)^n )/(n^3 (n+1)^2 ))

$${calculate}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\:\frac{\left(\mathrm{2}{n}+\mathrm{1}\right)\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{3}} \left({n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$

Question Number 64818    Answers: 2   Comments: 1

find ∫ (dx/(1+cosx +cos(2x)))

$${find}\:\int\:\:\:\frac{{dx}}{\mathrm{1}+{cosx}\:+{cos}\left(\mathrm{2}{x}\right)} \\ $$

Question Number 64812    Answers: 0   Comments: 2

how do i prove by induction? please

$${how}\:{do}\:{i}\:{prove}\:{by}\:{induction}? \\ $$$${please} \\ $$

Question Number 64805    Answers: 0   Comments: 0

∫log(((1+sinhx))/((1−sinhx)))tanhx dx

$$\int{log}\frac{\left(\mathrm{1}+{sinhx}\right)}{\left(\mathrm{1}−{sinhx}\right)}{tanhx}\:{dx} \\ $$

Question Number 64804    Answers: 1   Comments: 0

∫_0 ^(+∞) e^(−x^2 ) dx

$$\overset{+\infty} {\int}_{\mathrm{0}} {e}^{−{x}^{\mathrm{2}} } {dx} \\ $$

Question Number 64802    Answers: 1   Comments: 1

∫(cos^4 x+sin^4 x)/(cos2x+1)dx

$$\int\left({cos}^{\mathrm{4}} {x}+{sin}^{\mathrm{4}} {x}\right)/\left({cos}\mathrm{2}{x}+\mathrm{1}\right){dx} \\ $$

Question Number 64801    Answers: 0   Comments: 0

∫(cos^4 x+sin^4 x)/(cos2x+1)dx

$$\int\left({cos}^{\mathrm{4}} {x}+{sin}^{\mathrm{4}} {x}\right)/\left({cos}\mathrm{2}{x}+\mathrm{1}\right){dx} \\ $$

Question Number 64797    Answers: 0   Comments: 3

for Tawa, an old problem explained (1) x+y+z=α (2) x^2 +y^2 +z^2 =β (3) x^3 +y^3 +z^3 =γ α, β, γ are given (4) x^4 +y^4 +z^4 =p (5) x^5 +y^5 +z^5 =q find p, q we could try to solve the system but it′s hard to exactly solve it and then calculate the sums of the 4^(th) and 5^(th) powers. instead, let′s think of the shape of the solutions of a polynome of 3^(rd) degree. they might look like this: t_1 =a t_2 =b−(√c) t_3 =b+(√c) let′s try it! x=a y=b−(√c) z=b+(√c) ⇒ (1) a+2b=α (2) a^2 +2b^2 +2c=β (3) a^3 +2b^3 +6bc=γ (4) a^4 +2b^4 +12b^2 c+2c^2 −p=0 (5) a^5 +2b^5 +20b^3 c+10bc^2 −q=0 doesn′t look that bad. we can easily solve equation (1) for a and equation (2) for c (1) a=α−2b (2) c=−(a^2 /2)−b^2 +(β/2)=^([using (1)]) −3b^2 +2αb−(α^2 /2)+(β/2) now let′s insert these in (3), (4) and (5) (3) −24b^3 +24αb^2 −3(3α^2 −β)b+α^3 −γ=0 (4) −32αb^3 +32α^2 b^2 −4α(3α^2 −β)b+((3α^4 )/2)−α^2 β+(β^2 /2)−p=0 (5) −20(α^2 +β)b^3 +20α(α^2 +β)b^2 −(5/2)(3α^4 +2α^2 β−β^2 )b+α^5 −q=0 dividing by the constant factors of b^2 (3) b^3 −αb^2 +((3α^2 −β)/8)b−((α^3 −γ)/(24))=0 (4) b^3 −αb^2 +((3α^2 −β)/8)b+((2p−3α^4 +2α^2 β−β^2 )/(64α))=0 (5) b^3 −αb^2 +((3α^2 −β)/8)b+((q−α^5 )/(20(α^2 +β)))=0 only the red factors differ ⇒ they must have the same values! ⇒ −((α^3 −γ)/(24))=((2p−3α^4 +2α^2 β−β^2 )/(64α))=((q−α^5 )/(20(α^2 +β))) ⇒ p=(α^4 /6)−α^2 β+((4αγ)/3)+(β^2 /2) q=(α^5 /6)−((5α^3 β)/6)+((5α^2 γ)/6)+((5βγ)/6) solved without solving the same idea is trivial for 2 equations x+y=α x^2 +y^2 =β x^n +y^n =p with n>2∧n∈N put x=a−(√b) y=a+(√b) ⇒ 2a=α 2a^2 +2b=β ⇒ a=(α/2) b=(β/2)−(α^2 /4) we might as well solve the system and for 4 equations unfortunately we must solve the system w+x+y+z=α w^2 +x^2 +y^2 +z^2 =β w^3 +x^3 +y^3 +z^3 =γ w^4 +x^4 +y^4 +z^4 =δ but it still makes it easier to solve putting w=a−(√b) x=a+(√b) y=c−(√b) z=c+(√d)

$$\mathrm{for}\:\mathrm{Tawa},\:\mathrm{an}\:\mathrm{old}\:\mathrm{problem}\:\mathrm{explained} \\ $$$$\left(\mathrm{1}\right)\:\:{x}+{y}+{z}=\alpha \\ $$$$\left(\mathrm{2}\right)\:\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} =\beta \\ $$$$\left(\mathrm{3}\right)\:\:{x}^{\mathrm{3}} +{y}^{\mathrm{3}} +{z}^{\mathrm{3}} =\gamma \\ $$$$\alpha,\:\beta,\:\gamma\:\mathrm{are}\:\mathrm{given} \\ $$$$ \\ $$$$\left(\mathrm{4}\right)\:\:{x}^{\mathrm{4}} +{y}^{\mathrm{4}} +{z}^{\mathrm{4}} ={p} \\ $$$$\left(\mathrm{5}\right)\:\:{x}^{\mathrm{5}} +{y}^{\mathrm{5}} +{z}^{\mathrm{5}} ={q} \\ $$$$\mathrm{find}\:{p},\:{q} \\ $$$$ \\ $$$$\mathrm{we}\:\mathrm{could}\:\mathrm{try}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{the}\:\mathrm{system}\:\mathrm{but}\:\mathrm{it}'\mathrm{s}\:\mathrm{hard} \\ $$$$\mathrm{to}\:\mathrm{exactly}\:\mathrm{solve}\:\mathrm{it}\:\mathrm{and}\:\mathrm{then}\:\mathrm{calculate}\:\mathrm{the} \\ $$$$\mathrm{sums}\:\mathrm{of}\:\mathrm{the}\:\mathrm{4}^{\mathrm{th}} \:\mathrm{and}\:\mathrm{5}^{\mathrm{th}} \:\mathrm{powers}. \\ $$$$\mathrm{instead},\:\mathrm{let}'\mathrm{s}\:\mathrm{think}\:\mathrm{of}\:\mathrm{the}\:\mathrm{shape}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{solutions}\:\mathrm{of}\:\mathrm{a}\:\mathrm{polynome}\:\mathrm{of}\:\mathrm{3}^{\mathrm{rd}} \:\mathrm{degree}.\:\mathrm{they} \\ $$$$\mathrm{might}\:\mathrm{look}\:\mathrm{like}\:\mathrm{this}: \\ $$$${t}_{\mathrm{1}} ={a}\:\:{t}_{\mathrm{2}} ={b}−\sqrt{{c}}\:\:{t}_{\mathrm{3}} ={b}+\sqrt{{c}} \\ $$$$\mathrm{let}'\mathrm{s}\:\mathrm{try}\:\mathrm{it}! \\ $$$$ \\ $$$${x}={a} \\ $$$${y}={b}−\sqrt{{c}} \\ $$$${z}={b}+\sqrt{{c}} \\ $$$$\Rightarrow \\ $$$$\left(\mathrm{1}\right)\:\:{a}+\mathrm{2}{b}=\alpha \\ $$$$\left(\mathrm{2}\right)\:\:{a}^{\mathrm{2}} +\mathrm{2}{b}^{\mathrm{2}} +\mathrm{2}{c}=\beta \\ $$$$\left(\mathrm{3}\right)\:\:{a}^{\mathrm{3}} +\mathrm{2}{b}^{\mathrm{3}} +\mathrm{6}{bc}=\gamma \\ $$$$ \\ $$$$\left(\mathrm{4}\right)\:\:{a}^{\mathrm{4}} +\mathrm{2}{b}^{\mathrm{4}} +\mathrm{12}{b}^{\mathrm{2}} {c}+\mathrm{2}{c}^{\mathrm{2}} −{p}=\mathrm{0} \\ $$$$\left(\mathrm{5}\right)\:\:{a}^{\mathrm{5}} +\mathrm{2}{b}^{\mathrm{5}} +\mathrm{20}{b}^{\mathrm{3}} {c}+\mathrm{10}{bc}^{\mathrm{2}} −{q}=\mathrm{0} \\ $$$$ \\ $$$$\mathrm{doesn}'\mathrm{t}\:\mathrm{look}\:\mathrm{that}\:\mathrm{bad}.\:\mathrm{we}\:\mathrm{can}\:\mathrm{easily}\:\mathrm{solve} \\ $$$$\mathrm{equation}\:\left(\mathrm{1}\right)\:\mathrm{for}\:{a}\:\mathrm{and}\:\mathrm{equation}\:\left(\mathrm{2}\right)\:\mathrm{for}\:{c} \\ $$$$\left(\mathrm{1}\right)\:\:{a}=\alpha−\mathrm{2}{b} \\ $$$$\left(\mathrm{2}\right)\:\:{c}=−\frac{{a}^{\mathrm{2}} }{\mathrm{2}}−{b}^{\mathrm{2}} +\frac{\beta}{\mathrm{2}}\overset{\left[\mathrm{using}\:\left(\mathrm{1}\right)\right]} {=}−\mathrm{3}{b}^{\mathrm{2}} +\mathrm{2}\alpha{b}−\frac{\alpha^{\mathrm{2}} }{\mathrm{2}}+\frac{\beta}{\mathrm{2}} \\ $$$$\mathrm{now}\:\mathrm{let}'\mathrm{s}\:\mathrm{insert}\:\mathrm{these}\:\mathrm{in}\:\left(\mathrm{3}\right),\:\left(\mathrm{4}\right)\:\mathrm{and}\:\left(\mathrm{5}\right) \\ $$$$\left(\mathrm{3}\right)\:\:−\mathrm{24}{b}^{\mathrm{3}} +\mathrm{24}\alpha{b}^{\mathrm{2}} −\mathrm{3}\left(\mathrm{3}\alpha^{\mathrm{2}} −\beta\right){b}+\alpha^{\mathrm{3}} −\gamma=\mathrm{0} \\ $$$$\left(\mathrm{4}\right)\:\:−\mathrm{32}\alpha{b}^{\mathrm{3}} +\mathrm{32}\alpha^{\mathrm{2}} {b}^{\mathrm{2}} −\mathrm{4}\alpha\left(\mathrm{3}\alpha^{\mathrm{2}} −\beta\right){b}+\frac{\mathrm{3}\alpha^{\mathrm{4}} }{\mathrm{2}}−\alpha^{\mathrm{2}} \beta+\frac{\beta^{\mathrm{2}} }{\mathrm{2}}−{p}=\mathrm{0} \\ $$$$\left(\mathrm{5}\right)\:\:−\mathrm{20}\left(\alpha^{\mathrm{2}} +\beta\right){b}^{\mathrm{3}} +\mathrm{20}\alpha\left(\alpha^{\mathrm{2}} +\beta\right){b}^{\mathrm{2}} −\frac{\mathrm{5}}{\mathrm{2}}\left(\mathrm{3}\alpha^{\mathrm{4}} +\mathrm{2}\alpha^{\mathrm{2}} \beta−\beta^{\mathrm{2}} \right){b}+\alpha^{\mathrm{5}} −{q}=\mathrm{0} \\ $$$$\mathrm{dividing}\:\mathrm{by}\:\mathrm{the}\:\mathrm{constant}\:\mathrm{factors}\:\mathrm{of}\:{b}^{\mathrm{2}} \\ $$$$\left(\mathrm{3}\right)\:\:{b}^{\mathrm{3}} −\alpha{b}^{\mathrm{2}} +\frac{\mathrm{3}\alpha^{\mathrm{2}} −\beta}{\mathrm{8}}{b}−\frac{\alpha^{\mathrm{3}} −\gamma}{\mathrm{24}}=\mathrm{0} \\ $$$$\left(\mathrm{4}\right)\:\:{b}^{\mathrm{3}} −\alpha{b}^{\mathrm{2}} +\frac{\mathrm{3}\alpha^{\mathrm{2}} −\beta}{\mathrm{8}}{b}+\frac{\mathrm{2}{p}−\mathrm{3}\alpha^{\mathrm{4}} +\mathrm{2}\alpha^{\mathrm{2}} \beta−\beta^{\mathrm{2}} }{\mathrm{64}\alpha}=\mathrm{0} \\ $$$$\left(\mathrm{5}\right)\:\:{b}^{\mathrm{3}} −\alpha{b}^{\mathrm{2}} +\frac{\mathrm{3}\alpha^{\mathrm{2}} −\beta}{\mathrm{8}}{b}+\frac{{q}−\alpha^{\mathrm{5}} }{\mathrm{20}\left(\alpha^{\mathrm{2}} +\beta\right)}=\mathrm{0} \\ $$$$\mathrm{only}\:\mathrm{the}\:\mathrm{red}\:\mathrm{factors}\:\mathrm{differ}\:\Rightarrow\:\mathrm{they}\:\mathrm{must}\:\mathrm{have} \\ $$$$\mathrm{the}\:\mathrm{same}\:\mathrm{values}! \\ $$$$\Rightarrow\:−\frac{\alpha^{\mathrm{3}} −\gamma}{\mathrm{24}}=\frac{\mathrm{2}{p}−\mathrm{3}\alpha^{\mathrm{4}} +\mathrm{2}\alpha^{\mathrm{2}} \beta−\beta^{\mathrm{2}} }{\mathrm{64}\alpha}=\frac{{q}−\alpha^{\mathrm{5}} }{\mathrm{20}\left(\alpha^{\mathrm{2}} +\beta\right)} \\ $$$$\Rightarrow \\ $$$${p}=\frac{\alpha^{\mathrm{4}} }{\mathrm{6}}−\alpha^{\mathrm{2}} \beta+\frac{\mathrm{4}\alpha\gamma}{\mathrm{3}}+\frac{\beta^{\mathrm{2}} }{\mathrm{2}} \\ $$$${q}=\frac{\alpha^{\mathrm{5}} }{\mathrm{6}}−\frac{\mathrm{5}\alpha^{\mathrm{3}} \beta}{\mathrm{6}}+\frac{\mathrm{5}\alpha^{\mathrm{2}} \gamma}{\mathrm{6}}+\frac{\mathrm{5}\beta\gamma}{\mathrm{6}} \\ $$$$\mathrm{solved}\:\mathrm{without}\:\mathrm{solving} \\ $$$$ \\ $$$$\mathrm{the}\:\mathrm{same}\:\mathrm{idea}\:\mathrm{is}\:\mathrm{trivial}\:\mathrm{for}\:\mathrm{2}\:\mathrm{equations} \\ $$$${x}+{y}=\alpha \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\beta \\ $$$${x}^{{n}} +{y}^{{n}} ={p}\:\mathrm{with}\:{n}>\mathrm{2}\wedge{n}\in\mathbb{N} \\ $$$$\mathrm{put}\:{x}={a}−\sqrt{{b}}\:\:{y}={a}+\sqrt{{b}} \\ $$$$\Rightarrow \\ $$$$\mathrm{2}{a}=\alpha \\ $$$$\mathrm{2}{a}^{\mathrm{2}} +\mathrm{2}{b}=\beta \\ $$$$\Rightarrow\:{a}=\frac{\alpha}{\mathrm{2}}\:\:{b}=\frac{\beta}{\mathrm{2}}−\frac{\alpha^{\mathrm{2}} }{\mathrm{4}} \\ $$$$\mathrm{we}\:\mathrm{might}\:\mathrm{as}\:\mathrm{well}\:\mathrm{solve}\:\mathrm{the}\:\mathrm{system} \\ $$$$ \\ $$$$\mathrm{and}\:\mathrm{for}\:\mathrm{4}\:\mathrm{equations}\:\mathrm{unfortunately}\:\mathrm{we}\:\boldsymbol{\mathrm{must}} \\ $$$$\mathrm{solve}\:\mathrm{the}\:\mathrm{system} \\ $$$${w}+{x}+{y}+{z}=\alpha \\ $$$${w}^{\mathrm{2}} +{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} =\beta \\ $$$${w}^{\mathrm{3}} +{x}^{\mathrm{3}} +{y}^{\mathrm{3}} +{z}^{\mathrm{3}} =\gamma \\ $$$${w}^{\mathrm{4}} +{x}^{\mathrm{4}} +{y}^{\mathrm{4}} +{z}^{\mathrm{4}} =\delta \\ $$$$\mathrm{but}\:\mathrm{it}\:\mathrm{still}\:\mathrm{makes}\:\mathrm{it}\:\mathrm{easier}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{putting} \\ $$$${w}={a}−\sqrt{{b}}\:\:{x}={a}+\sqrt{{b}}\:\:{y}={c}−\sqrt{{b}}\:\:{z}={c}+\sqrt{{d}} \\ $$

Question Number 64791    Answers: 0   Comments: 1

Question Number 64788    Answers: 0   Comments: 2

two consercutive integers between which a root of the equation lie are: x^2 +3x+2=0

$${two}\:{consercutive}\:{integers}\:{between}\:{which}\:{a}\:{root}\:{of}\:{the}\:{equation}\:{lie}\:{are}: \\ $$$${x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{2}=\mathrm{0} \\ $$

Question Number 64785    Answers: 1   Comments: 0

Question Number 64778    Answers: 1   Comments: 2

Question Number 64775    Answers: 1   Comments: 1

Question Number 64773    Answers: 1   Comments: 0

sin^(−1) (1/(√(5)))+cot^(−1) 3

$$\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{1}/\sqrt{\left.\mathrm{5}\right)}+\mathrm{cot}^{−\mathrm{1}} \mathrm{3}\right. \\ $$

Question Number 64762    Answers: 1   Comments: 1

Given that g(x)=(2/((1+x)(1+3x^2 )) a) express g(x) in partial fractions. b) evaluate ∫_0 ^1 g((x) dx.

$$\mathrm{Given}\:\mathrm{that}\:\mathrm{g}\left(\mathrm{x}\right)=\frac{\mathrm{2}}{\left(\mathrm{1}+\mathrm{x}\right)\left(\mathrm{1}+\mathrm{3x}^{\mathrm{2}} \right.} \\ $$$$\left.\mathrm{a}\right)\:\mathrm{express}\:\mathrm{g}\left(\mathrm{x}\right)\:\mathrm{in}\:\mathrm{partial}\:\mathrm{fractions}. \\ $$$$\left.\mathrm{b}\right)\:\mathrm{evaluate}\:\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{g}\left(\left(\mathrm{x}\right)\:\mathrm{dx}.\right. \\ $$

Question Number 64759    Answers: 0   Comments: 0

∫((ln(ln(x)))/((ln(x))^n )) dx , n≠1

$$\int\frac{{ln}\left({ln}\left({x}\right)\right)}{\left({ln}\left({x}\right)\right)^{{n}} }\:{dx}\:\:\:,\:\:\:{n}\neq\mathrm{1} \\ $$

Question Number 64758    Answers: 2   Comments: 0

Solve: x^4 + 5x^3 − 4x^2 + 7x − 1 = 0

$$\mathrm{Solve}:\:\:\:\:\:\mathrm{x}^{\mathrm{4}} \:+\:\mathrm{5x}^{\mathrm{3}} \:−\:\mathrm{4x}^{\mathrm{2}} \:+\:\mathrm{7x}\:−\:\mathrm{1}\:\:=\:\:\mathrm{0} \\ $$

Question Number 64751    Answers: 0   Comments: 1

Find the force parallel to the slope required to move a body of mass 2kg up a slope inclined at 30° to the horizontal with an acceleration oc 2m/s^2 if the frictional force is 10N.

$${Find}\:{the}\:{force}\:{parallel}\:{to}\:{the}\:{slope} \\ $$$${required}\:{to}\:{move}\:{a}\:{body}\:{of}\:{mass}\:\mathrm{2}{kg} \\ $$$${up}\:{a}\:{slope}\:{inclined}\:{at}\:\mathrm{30}°\:{to}\:{the} \\ $$$${horizontal}\:{with}\:{an}\:{acceleration}\:{oc}\:\mathrm{2}{m}/{s}^{\mathrm{2}} \\ $$$${if}\:{the}\:{frictional}\:{force}\:{is}\:\mathrm{10}{N}. \\ $$

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