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3 years ago

L 6 K 3 6 6 Find the area of ALKJ. 8 Please help

Mathematics

1 answer:

LenKa [72]3 years ago

3 0

Answer:

37.5

Step-by-step explanation:

KM^2=LM*MJ(consequence from the similarity of KLM and JLK )

=>36=LM*8

LM=36/8=4.5

=>S=6/2*(8+4.5)=3*12.5=37.5

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Answer:

y = mx + b

Step-by-step explanation:

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3 years ago

Help worth a lot of pounts

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Answer:

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Step-by-step explanation:

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3 years ago

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Find the angles between -pi and pi that satisfy:<br><br> (−√3) sin(v)+cos(v)=√3

MAXImum [283]

Observe that

$\sin\left(\dfrac\pi6-v\right)=\sin\dfrac\pi6\cos v-\cos\dfrac\pi6\sin v=\dfrac12\cos v-\dfrac{\sqrt3}2\sin v$

In the original equation, divide both sides by $\dfrac12$ :

$-\sqrt3\sin v+\cos v=\sqrt3\implies\dfrac12\cos v-\dfrac{\sqrt3}2\sin v=\dfrac{\sqrt3}2$

$\implies\sin\left(\dfrac\pi6-v\right)=\dfrac{\sqrt3}2$

Next,

$\sin x=\dfrac{\sqrt3}2\implies x=\dfrac\pi3+2n\pi,x=\dfrac{2\pi}3+2n\pi$

where $n$ is any integer. Then

$\sin\left(\dfrac\pi6-v\right)\implies v=-\dfrac\pi6-2n\pi,v=-\dfrac\pi2-2n\pi$

Fix $n=0$ to ensure $-\pi$ , so that

$v=-\dfrac\pi6,v=-\dfrac\pi2$

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4 years ago

A candle in the shape of a circular cone has a base of radius r and a height of h that is the same length as the radius. which e

ladessa [460]

Answer:

$\frac{r(1-\sqrt{2})}{-3}$

Step-by-step explanation:

Volume of cone = $\frac{1}{3} \pi r^{2} h$

Since we are given that a circular cone has a base of radius r and a height of h that is the same length as the radius

= $\frac{1}{3} \pi r^{2} \times r$

= $\frac{1}{3} \pi r^{3}$

Surface area of cone including 1 base = $\pi r^{2} +\pi\times r \times \sqrt{r^2+h^2}$

Since r = h

So, area = $\pi r^{2} +\pi\times r \times \sqrt{r^2+r^2}$

= $\pi r^{2} +\pi\times r \times \sqrt{2r^2}$

= $\pi r^{2} +\pi\times r^2 \times \sqrt{2}$

Ratio of volume of cone to its surface area including base :

$\frac{\frac{1}{3} \pi r^{3}}{\pi r^{2} +\pi\times r^2 \times \sqrt{2}}$

$\frac{\frac{1}{3}r}{1+\sqrt{2}}$

$\frac{r}{3(1+\sqrt{2})}$

Rationalizing

$\frac{r}{3(1+\sqrt{2})} \times \frac{1-\sqrt{2}}{1-\sqrt{2}}$

$\frac{r(1-\sqrt{2})}{-3}$

Hence the ratio the ratio of the volume of the candle to its surface area(including the base) is $\frac{r(1-\sqrt{2})}{-3}$

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3 years ago

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Solve the triangle m<N= 118°, m<P= 33° and m=15. Round to the nearest tenth. Please show work.

kirill [66]

Answer:

M = 29

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Step-by-step explanation:

M = 180 -33-118 = 29

We can use the rule of sines

sin A sin B sin C

------------ = ---------- = ------------

a b c

sin 118 sin 29

------------ = ----------

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Using cross products

15 sin 118 = n sin 29

Divide by sin 29

15 sin 118 / sin 29 = n

27.31838095 = n

sin 33 sin 29

------------ = ----------

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Using cross products

15 sin 33 = p sin 29

Divide by sin 29

15 sin 33 / sin 29 = p

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3 years ago