All categories

3 years ago

JKL is a straight line.

Mathematics

1 answer:

kirza4 [7]3 years ago

3 0

Answer:

x = 32°

Step-by-step explanation:

∆KLM is an isosceles triangle because it has two equal sides, KL & KM. Therefore, the angles opposite to each of the two equal sides, which are referred to as the base angles are congruent to each other.

m<KML = m<KLM = 58°

m<MKL = 180 - (58 + 58) (Sum of triangle)

m<MKL = 64°

m<JKM = 180 - m<MKL (linear pair theorem)

m<JKM = 180 - 64 (Substitution)

m<JKM = 116°

∆JKM is also an isosceles triangle with two equal sides. Therefore, it's based angles (x & <J) would also be equal to each other.

Thus:

x = ½(180 - m<JKM)

x = ½(180 - 116) (Substitution)

x = 32°

You might be interested in

Determine if the following system of equations has no solutions, infinitely many

mylen [45]

This are similar answers to your questions.

8 0

2 years ago

A. If the side ratio is 5:13, then the area ratio is ?

frez [133]

25:169 can be if is a rectangle ( times by the number itself )

or is is a square times by 2 ( 10/26)

8 0

3 years ago

Evaluate the surface integral. s x2 + y2 + z2 ds s is the part of the cylinder x2 + y2 = 4 that lies between the planes z = 0 an

Leya [2.2K]

Parameterize the lateral face $T_1$ of the cylinder by

$\mathbf r_1(u,v)=(x(u,v),y(u,v),z(u,v))=(2\cos u,2\sin u,v$

where $0\le u\le2\pi$ and $0\le v\le3$ , and parameterize the disks $T_2,T_3$ as

$\mathbf r_2(r,\theta)=(x(r,\theta),y(r,\theta),z(r,\theta))=(r\cos\theta,r\sin\theta,0)$
$\mathbf r_3(r,\theta)=(r\cos\theta,r\sin\theta,3)$

where $0\le r\le2$ and $0\le\theta\le2\pi$ .

The integral along the surface of the cylinder (with outward/positive orientation) is then

$\displaystyle\iint_S(x^2+y^2+z^2)\,\mathrm dS=\left\{\iint_{T_1}+\iint_{T_2}+\iint_{T_3}\right\}(x^2+y^2+z^2)\,\mathrm dS$
$=\displaystyle\int_{u=0}^{u=2\pi}\int_{v=0}^{v=3}((2\cos u)^2+(2\sin u)^2+v^2)\left\|{{\mathbf r}_1}_u\times{{\mathbf r}_2}_v\right\|\,\mathrm dv\,\mathrm du+\int_{r=0}^{r=2}\int_{\theta=0}^{\theta=2\pi}((r\cos\theta)^2+(r\sin\theta)^2+0^2)\left\|{{\mathbf r}_2}_r\times{{\mathbf r}_2}_\theta\right\|\,\mathrm d\theta\,\mathrm dr+\int_{r=0}^{r=2}\int_{\theta=0}^{\theta=2\pi}((r\cos\theta)^2+(r\sin\theta)^2+3^2)\left\|{{\mathbf r}_3}_r\times{{\mathbf r}_3}_\theta\right\|\,\mathrm d\theta\,\mathrm dr$
$=\displaystyle2\int_{u=0}^{u=2\pi}\int_{v=0}^{v=3}(v^2+4)\,\mathrm dv\,\mathrm du+\int_{r=0}^{r=2}\int_{\theta=0}^{\theta=2\pi}r^3\,\mathrm d\theta\,\mathrm dr+\int_{r=0}^{r=2}\int_{\theta=0}^{\theta=2\pi}r(r^2+9)\,\mathrm d\theta\,\mathrm dr$
$=\displaystyle4\pi\int_{v=0}^{v=3}(v^2+4)\,\mathrm dv+2\pi\int_{r=0}^{r=2}r^3\,\mathrm dr+2\pi\int_{r=0}^{r=2}r(r^2+9)\,\mathrm dr$
$=136\pi$

7 0

3 years ago

HELP ME PLEASE, I NEED THIS BY TONIGHT

saul85 [17]

The first option and the third option are correct.

(Although, it's probably way too late now...)

7 0

3 years ago

Find the perimeter of the triangle. Please help! That one is 2x+5 if you cannot see.

Paul [167]

2x3x5=30
3 is from the amount of sides in the triangle

6 0

3 years ago