Perimeter squares are:

\(9\times4=36\left(cm^2\right)\)

           Answer: \(36\left(cm^2\right)\)

\(78129\)  \(and\)   \(7\)   

\(\widehat{xOy}=50^o;\widehat{xOz}=100^o;\widehat{xOt}=130^o\)

Solution:

\(\widehat{BOI}=\dfrac{1}{4}=\widehat{AOB}=\dfrac{1}{4}\times60^o=15^o.\)

Because OI rays are between 2 OA rays, the OB should \(\widehat{AOI}+\widehat{BOI}=\widehat{AOB}\)

Inferred \(\widehat{AOI}+15^o=60^o\) or  \(\widehat{AOI}=45^o\)To

Two corners \(xOy\) and \(yOy'\)  compensated \(\widehat{xOy}+\widehat{yOy'}=180^o\).

Imagine: \(\widehat{yOy'}=180^o-\widehat{xOy}=180^o-120^o=60^o\).

So \(\widehat{yOy'}=60^o\)

The area of the shaded region is:

 \(48\times24=1152\left(m^2\right)\)

               Answer: \(1152m^2\)

mk ko nhanh như bn khác nhưng hãy k mk! *_<

Change 2m4dm = 24 dm

Fold the cloth into 4 equal parts and cut out a portion 

yes., you can make friend with me :vv

I'm also an A.R.M.Y .<3<3 bts

a, \(A=\left\{x\in N;x=7.q +3;q\in N;x\le150\right\}\) Inferred

\(A=\left\{3;10;17;24;.........;143;150\right\}\)

b, easy to see numbers \(3;10;17;24;....;143;150\)is the sequence of numbers plus  \(u_1=3;d=7.\)

The term number of the sequence \(\left(1\right)\)was:

\(n=\dfrac{u_n-u_1}{d}+1=\dfrac{150-3}{7}+1=22\) (term)

Sum the numbers of the sequence \(\left(1\right)\)was:

\(S_n=\dfrac{\left(u_n+u_1\right)\times n}{2}=\dfrac{\left(3+150\right)\times22}{2}=1683.\)

ko viết bằng T.A à??????

we put \(\left(a,b\right)=d\)  inferred \(a=dm;b=d.n\) so in that \(\left(m,n\right)=1.\)

suppose \(a\le b\) then \(m\le n.\)

we have: \(ab=dm.dn=d^2m.n.\)

              \(\left[a,b\right]=\dfrac{ab}{\left(a;b\right)}=\dfrac{d^2m.n}{d}=d.m.n\)

According to the post: \(\left[a,b\right]=210\) so \(d.m.n=210.\)

In that, \(d=\dfrac{ab}{\left[a,b\right]}=\dfrac{2940}{210}=14\) . So \(mn=\dfrac{210}{10}=15\)

We have following list:

\(a,\) \(37^5\div37^3=37^{5-3}=37^2=1369\)

\(b,\) \(\left(x+3\right)^7\div\left(x+3\right)^5=\left(x+3\right)^{7-5}=\left(x+3\right)^2\)

\(c,\) \(a^{10}\div a^{10}=a^{10-10}=a^0=1\)

a,\(we\) \(have:\)\(516=516^n\div516^{13}=516^{n-13}\)

  \(but\)\(516=516^1.\) \(So\) \(that:\)  \(516^1=516^{n-13}.\) \(Inferred\): \(n-13=1.\)

\(so :n=14\)

b,\(3427^2=3427^6\div3427^n=3427^{6-n}.\) \(So\) \(that\)\(6-n=2\) \(or\) \(n=6-2.\) \(so : n=4\)

c,\(we\) \(know:\) \(64=8^2\) \(and\) \(8=8^1\) \(so\) \(8^2=64=8^n\div8^1=8^{n-1}.\)

\(so\) \(that:\) \(n-1=2.\) \(So\) \(n=3\)

you ask that quesion again?

I just guess is that pictute has 13 polygon

a,\(219-7\left(x+1\right)=100\)

\(Inferred: 7\left(x+1\right)=219-100 or7\left(x+1\right)=119. \)

\(So:x+1=119\div7=17.So:x=6\)

b,\(\left(3x-6\right)\times3=3^4\)

\(Inferred : 3x-6=3^4\times3,or:3x-6=27.\)

\(So:3x=27+6\) \(or\) \(3x=33.\)

\(so\) \(end\) \(x=11\)

\(a,5^3\times5^7=5^{3+7}=5^{10}\)

\(102^{11}\times102^5=102^{11+5}=102^{16}\)

\(19^3\times19^7\times19^5=19^{3+7+5}=19^{15}\)

\(b,x^m\times x^n\times x^p\times x^q=x^{m+n}\times x^p\times x^q=x^{m+n+p}\times x^q=x^{m+n+p+q}\)

Z ={.....-3,-2.-1,0,1,2,3...}

a, \(\left[\left(-13\right)+\left(-15\right)\right]+\left(-8\right)\)

    \(=\left(-28\right)+\left(-8\right)\)

    \(=-36\)

b, \(-\left(-129\right)+\left(-119\right)-301+12\)

   \(=129-119-301+12\)

   \(=10-301+12\)

  \(=\left(-291\right)+12\)

  \(=-279\)

c, \(500-\left(-200\right)-210-100\)

  \(=300-\left(210+100\right)\)

  \(=300-310\)

  \(=-10\)

d, \(777-\left(-111\right)-\left(-222\right)+20\)

    \(=777+111+222+20\)

    \(=888+242\)

    \(=1130\)

\(64cm^2\)