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brilliants [131]
2 years ago
13

\sqrt{6}\:" alt="\sqrt{6}\:\times \sqrt{6}\:" align="absmiddle" class="latex-formula">
Mathematics
2 answers:
777dan777 [17]2 years ago
4 0

Answer:

\sqrt{6} \:  \times  \sqrt{6}  =  \\  \\   \sqrt[2]{6 }   = 6 \: is \: the \: answer

I am Lyosha [343]2 years ago
3 0

Answer:

6

Step-by-step explanation:

\sqrt{6}  *\sqrt{6}

\sqrt{6*6} \\\sqrt{6^2}

6

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PLS HELP <br> NO LINKS OR I WILL REPORT <br> WILL MARK BRAINLIEST
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2 years ago
Help me out please with domain and range and the last question please (algebra 2) no links please I will report
baherus [9]

Answer:

domain = ( -infinity, infinity) = all real numbers

range = ( -1, infinity)

3 0
2 years ago
In a study of preferences in leafcutter ants, a researcher presented 20 randomly chosen ant colonies with leaves from the two mo
Rufina [12.5K]

Answer:

uyjytyhy

Step-by-step explanation:

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8 0
3 years ago
Evaluate the line integral for x^2yds where c is the top hal fo the circle x62 _y^2 = 9
natulia [17]
Parameterize C by

\mathbf r(t)=\langle x(t),y(t)\rangle=\langle3\cos t,3\sin t\rangle

where 0\le t\le\pi. Then the line integral is

\displaystyle\int_Cx^2y\,\mathrm dS=\int_{t=0}^{t=\pi}x(t)^2y(t)\left\|\frac{\mathrm d\mathbf r(t)}{\mathrm dt}\right\|\,\mathrm dt
=\displaystyle\int_{t=0}^{t=\pi}(3\cos t)^2(3\sin t)\sqrt{(-3\sin t)^2+(3\cos t)^2}\,\mathrm dt
=\displaystyle3^4\int_{t=0}^{t=\pi}\cos^2t\sin t\,\mathrm dt

Take u=\cos t, then

=\displaystyle-3^4\int_{u=1}^{u=-1}u^2\,\mathrm du
=\displaystyle3^4\int_{u=-1}^{u=1}u^2\,\mathrm du
=\displaystyle2\times3^4\int_{u=0}^{u=1}u^2\,\mathrm du
=54
6 0
3 years ago
12÷3×15(15-6)+3.use order of operations to solve
dlinn [17]
12/3=4
4*15=60
60*(9)=540
540+3=543

4 0
3 years ago
Read 2 more answers
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