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tatuchka [14]
3 years ago
6

Which is equivalent to 243 2/5 ? 3 6 9 12

Mathematics
2 answers:
Vlada [557]3 years ago
5 0

Answer:

The answer is 9

Step-by-step explanation:

we have

(243)^{\frac{2}{5}}

we know that

a^{\frac{n}{m}}=\sqrt[m]{a^{n}}

so

(243)^{\frac{2}{5}}=\sqrt[5]{243^{2}}

Remember that

243=3^{5}

Substitute

\sqrt[5]{243^{2}}=\sqrt[5]{(3^{5})^{2}}\\ \\= \sqrt[5]{(3^{10})}\\ \\=3^{\frac{10}{5}}\\ \\= 3^{2} \\ \\=9

NeTakaya3 years ago
3 0
9 Good luckkkkkkkkkkkkkkkkkkkkkkkkkkkkk

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Let L be the line with parametric equations x=5+t,y=6,z=−2−3t. Find the vector equation for a line that passes through the point
scZoUnD [109]

Answer:

The required equations are

(-5 \hat i + 7 \hat j + 8 \hat k )+\lambda \left((10+\frac {3}{\sqrt {10}})\hat i -\hat j +(6- \frac {9}{\sqrt {10}})\hat k\right)=0 and

(-5 \hat i + 7 \hat j + 8 \hat k )+\lambda \left((10-\frac {3}{\sqrt {10}})\hat i -\hat j +(6+ \frac {9}{\sqrt {10}})\hat k\right)=0.

Step-by-step explanation:

The given parametric equation of the line, L, is x=5+t, y=6, z=-2-3t, so, an arbitrary point on the line is R(x,y,z)=R(5+t, 6, -2-3t)

The vector equation of the line passing through the points P(-5,7,-8) and R(5+t, 6, -2-3t) is

\vec P + \lambda \vec{(PR)}=0

\Rightarrow (-5 \hat i + 7 \hat j - 8 \hat k )+\lambda \left((5+t-(-5))\hat i + (6-7)\hat j +(-2-3t-8)\hat k\right)=0

\Rightarrow (-5 \hat i + 7 \hat j - 8 \hat k )+\lambda \left((10+t)\hat i -\hat j +(6-3t)\hat k\right)=0\;\cdots (i)

The given equation can also be written as

\frac {x-5}{1}=\frac {v-6}{0}=\frac{z+2}{-3}=t \; \cdots (ii)

The standard  equation of the line passes through the point P_0(x_0,y_0,z_0) and having direction\vec v= a_1 \hat i +a_2 \hat j +a_3 \hat k is

\frac {x-x_0}{a_1}=\frac {y-y_0}{a_2}=\frac{z-z_0}{a_3}=t \;\cdots (iii)

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|t \vec v|=d\;\cdots (iv)

Comparing equations (ii) and (iii)

The line is passing through the point P_0 (5,6,-2) having direction \vec v=\hat i -3 \hat k.

Note that the point Q(5,6,-2) is the same as P_0 obtained above.

Now, the value of the parameter, t, for point R at distance d=3 from the point Q(5,6,-2) can be determined by equation (iv), we have

|t(\hat i -3 \hat k)|=3

\Rightarrow t^2|(\hat i -3 \hat k)|^2=9

\Rightarrow 10t^2=9

\Rightarrow t^2=\frac {9}{10}

\Rightarrow t=\pm \frac {3}{\sqrt {10}}

Put the value of t in equation (i), the required equations are as follows:

For t= \frac {3}{\sqrt {10}}

(-5 \hat i + 7 \hat j - 8 \hat k )+\lambda \left((10+\frac {3}{\sqrt {10}})\hat i -\hat j +\left(6-3\times \frac {3}{\sqrt {10}})\hat k\right)=0

\Rightarrow (-5 \hat i + 7 \hat j - 8 \hat k )+\lambda \left((10+\frac {3}{\sqrt {10}})\hat i -\hat j +(6- \frac {9}{\sqrt {10}})\hat k\right)=0

and for t= -\frac {3}{\sqrt {10}},

(-5 \hat i + 7 \hat j - 8 \hat k )+\lambda \left((10+\left (-\frac {3}{\sqrt {10}}\right))\hat i -\hat j +(6-3\times \left(-\frac {3}{\sqrt {10}}\right)\hat k\right)=0

\Rightarrow  (-5 \hat i + 7 \hat j - 8 \hat k )+\lambda \left((10-\frac {3}{\sqrt {10}})\hat i -\hat j +(6+ \frac {9}{\sqrt {10}})\hat k\right)=0

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