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2 years ago

The diameter of a circle is 8 cm. find it's area in terms of pi

Mathematics

2 answers:

nydimaria [60]2 years ago

3 0

Answer:

d=8

r=d/2

r=4

a=πr^2

=22/7×4×4

=50.28

sineoko [7]2 years ago

3 0

The answer is 50.27cm to the second power

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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

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

Here, The value of the parameter, $t$ , of any point $R$ at a distance $d$ from the point, $P_0$ , can be determined by

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

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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3 years ago

The graph represents function 1, and the equation represents function 2:

Veseljchak [2.6K]

If there was a graphical representation, I would be happy to assist you.

6 0

3 years ago

Find the area of the parallelogram.

kotegsom [21]

You can see that the base is 5 units.
The height is 3.
The area of a parallelogram is
A = B * H
A = 5 * 3
A = 15 units^2

Hope this helps!

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3 years ago

A wildlife conservancy maintains a ratio of 2 squirrels for every 8 birds how many birds would there be if there were 15 squirre

GrogVix [38]

Answer: 16 birds

Step-by-step explanation:

Acording to the info we know that: for every 1 squirrel there are 4 birds

=> the number of the birds if there were 15 squirrels: 15*4=60 (birds)

7 0

3 years ago

Please help i need help

vodomira [7]

Answer:

fractions

its right answer

8 0

2 years ago

Read 2 more answers

The diameter of a circle is 8 cm. find it's area in terms of pi​

The diameter of a circle is 8 cm. find it's area in terms of pi