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3 years ago

Please help me i don’t know this

Mathematics

1 answer:

scZoUnD [109]3 years ago

3 0

Answer: Surface Area = 16in.

Step-by-step explanation:

So we need to find the area of each side and add them up.

First we find the area of one of the triangular sides: 2*3/2 = 3

Then we multiply it by 4 to accommodate the 4 sides that are the same size: 12

Then we do 2*2 = 4 for the base. We then add those up.

4 + 12 = 16

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Vector V is in standard position and makes an angle of 50° with the positive x-axis. Its magnitude is 18. Write V in component f

Umnica [9.8K]

Answer:

11.57 units east, 13.79 units upwards

11.57i + 13.79j

Step-by-step explanation:

V = 18 units at an angle of 50° to the positive x-axis

In component Form, this is how each component is calculated

a = 18cos50° = 11.57 units east

b = 18sin50° = 13.79 units upwards

check if your answer is correct

$\sqrt{11.57^{2} +13.79^{2} } = 18.0$

In vector component form ai + bj = 11.57i + 13.79j

7 0

3 years ago

If x varies inversely as y and x = 24 when y = 4, find x when y = 12.

ANTONII [103]

what that makes no sense

8 0

3 years ago

X+1/2(x-4)=1/2x+6(x-2)

Andrej [43]

Answer:

B) 1

Step-by-step explanation:

x+1/2(x-4)=1/2x+6(x-2)

x+1/2x-4/2=1/2x+6x-12

x+1/2x-1/2x-2=6x-12

x-2=6x-12

x-6x-2=-12

-5x-2=-12

-5x=-12+2

-5x=-10

5x=10

x=10/5

x=2

8 0

3 years ago

Pleaseeee help I already have the graph.

vekshin1

Answer: 5x + 2y = 8 = x=-2/5y+8/5

y = 3x - 7= x= 1/3y+7/3

Step-by-step explanation:

7 0

3 years ago

Find the sum \[\sum_{k = 1}^{2004} \frac{1}{1 + \tan^2 (\frac{k \pi}{2 \cdot 2005})}.\]

vampirchik [111]

Answer:1002

Step-by-step explanation:

$\Rightarrow \sum_{k=1}^{2004}\left ( \frac{1}{1+\tan ^2\left ( \frac{k\pi }{2\cdot 2005}\right )}\right )$

and $1+\tan ^2\theta =\sec^2\theta$

and $\cos \theta =\frac{1}{\sec \theta }$

$\Rightarrow \sum_{k=1}^{2004}\left ( \cos^2\frac{k\pi }{2\cdot 2005}\right )$

as $\cos ^2\theta =\cos ^2(\pi -\theta )$

Applying this we get

$\Rightarrow \sum_{1}^{1002}\left [ \cos^2\left ( \frac{k\pi }{2\cdot 2005}\right )+\cos^2\left ( \frac{(2005-k)\pi }{2\cdot 2005}\right )\right ]$

every $\theta$ there exist $\pi -\theta$

such that $\sin^2\theta +\cos^2\theta =1$

therefore

$\Rightarrow \sum_{1}^{1002}\left [ \cos^2\left ( \frac{k\pi }{2\cdot 2005}\right )+\cos^2\left ( \frac{(2005-k)\pi }{2\cdot 2005}\right )\right ]=1002$

6 0

3 years ago