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Pavel [41]
3 years ago
9

Is a cub a polygon ?why or why not

Mathematics
1 answer:
Vladimir79 [104]3 years ago
6 0
A cube<span> is </span>not<span> a </span>polygon<span>. </span>Polygons<span> are flat, plane figures. A </span>cube<span> is a 3- di mensional solid figure.</span>
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Solve for x and y simultaneously:<br> X-2y=3<br> 4xsquared-5xy+6y=3
Mkey [24]
X - 2y = 3
<span>4x^2 - 5xy + 6y = 3
lets solve for x the first and substitute in the second:
x = 3 + 2y
4(</span>3 + 2y)^2 - 5(3 + 2y)y + 6y = 3
4(9 + 12y + 4y^2) - 15y - 10y^2 = 3
36 + 48y +16y^2<span> - 15y - </span><span>10y^2 = 3
6y^2 + 33y + 33 = 0
we can solve using the general quadratic formula:
y = (-33 +- </span>√(33^2 - 4*6*33)<span>)/12
</span>y = (-33 +- √(297)<span>)/12
</span>so there are 2 solutions for y:
y1 = (-33 + √(297)<span>)/12
</span>y2 = (-33 - √(297)<span>)/12
</span>pick one and then substitute the y value in the first equation to find x
4 0
3 years ago
Read 2 more answers
Y=4(x+1)^-4 HELP I AM STUCK!
Feliz [49]

Answer:

y=4x+.0004

Step-by-step explanation:

y=4(x+1)^-4

distribute your parenthesis

y=4x+4^-4

do your exponents

y=4x+.004

and thats as far as the problem can be simplified

hope this helps  

5 0
3 years ago
What are three names for 3.09
mixas84 [53]

Answer:

1.Three and Nine Hundredths    2. 3 9/100      3. 390%

Step-by-step explanation:


7 0
3 years ago
How many 10-digit ternary strings are there that contain exactly two 0s, three 1s, and five 2s?
Svet_ta [14]

There are \dbinom{10}2 ways of picking 2 of the 10 available positions for a 0. 8 positions remain.

There are \dbinom83 ways of picking 3 of the 8 available positions for a 1. 5 positions remain, but we're filling all of them with 2s, and there's \dbinom55=1 way of doing that.

So we have

\dbinom{10}2\dbinom83\dbinom55=\dfrac{10!}{2!(10-2)!}\dfrac{8!}{3!(8-3)!}\dfrac{5!}{5!(5-5)!}=2520

The last expression has a more compact form in terms of the so-called multinomial coefficient,

\dbinom{10}{2,3,5}=\dfrac{10!}{2!3!5!}=2520

5 0
3 years ago
De moirve's <br> (√3-i ÷ √3+i)^6 = 1
bulgar [2K]

(√3 - <em>i </em>) / (√3 + <em>i</em> ) × (√3 - <em>i</em> ) / (√3 - <em>i</em> ) = (√3 - <em>i</em> )² / ((√3)² - <em>i</em> ²)

… = ((√3)² - 2√3 <em>i</em> + <em>i</em> ²) / (3 - <em>i</em> ²)

… = (3 - 2√3 <em>i</em> - 1) / (3 - (-1))

… = (2 - 2√3 <em>i</em> ) / 4

… = 1/2 - √3/2 <em>i</em>

… = √((1/2)² + (-√3/2)²) exp(<em>i</em> arctan((-√3/2)/(1/2))

… = exp(<em>i</em> arctan(-√3))

… = exp(-<em>i</em> arctan(√3))

… = exp(-<em>iπ</em>/3)

By DeMoivre's theorem,

[(√3 - <em>i </em>) / (√3 + <em>i</em> )]⁶ = exp(-6<em>iπ</em>/3) = exp(-2<em>iπ</em>) = 1

7 0
3 years ago
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