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algol [13]
2 years ago
10

What is the total surface area of the figure below? Only enter the number part of your answer. Do not include the units.

Mathematics
1 answer:
Leya [2.2K]2 years ago
4 0

Answer:

193.2

Step-by-step explanation:

To solve the triangle:

5.2 × 6= 31.2

You would normally find half of that since it's a triangle but there are two triangles, so each would equal half of 31.2 but if you add both halves that just equals 31.2.

Square one:

9 × 6= 54

All three squares are equal since they have the same measurements.

 54 + 54 + 54= 162

162 + 31.2= 193.2                                                              

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A student writes 5y*3 to model the relationship the sum of 5y and 3. explain the error
mezya [45]
So,

5y*3 is the open phrase the student uses to model "the sum of 5y and 3".

"The sum of" means addition.  The student put 5y*3, while the sum of 5y and 3 is actually 5y + 3.
4 0
3 years ago
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Consider the parabola r​(t)equalsleft angle at squared plus 1 comma t right angle​, for minusinfinityless thantless thaninfinity
kodGreya [7K]

Given:-   r(t)=< at^2+1,t>  ; -\infty < t< \infty , where a is any positive real number.

Consider the helix parabolic equation :  

                                              r(t)=< at^2+1,t>

now, take the derivatives we get;

                                            r{}'(t)=

As, we know that two vectors are orthogonal if their dot product is zero.

Here,  r(t) and r{}'(t)  are orthogonal i.e,   r\cdot r{}'=0

Therefore, we have ,

                                  < at^2+1,t>\cdot < 2at,1>=0

< at^2+1,t>\cdot < 2at,1>=

                                              =2a^2t^3+2at+t

2a^2t^3+2at+t=0

take t common in above equation we get,

t\cdot \left (2a^2t^2+2a+1\right )=0

⇒t=0 or 2a^2t^2+2a+1=0

To find the solution for t;

take 2a^2t^2+2a+1=0

The numberD = b^2 -4ac determined from the coefficients of the equation ax^2 + bx + c = 0.

The determinant D=0-4(2a^2)(2a+1)=-8a^2\cdot(2a+1)

Since, for any positive value of a determinant is negative.

Therefore, there is no solution.

The only solution, we have t=0.

Hence, we have only one points on the parabola  r(t)=< at^2+1,t> i.e <1,0>




                                               




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3 years ago
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The correct answer is

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