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Dovator [93]
1 year ago
14

50 points!!!!!!!!!!!!

Mathematics
2 answers:
igomit [66]1 year ago
8 0

Answer: the first one .00000000058

Step-by-step explanation:

AVprozaik [17]1 year ago
5 0

Answer:

0.00000000058

Step-by-step explanation:

Move the decimal by 10 places to the right

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NEED HELP ASAP!!!!
Genrish500 [490]

Answer:

B) Jill omitted a factor pair, which affected the GCF and factored expression.

Step-by-step explanation:

Jill is factoring the expression 13xy-52y. Her work is shown below.  

Factors of 13xy: 1, 13, x, y  

Factors of 52y: 1, 2, 26, 52, y       OOPSIE  she forgot 4,13

GCF: y

Factored expression: y(13x-52)   13y (x-4)

4 0
3 years ago
Read 2 more answers
What's the surface area?<br><br>(EXPLAIN YOUR ANSWER)
dedylja [7]
714,110ft



Hope this helpsss
6 0
3 years ago
I have no clue how to do this
gregori [183]
Me either just ask your teacher or use the app Socratic
8 0
4 years ago
AP calculus HW Need help on #61
Yuki888 [10]
Expanding the limit, we get (x^2+2x∆x+∆x^2-2x-2∆x+1-x^2+2x-1)/<span>∆x

Crossing the 1s , the 2xs, and the x^2s out, we get

(2x</span>∆x+∆x^2-2∆x)/<span>∆x

Dividing the </span><span>∆x, we get 
2x+</span><span>∆x-2. 

Making the limit of </span><span>∆x=0, we get 2x-2.</span>
7 0
4 years ago
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>




                                               




6 0
3 years ago
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