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murzikaleks [220]
4 years ago
9

Find the equation of the perpendicular bisector between two points (2,7) and (-6, -1)

Mathematics
1 answer:
Veseljchak [2.6K]4 years ago
4 0

Answer:

well it fjgkgjg

Step-by-step explanation:

jdjbdhdhdhdhdhxyxyxyxyd

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The table gives the grouped frequency distribution for the lengths of the electrical cords of 80 kettles.
Verizon [17]

Answer: First box 44, second box 77 and third box 80❤️

Step-by-step explanation:

4 0
3 years ago
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Janet drives from Clarkson to Humbolt in 2 hours. Suppose Janet drives for 10 hours. If she maintains the same driving rate, can
N76 [4]
Triangle OPQ is shown below with line RS passing through points R and S:
Triangle OPQ is shown with point R lying on side OP and point S lying on side OQ Line RS is perpendicular to side OQ.
If triangle OPQ is dilated about the center of the triangle to create triangle O'P'Q', what can you conclude about segments RS and R'S'? (6 points)
Segment O'Q' is perpendicular to segment R'S'.  Line RS is parallel to line R'S'.  Line RS is perpendicular to line R'S'.  Point P' passes through line RS.
4 0
3 years ago
Are the equations m + 3 = -5 and m = -2 equivalent?​
guajiro [1.7K]
M + 3 = 5
-3 -3

m = -2
m= -2

yes the equations are equivalent
5 0
3 years ago
Two types of sandwiches were made for a tea party. 55% of the sandwiches were cheese sandwiches and the rest were chicken sandwi
DedPeter [7]

To solve this problem, the formula can be used

I = P / T

Where I is the fraction of the sample

P is the amount of the sample

T is the total amount of sample

Since P = 252 chicken sandwiches and I = 0.55

And solve for T

T = 252 / 0.52

<span>T = 458 sandwiches in total</span>

6 0
3 years ago
I need help with part b. I feel like there’s a catch, I want to do the first derivative test, however, I feel like there is a be
Sladkaya [172]

Answer:

The fifth degree Taylor polynomial of g(x) is increasing around x=-1

Step-by-step explanation:

Yes, you can do the derivative of the fifth degree Taylor polynomial, but notice that its derivative evaluated at x =-1 will give zero for all its terms except for the one of first order, so the calculation becomes simple:

P_5(x)=g(-1)+g'(-1)\,(x+1)+g"(-1)\, \frac{(x+1)^2}{2!} +g^{(3)}(-1)\, \frac{(x+1)^3}{3!} + g^{(4)}(-1)\, \frac{(x+1)^4}{4!} +g^{(5)}(-1)\, \frac{(x+1)^5}{5!}

and when you do its derivative:

1) the constant term renders zero,

2) the following term (term of order 1, the linear term) renders: g'(-1)\,(1) since the derivative of (x+1) is one,

3) all other terms will keep at least one factor (x+1) in their derivative, and this evaluated at x = -1 will render zero

Therefore, the only term that would give you something different from zero once evaluated at x = -1 is the derivative of that linear term. and that only non-zero term is: g'(-1)= 7 as per the information given. Therefore, the function has derivative larger than zero, then it is increasing in the vicinity of x = -1

6 0
3 years ago
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