What is your favorite place to get wings and what type of wings do you like?

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Answer 1

Answer: american deli and honey hot

Related Questions

The area of a circle of radius r is given by A=\pi r^2A=πr 2 and its circumference is given by C=2\pi rC=2πr. At a certain point in time, the radius of the circle is r=8r=8 inches and the area of the circle is changing at a rate of \frac{dY}{dt}=\pi\sqrt{2} dt dA =π 2 square inches per second. How fast is the radius of the circle changing at this time
Evaluate the factorial expression.left parenthesis 6 minus 2 right parenthesis exclamation mark
HELP ME FIND THE SLOPE PLS
3x−8y+44=0 7x=12y−56
114% is equivalent to what fraction in reduced terms?

A triangle with one obtuse angle must also have two acute sides

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that would be true :D hope i helped you

That is true :))))) good job

Evaluate $\rm (3^2+1)/(3^2-1)+(5^2+1)/(5^2-1)+(7^2+1)/(7^2-1)+\ldots+(101^2+1)/(101^2-1) =$
With step by step explanation !

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It's easier to deal with the symbolic sum (in sigma notation),

$\displaystyle\sum_(k=1)^(50)((2k+1)^2+1)/((2k+1)^2-1)$

Expanding the terms in the fraction, computing the quotient, and decomposing into partial fractions gives

$((2k+1)^2+1)/((2k+1)^2-1) = (4k^2 + 4k + 2)/(4k^2 + 4k)$

$=\frac12*(2k^2 + 2k + 1)/(k^2 + k)$

$=\frac12\left(2+\frac1{k(k+1)}\right)$

$=\frac12\left(2 + \frac1k - \frac1{k+1}\right)$

and it's the latter two terms that reveal a telescoping pattern.

In case you need more details about the partial fraction decomposition, we are looking for coefficients a and b such that

$\frac1{k(k+1)}=\frac ak+\frac b{k+1}$

$1 = a(k+1) +bk =(a+b)k+a$

which gives a = 1, and a + b = 0 so that b = -1.

Our sum has been rearranged as

$\displaystyle\frac12\sum_(k=1)^(50)\left(2+\frac1k-\frac1{k+1}\right)=\sum_(k=1)^(50)1+\frac12\sum_(k=1)^(50)\left(\frac1k-\frac1{k+1}\right)=50+\frac12\sum_(k=1)^(50)\left(\frac1k-\frac1{k+1}\right)$

The remaining telescoping sum is

1/2 [(1/1 - 1/2) + (1/2- 1/3) + (1/3- 1/4) + … + (1/48- 1/49) + (1/49- 1/50) + (1/50 - 1/51)]

and you can see how there are pairs of numbers that cancel, so that the sum reduces to

1/2 [1/1 - 1/51] = 1/2 [1 - 1/51] = 1/2 × 50/51 = 25/51

So, our original sum ends up being

$\displaystyle\sum_(k=1)^(50)((2k+1)^2+1)/((2k+1)^2-1) = 50 + (25)/(51) = \boxed{(2575)/(51)}$

The pair of points (6,y) and (10,-1) lie on a line with a slope 1(over)4. What is the value of y?

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The formula for Slope is m = $(y_2 - y_1)/(x_2 - x_1)$ where m is the slope and the x's and y's are your given coordinates. So, plug the given information into the formula and solve for y.

m = $(y_2 - y_1)/(x_2 - x_1)$ Plug in the given values
$(1)/(4)$ = $(-1 - y)/(10 - 6)$ Cross multiply
10 - 6 = -4 - 4y Subtract
4 = -4 - 4y Add 4 to both sides
8 = -4y Divide both sides by -4
-2 = y Swich the sides to make it easier to read
y = -2

Check your answer by plugging -2 back into the equation with the other values.

[tex] \frac{1}{4} [/tex] = $(-1 - (-2))/(10 - 6)$ Simplify the double negative
$(1)/(4)$ = $(-1 + 2)/(10 - 6)$ Simplify the numerator and denominator
$(1)/(4)$ = $(1)/(4)$

Since both sides equal each other, y = -2.

Use the formula to find the slope which is m= y2-y1 over x2-x2. The equation would look like 1/4= -1 -y over 10-6. First, you do the subtraction in the denominator and you get 4. The equation would look like 1/4=-1-y over 4.

Then, you multiply the 4 to both side in order to get rid of the fraction from the equation. You would then have 1=-1-y. You add -1 to both side and you would have 2=-y. Since the variable cannot be negative, you divide -1 from both side and get -2=y. Therefore the answer is y=-2.

Hope this helps.

Hi can someone help me with this

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6-3y-9y+5.7
3y-9y+6+5.7
6y+11.7

You can’t do anything further bc y doesn’t equal anything

Answer:

=−12y+11.7 I believe it's that but I am not completely sure

I believe it is

Let set C = {1, 2, 3, 4, 5, 6, 7, 8} and set D = {2, 4, 6, 8}.Which notation shows the relationship between set C and set D?

A: C ∪ D
B: C ∩ D
C: D ⊆ C
D: C ⊆ D

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1. If it is intersection then it SHOULD be included in both the sets right? Now we know that odd numbers from 1-100 but the second set are multiples of 5 from 50-150! So we mainly need to look for common numbers which are ODD and are a MULTIPLE OF 5 BETWEEN 50 - 100!! So A={51,53,57,59,61......99} B={55,60,65,70.......95} [We stop till 100 because set A has no such element] So what is A ∩ B here? A ∩ B = {All odd numbers and multiples of 5 between 50 - 100}

10,infinity
Interval notation-
Inequality notation-