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Alik [6]
3 years ago
8

Given 2 lines cut by a transversal, what would the m<7 have to be if m <4=85 for the lines to be parallel and what is the

relationship between <7 and <4​

Mathematics
1 answer:
nlexa [21]3 years ago
8 0

Answer:

85º

Step-by-step explanation:

<7=<4 since we know the lines are parallel, and intersected at the same angle

Therefore, <7=85º.

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What is the explicit rule for the sequence 11 22 44 88​
sertanlavr [38]

Answer: an = 11(2)^n-1

7 0
3 years ago
PLZZ HELP IM REALLY NOT GOOD AT THIS. Missy Elliot loves to shop at Neiman Marcus last Saturday she bought pants and shirts for
wel

Answer:

17 pants and 9 shirts

Step-by-step explanation:

questions like this require you to guess numbers first then adjust them as needed. I originally guessed 20 pants and 6 shirts, however the number was too high, so I put less pants, since they're the higher number. if the number was too low you would use less shirts since they're the lower number. after a couple of tries you'll get

(225 × 17) + (125 × 9) = 4950

but seriously who spends almost 5000 dollars on clothes from one store in one outing, that's so much money

5 0
3 years ago
Plz help i dont get this at all i will give brainlyest
riadik2000 [5.3K]

6+1/343+36+1/81−27−36+1/81-27-243+1/256+1/144+1/8+1/5

6 0
3 years ago
Read 2 more answers
What is the solution of |x – 2| &gt; –3
Amanda [17]
Answer: The solution is the set of all real numbers (there are infinitely many solutions).

The reason why this is the case is because |x| is never negative. The smallest it can get is 0, which is larger than -3. That applies to |x-2| as well. So |x-2| is ALWAYS larger than -3 no matter what you pick for x. The smallest |x-2| can get is 0 and that happens when x = 2. Otherwise, the result is some positive value which is larger than -3.

So that's why |x-2| > -3 has infinitely many solutions. We can replace x with any real number we want, and the inequality would be true. 
6 0
3 years ago
How to express -8+4-2+1 in sigma notation
dlinn [17]
It's pretty clear what the pattern is - 2^(k-1) * (-1)^(k-1) would work backwards, but we have to work forwards to get lower in this case, so we could start at -3 (as long as we put the k-1 in absolute value to make sure it's not 1/8), so we could write it as ∑ (k=1, goes to 4) 2^(|k-4|)*(-1)^(k-4) 

It doesn't matter for k-4 because 1/-1 and 1/1 are the same thing as -1 and 1 respectively
3 0
3 years ago
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