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TEA [102]
3 years ago
12

I need help with question 1 please. I’ll mark brainliest

Mathematics
2 answers:
STatiana [176]3 years ago
5 0

Answer:

neHfzrusufsfisjfsjgdjtxktxkthkdkgdkgigcktxiyxkgxtixitzkfxkfxit

Step-by-step explanation:

mekhhggggggggggty

Dvinal [7]3 years ago
4 0
It is x^2-2x-9 since u just subtract both equations
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A square and a rhombus have the following in common except
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A square is a rhombus, but a rhombus is not always a square
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B. (3.9) and (4.6)<br> The rate of change between these two points is
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Answer: 0.7 is the rate change between the two points

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Mai has to decide between two cafeteria meal plans. Under plan A, each meal cost $2.50. Under plan B, one month of meals cost $3
BaLLatris [955]

Answer:

If he eats more then 12 lunches in a month he should take Plan B, if he eats less then 12 lunches a month he should take Plan A.

Step-by-step explanation:

2.5x12=30

If he is on Plan A and buys 13 lunches it would cost more money then to be on Plan B.  If he buys less then it would cost him less money to be on Plan A since he would be paying under $30.

4 0
3 years ago
Point A, located at (-5, 1) is translated. Point A' is located at (6, 1). What was the translation?
earnstyle [38]

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Step-by-step explanation:

11 units right

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4 0
3 years ago
Read 2 more answers
Find all values of x for which the series converges. (Enter your answer using interval notation.) [infinity] 9 x − 7 9 n n = 0 F
makkiz [27]

Answer:

The series converges to   $ \frac{1}{1-9x} $     for   $ \frac{-1}{9} < x < \frac{1}{9} $                

Step-by-step explanation:

Given the series is     $ \sum_{n=0}^{\infty} 9^n x^n $

We have to find the values of x for which the series converges.

We know,

$ \sum_{n=0}^{\infty} ar^{n-1} $ converges to  (a) / (1-r) if r < 1

Otherwise the series will diverge.

Here, $ \sum_{n=0}^{\infty} 9^n x^n =  \sum_{n=0}^{\infty} (9x)^{n}  $ is a geometric series with |r| = | 9x |

And it converges for |9x| < 1

Hence, the given series gets converge for $ \frac{-1}{9} < x < \frac{1}{9} $

And geometric series converges to $ \frac{a}{1-r} $

Here, a = 1 and r = 9x

Therefore, $ \frac{a}{1-r} = \frac{1}{1-9x} $

Hence, the given series converges to   $ \frac{1}{1-9x} $     for   $ \frac{-1}{9} < x < \frac{1}{9} $

4 0
3 years ago
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