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Sladkaya [172]
3 years ago
5

Barnard’s Star is approximately 35,133 billion miles from Earth. What is 35,133 billion in scientific notation?

Mathematics
1 answer:
julsineya [31]3 years ago
4 0
C


Mark brainliest please


Hope this helps you
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Which equation represents y = -x2 - 2x + 3 in vertex form?
Georgia [21]

Answer:

y = -(x + 1)² + 4

Step-by-step explanation:

y = -x² − 2x + 3

First, factor the leading coefficient from the first two terms.

y = -(x² + 2x) + 3

Complete the square.

y = -(x² + 2x + 1 − 1) + 3

Simplify and factor.

y = -(x² + 2x + 1) + 1 + 3

y = -(x + 1)² + 4

6 0
4 years ago
a small acting club has 6 members. 3 of the members are to be chosen for a trip to see a Broadway play. how many different 3-mem
nignag [31]
A group of k elements can be chosen from a group of n elements in \frac{n!}{k!(n-k)!} ways.

(^6 _3)=\frac{6!}{3!(6-3)!}=\frac{6!}{3! \times 3!}=\frac{3! \times 4 \times 5 \times 6}{3! \times 6}=4 \times 5=20

There are 20 different 3-member groups.
8 0
3 years ago
Solve the following quadratic equation by factoring.<br> f (x) = -x^2-11x – 28
spayn [35]

Answer:

(-x-7)(x+4)

Step-by-step explanation:

5 0
3 years ago
How do i do 3 part a ?
WINSTONCH [101]


In general the binomial expansion is


(a+b)^n = {n \choose 0} a^0 b^n + {n \choose 1} a^1 b^{n-1} + {n \choose 2} a^2 b^{n-2} + ... + {n \choose n} a^n b^0


So in our case, because we want ascending powers of x we'll write,


(-3x + 1)^{11} =  {11 \choose 0} (-3x)^0 1^{11} + {11 \choose 1} (-3x)^{1} 1^{10} + {11 \choose 2} (-3x)^{2} 1^9  + {11 \choose 3 } (-3x)^3 1^8 + ...


We need to calculate the binomial coefficients:


{11 \choose 0}  = 1


{11 \choose 1}  = 11


{11 \choose 2}  = \dfrac{11 \times 10}{2} = 55


{11 \choose 3}  = \dfrac{11 \times 10 \times 9}{3 \times 2} = 165


(-3x+1)^{11} =  1 (-3x)^0 1^{11}  + 11(-3x)^{1} 1^{10}  + 55 (-3x)^2 1^{9}  + 165 (-3x)^3 1^8 + ...


(1-3x)^{11} =  1 -33 x + 495 3x^2 - 4455 x^3+ ...



6 0
4 years ago
Find the values of x (if any) at which f(x)= x+1/x^2-1 is not continuous. If so, is the discontinuity removable?
Mkey [24]

f(x)=\dfrac{x+1}{x^2-1} has two points of discontinuity at x=\pm1, where x^2-1=0.

We have

\dfrac{x+1}{x^2-1}=\dfrac{x+1}{(x+1)(x-1)}

so as long as x\neq-1, we can cancel x+1 and be left with \dfrac1{x-1}. Then as x\to-1, we have

\displaystyle\lim_{x\to-1}f(x)=\lim_{x\to-1}\frac1{x-1}=-\frac12

The limit exists, so the discontinuity at -1 is removable (so definitely not C or D).

Meanwhile,

\displaystyle\lim_{x\to1}f(x)=\lim_{x\to1}\frac1{x-1}

does not exist, so the discontinuity at 1 is non-removable, making A the correct answer.

8 0
3 years ago
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