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rodikova [14]
3 years ago
11

PLEASE HELP ASAAAPPPPPP

Mathematics
1 answer:
Ad libitum [116K]3 years ago
7 0
A)  276 / 138 = 2

B) 26/2 = 13,  13/2 = 6.5

Answer is B

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A birthday celebration meal is $ 46.80
kolezko [41]
10% of 46.80 =4.68
46.80 + 4.68 = 51.48

Hope this helps :D
3 0
2 years ago
Use the expression 9(7 + 2x) to answer the following:
oksian1 [2.3K]

Part A: Describe the two factors in this expression. (4 points) The factors are (1) the constant coefficient 9 and (2) the binomial (7+2x).

Part B: How many terms are in each factor of this expression? (4 points) The first factor (multiplicand), 9, has one term. The second factor (multiplicand), (7+2x), has two terms (and is thus called a binomial).

Part C: What is the coefficient of the variable term? (2 points) The only such coefficient is 2.

4 0
3 years ago
If C is the center of the circle, find the measure of KM
stiv31 [10]

Your answer is 160. Sorry about that answer above I laughed reading it hahaha.

6 0
3 years ago
Read 2 more answers
What will you get when you add two integers
dybincka [34]
Well, it depends if its positive or negative. 2 positive integers will equal a positive, 2 negatives will equal a negative. A negative and a positive will depend.
8 0
3 years ago
Read 2 more answers
Define fn : [0,1] --> R by the
sasho [114]

Answer:

The sequence of functions \{x^{n}\}_{n\in \mathbb{N}} converges to the function

f(x)=\begin{cases}0&0\leq x.

Step-by-step explanation:

The limit \lim_{n\to \infty }c^{n} exists and converges to zero whenever \lvert c \rvert. But, if c=1 the sequence \{c^{n}\} is constant and all its terms are equal to 1, then converges to 1. Using this result, consider the sequence of functions \{f_{n}\} defined on the interval [0,1] by f_{n}(x)=x^{n}. Then, for all 0\leq x we have that \lim_{n\to \infty}x^{n}=0. Now, if x=1, then \lim_{n\to \infty }x^{n}=1. Therefore, the limit function of the sequence of functions is

f(x)=\begin{cases}0&0\leq x.

To show that the convergence is not uniform consider 0. For any n>1 choose x\in (0,1)  such that \varepsilon^{1/n}. Then

\varepsilon

This implies that the convergence is not uniform.

8 0
3 years ago
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