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sineoko [7]
3 years ago
8

PLEASE HELP Which of the following ordered pairs represents a function?​

Mathematics
1 answer:
nordsb [41]3 years ago
7 0

Answer:

The Answer is b, as the X value does not repeat.

Step-by-step explanation:

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The price of a stove is s dollars. Pedro makes a 10% down payment for a two-year installment purchase. The monthly payment is m
Sever21 [200]

Answer:

0.2

Step-by-step explanation:

8 0
3 years ago
On Friday night 225 people saw the dinosaur exhibit at the museum this amount represents 20% of the people who visited the museu
Igoryamba
20 %  = 225
1%  =  225 / 20
100% =  225 * 100 / 20
           =   1125  answer
8 0
3 years ago
2x + 3y = 3 and x + 6y = -3
jekas [21]

Answer:

Step-by-step explanation:

Multiplying x+6y=-3 with 2 so its equal to first eqn,

Eliminating,

2x + 3y = 3        (1)

2x +12y = -6       (2)                  (because subtracting)

------------------

      -9y =9

     ∴y=-1

eq. y in 1

2x -3=3

2x=3+3

x=6/2

=3

Pls mark me brainliest

4 0
3 years ago
Write an equation in standard form using integers <br><br> Y= - x<br> —<br> 5
kykrilka [37]

Answer:

x + y = - 5

Step-by-step explanation:

The equation of a line in standard form is

Ax + By = C ( A is a positive integer and B, C are integers )

Given

y = - x - 5 ( add x to both sides )

x + y = - 5 ← in standard form

8 0
3 years ago
Prove or disprove (from i=0 to n) sum([2i]^4) &lt;= (4n)^4. If true use induction, else give the smallest value of n that it doe
ddd [48]

Answer:

The statement is true for every n between 0 and 77 and it is false for n\geq 78

Step-by-step explanation:

First, observe that, for n=0 and n=1 the statement is true:

For n=0: \sum^{n}_{i=0} (2i)^4=0 \leq 0=(4n)^4

For n=1: \sum^{n}_{i=0} (2i)^4=16 \leq 256=(4n)^4

From this point we will assume that n\geq 2

As we can see, \sum^{n}_{i=0} (2i)^4=\sum^{n}_{i=0} 16i^4=16\sum^{n}_{i=0} i^4 and (4n)^4=256n^4. Then,

\sum^{n}_{i=0} (2i)^4 \leq(4n)^4 \iff \sum^{n}_{i=0} i^4 \leq 16n^4

Now, we will use the formula for the sum of the first 4th powers:

\sum^{n}_{i=0} i^4=\frac{n^5}{5} +\frac{n^4}{2} +\frac{n^3}{3}-\frac{n}{30}=\frac{6n^5+15n^4+10n^3-n}{30}

Therefore:

\sum^{n}_{i=0} i^4 \leq 16n^4 \iff \frac{6n^5+15n^4+10n^3-n}{30} \leq 16n^4 \\\\ \iff 6n^5+10n^3-n \leq 465n^4 \iff 465n^4-6n^5-10n^3+n\geq 0

and, because n \geq 0,

465n^4-6n^5-10n^3+n\geq 0 \iff n(465n^3-6n^4-10n^2+1)\geq 0 \\\iff 465n^3-6n^4-10n^2+1\geq 0 \iff 465n^3-6n^4-10n^2\geq -1\\\iff n^2(465n-6n^2-10)\geq -1

Observe that, because n \geq 2 and is an integer,

n^2(465n-6n^2-10)\geq -1 \iff 465n-6n^2-10 \geq 0 \iff n(465-6n) \geq 10\\\iff 465-6n \geq 0 \iff n \leq \frac{465}{6}=\frac{155}{2}=77.5

In concusion, the statement is true if and only if n is a non negative integer such that n\leq 77

So, 78 is the smallest value of n that does not satisfy the inequality.

Note: If you compute  (4n)^4- \sum^{n}_{i=0} (2i)^4 for 77 and 78 you will obtain:

(4n)^4- \sum^{n}_{i=0} (2i)^4=53810064

(4n)^4- \sum^{n}_{i=0} (2i)^4=-61754992

7 0
4 years ago
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