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KatRina [158]
3 years ago
6

Which of the following points is collinear with (2, 1) and (3, 3)?

Mathematics
1 answer:
TiliK225 [7]3 years ago
8 0
A the origin is the correct answer
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Pls help with this math problem
sukhopar [10]

Answer:

y = 1500 - 63x. See the attached for a graph.

Step-by-step explanation:

Each withdrawal will total $63, which includes $60 cash and a $3 fee. Then x withdrawals will total 63x dollars. This amount is subtracted from the initial account balance of 1500 to give the amount remaining.

If y represents the account balance, the equation can be written as

... y = 1500 -63x

The graph will have a y-intercept of 1500, and a slope of -63.

7 0
3 years ago
Which of the following numbers has a value that is between 10% and 1/9? (a) 0.151 (b) 0.112 (c) 0.108 (d) 0.019
Westkost [7]
First you have to change each one into a decimal.

10% = 0.10
1/9 = 0.11

So the answer has to be between 0.10 and 0.11.

A is more than 0.11.
B is more than 0.11.
C is between 0.10 and 0.11.
D is less than 0.10.

The answer is C. 0.108.
6 0
3 years ago
Prove or disprove (from i=0 to n) sum([2i]^4) <= (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
3 years ago
At a craft festival tuan spend $12 for food $19.50 for a small painting and $6 for a straw hat tuan had left $4 how much did tua
NeX [460]
From my info it is 13.5
8 0
3 years ago
9x^2 - 8x - 1 = 0<br> What are the x intercepts?
sasho [114]
9x^2  - 8x - 1 = 0

(9x + 1(x - 1) = 0

x = -1/9 , 1  

x intercepts are -1/9 and 1.
4 0
2 years ago
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