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Darya [45]
3 years ago
5

Please help 100 points. Picture in the desc

Mathematics
2 answers:
zvonat [6]3 years ago
8 0
Given: (x1 , y1)= (2,2) and (X2,y2)= (-1,-2)

then using slope formula:
∆y/∆x= (y2-y1 / x2-x1)= (-2-2 / -1-2)= (-4/-3)= (4/3)

hence 4/3 is the slope of the line
DENIUS [597]3 years ago
4 0

rise over run

(-1,2) (2,2)

2-2/2-(-1)

2-2/2+1 equals 0/3

or alternatively

2-2/-1 - 2 which would be -0/3.

Whichever point is the first helps determine this alot

rise/run=rise over run= y2 - y1/ x2 - x1

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10 POINTS PLEASE HELP
ICE Princess25 [194]

Volume= whl

W= 4 ft

H= 2ft

L= 5ft

Multiply all (40), then multiply by the three times it was full.

Answer: 120ft^3

4 0
3 years ago
Someone please help
Alex777 [14]

Answer:

put it in desmos


Step-by-step explanation:


8 0
3 years ago
Given that a*b = 2a - 3b, then 2*(-3) =<br><br><br>​
belka [17]

Answer:

2*(-3)= -6

Step-by-step explanation:

I do not see how "a*b=2a-3b" would change the fact 2 times negative 3 is -6

6 0
3 years ago
A company charges $7 for a T-shirt and ships any order for $15. A school principal ordered a number of T-shirts for the school s
rosijanka [135]
7x + 15 = 1520
It’s $7 times the unknown number of shirts (x). Plus $15 to ship. All has to equal the total of $1,520.
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
3 years ago
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