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polet [3.4K]
3 years ago
7

Y = 3x^5 - 10x^3slove the minumon points ​

Mathematics
1 answer:
olya-2409 [2.1K]3 years ago
6 0
I believe the answer is 1.414214
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Suppose there is a pile of quarters dimes and pennies with a total value of $1.07 how much of each coin can be present without b
Korolek [52]
Hello,

Very nice as problem.

2 solutions:
1 quater,8 dimes, 2 pennies
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3 quaters,3 dimes, 2 pennies

since
107=( 0, 0, 107) but : 100= 0*25+ 0*10+ 100
107=( 0, 1, 97) but : 100= 0*25+ 1*10+ 90
107=( 0, 2, 87) but : 100= 0*25+ 2*10+ 80
107=( 0, 3, 77) but : 100= 0*25+ 3*10+ 70
107=( 0, 4, 67) but : 100= 0*25+ 4*10+ 60
107=( 0, 5, 57) but : 100= 0*25+ 5*10+ 50
107=( 0, 6, 47) but : 100= 0*25+ 6*10+ 40
107=( 0, 7, 37) but : 100= 0*25+ 7*10+ 30
107=( 0, 8, 27) but : 100= 0*25+ 8*10+ 20
107=( 0, 9, 17) but : 100= 0*25+ 9*10+ 10
107=( 0, 10, 7) but : 100= 0*25+ 10*10+ 0
107=( 1, 0, 82) but : 100= 1*25+ 0*10+ 75
107=( 1, 1, 72) but : 100= 1*25+ 1*10+ 65
107=( 1, 2, 62) but : 100= 1*25+ 2*10+ 55
107=( 1, 3, 52) but : 100= 1*25+ 3*10+ 45
107=( 1, 4, 42) but : 100= 1*25+ 4*10+ 35
107=( 1, 5, 32) but : 100= 1*25+ 5*10+ 25
107=( 1, 6, 22) but : 100= 1*25+ 6*10+ 15
107=( 1, 7, 12) but : 100= 1*25+ 7*10+ 5
107=( 1, 8, 2) is good
107=( 2, 0, 57) but : 100= 2*25+ 0*10+ 50
107=( 2, 1, 47) but : 100= 2*25+ 1*10+ 40
107=( 2, 2, 37) but : 100= 2*25+ 2*10+ 30
107=( 2, 3, 27) but : 100= 2*25+ 3*10+ 20
107=( 2, 4, 17) but : 100= 2*25+ 4*10+ 10
107=( 2, 5, 7) but : 100= 2*25+ 5*10+ 0
107=( 3, 0, 32) but : 100= 3*25+ 0*10+ 25
107=( 3, 1, 22) but : 100= 3*25+ 1*10+ 15
107=( 3, 2, 12) but : 100= 3*25+ 2*10+ 5
107=( 3, 3, 2) is good
107=( 4, 0, 7) but : 100= 4*25+ 0*10+ 0



4 0
3 years ago
Consider a single spin on the spinner shown below. A circle is split into sections 1, 2, 4, and 3. A spinner is pointing at numb
vladimir2022 [97]

Answer:

The answer is "Options A, B, and E represent mutually exclusive events".

Step-by-step explanation:

Two occurrences that can happen immediately called mutually incompatible. Let's now glance at our options and figure out where the statements are mutually incompatible events.

In Option A: You could see that landing on an unwanted portion and arriving on 2 are events that are locally incompatible, even as undesirable portion contains 3 and 4, and 2 were shaded.

In Option B: Arriving on a shaded part and falling on 3 are also mutually incompatible because there are 3 on a windows azure.

In Option C: A darkened portion and an increasing amount can land while 2 would be an even number as well as on the shaded portion. That number is very much the same.

In Option D: At the same time as 4 is greater than 3 and it is situated upon an undistressed section, landing and attracting a number larger than 3 can happen.

In Option E: Landing on a shaded part and landing on even a shaded part is an excluding event, since shaders may either be shaded or unlit.

3 0
3 years ago
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Determine which system below will produce infinitely many solutions.
mariarad [96]
A system that will produce infinitely many solutions will have the property where all the equations overlap. That means they will have the same slope and y-intercept.

An easy way to determine if they will produce infinitely many solutions is to put all the equations into y=mx+b form. If everything is the same, you know it will have infinite solutions.


4 0
3 years ago
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A person invests 2000 dollars in a bank. The bank pays 4.75% interest compounded
grin007 [14]

Answer:

t = 2.7

Step-by-step explanation:

1. Use formula provided

2. A = 3500 B = 2000 r = 4.75 n = 12months/4 = 3 months t = ?

3. Solve

3500=2000(1+4.75/3)^3t

3t=logbase(31/12)*1500

t=logbase(31/12)*1500 / 3

t ≅2.568

4 0
3 years ago
Read 2 more answers
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