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strojnjashka [21]
3 years ago
11

Midpoint of (-5, 5) (2, -5)

Mathematics
1 answer:
Charra [1.4K]3 years ago
3 0

Answer:

(-3/2,0)

Step-by-step explanation:

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Simplify completely quantity 5 x squared minus 21 x minus 20 all over quantity 5 x squared minus 16 x minus 16.
Morgarella [4.7K]
(5x^2-21x-20)/(5x^2-16x-16)
((x-5)(5x+4))/((x-4)(5x+4))
(x-5)/(x-4)
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The engine life for a certain brand of lawnmower is modeled by the function f(h)=50-1/3h where h represents hours used. What doe
Nadusha1986 [10]

Answer:

C)

We are given the function f(h)=50-(1/3)h

Step-by-step explanation:

Which means that the initial life of the lawn mower is 50 years with every year, the engine life of the mower is reduced by 1/3 of the time passed

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2 years ago
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It is not linear, as the 1st hour charged $25, and the 2nd hour charged $40, while the third hour charged $60, that's a $15 difference then a $20
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Multiplying a trinomial by a trinomial follows the same steps as multiplying a binomial by a trinomial. Determine the degree and
FinnZ [79.3K]

Answer: Degree of polynomial (highest degree) =4

Maximum possible terms =9

Number of terms in the product = 5

Step-by-step explanation:

A trinomial is a polynomial with 3 terms.

The given product of trinomial: (x^2 + x + 2)(x^2 - 2x + 3)

By using distributive property: a(b+c+d)= ab+ac+ad

(x^2 + x + 2)(x^2 - 2x + 3)=(x^2 + x + 2) x^2+(x^2 + x + 2) (-2x)+(x^2 + x + 2)(3)\\\\=x^2(x^2)+x(x^2)+2(x^2)+x^2 (-2x)+x (-2x)+2 (-2x)+x^2 (3)+x (3)+2 (3)\\\\\\=x^4+x^3+2x^2-2x^3-2x^2-4x+3x^2+3x+6

Maximum possible terms =9

Combine like terms

x^4+x^3-2x^3+3x^2-4x+3x+6\\\\=x^4-x^3+3x^2-x+6

Hence, \left(x^2\:+\:x\:+\:2\right)\left(x^2\:-\:2x\:+\:3\right)=x^4-x^3+3x^2-x+6

Degree of polynomial (highest degree) =4

Number of terms = 5

6 0
3 years ago
Prove by contradiction that the interval (a,<br> b.has no minimum element.
Snowcat [4.5K]

Assume (a,b) has a minimum element m.


m is in the interval so a < m < b.


a < m


Adding a to both sides,


2a < a + m


Adding m to both sides of the first inequality,


a + m < 2m


So


2a < a+m < 2m


a < (a+m)/2 < m < b


Since the average (a+m)/2 is in the range (a,b) and less than m, that contradicts our assumption that m is the minimum. So we conclude there is no minimum since given any purported minimum we can always compute something smaller in the range.


8 0
4 years ago
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