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lesya692 [45]
3 years ago
7

-2.3 + 1.6y = 6.1 solve for y please help fast!!

Mathematics
1 answer:
Elan Coil [88]3 years ago
8 0

Answer:

The answer is y = 5.25

Step-by-step explanation:

-2.3 + 1.6y = 6.1

1.6y = 6.1 + 2.3

1.6y = 8.4

y = 8.4/1.6

y = 5.25

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Find the zeros of the following polynomial functions, with their multiplicities. (a) f(x)= (x +1)(x − 1)(x² +1) (b) g(x) = (x −
larisa [96]

Answer:

a) zeros of the function are x = 1 and, x = -1

b) zeros of the function are x = 2 and, x = 4

c) zeros of the function are x = \frac{3}{2}

d) zeros of the function are x = \frac{-4}{3}  and, x = 17

Step-by-step explanation:

Zeros of the function are the values of the variable that will lead to the result of the equation being zero.

Thus,

a) f(x)= (x +1)(x − 1)(x² +1)

now,

for the (x +1)(x − 1)(x² +1) = 0

the condition that must be followed is

(x +1) = 0 ..........(1)

or

(x − 1) = 0 ..........(2)

or

(x² +1) = 0 ...........(3)

considering the equation 1, we have

(x +1) = 0

or

x = -1

for

(x − 1) = 0

x = 1

and,

for (x² +1) = 0

or

x² = -1

or

x = √(-1)         (neglected as it is a imaginary root)

Thus,

zeros of the function are x = 1 and, x = -1

b) g(x) = (x − 4)³(x − 2)⁸

now,

for the (x − 4)³(x − 2)⁸ = 0

the condition that must be followed is

(x − 4)³ = 0 ..........(1)

or

(x − 2)⁸ = 0 ..........(2)

considering the equation 1, we have

(x − 4)³ = 0

or

x -4 = 0

or

x = 4

and,

for (x − 2)⁸ = 0

or

x - 2 = 0

or

x = 2        

Thus,

zeros of the function are x = 2 and, x = 4

c) h(x) = (2x − 3)⁵

now,

for the (2x − 3)⁵ = 0

the condition that must be followed is

(2x − 3)⁵ = 0

or

2x - 3 = 0

or

2x = 3

or

x = \frac{3}{2}

Thus,

zeros of the function are x = \frac{3}{2}

d)   k(x) =(3x +4)¹⁰⁰(x − 17)⁴

now,

for the (3x +4)¹⁰⁰(x − 17)⁴ = 0

the condition that must be followed is

(3x +4)¹⁰⁰ = 0 ..........(1)

or

(x − 17)⁴ = 0 ..........(2)

considering the equation 1, we have

(3x +4)¹⁰⁰ = 0

or

(3x +4) = 0

or

3x = -4

or

x = \frac{-4}{3}

and,

for (x − 17)⁴ = 0

or

x - 17 = 0

or

x = 17        

Thus,

zeros of the function are x = \frac{-4}{3}  and, x = 17

7 0
3 years ago
PLEASE HELP ME OUT PLEASE!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
charle [14.2K]
If they sold 10 hammers and each hammer sells together with a package of nails for $12, that would be $120 total. Now subtract that from $950.50 and you get $830.50. Divide that by the remaining packages of nails sold (110 because you already sold 10 together with the hammers) and you get $7.55. Each package of nails sold for $7.55. Hope that helps! 
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Rhonda decides to pay a $300 fee in 4 payments. The first two payments are each 15% of the original fee. The third payment is 25
SCORPION-xisa [38]

Answer:

Step-by-step explanation:

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4 years ago
A photograph in a catalog has a scale of 2 in: 5 ft. A tent in the photograph is 3 inches long.
VladimirAG [237]

Answer:

7.5 ft

Step-by-step explanation:

divide 5 by 2 -- 5 ÷ 2 = 2.5

now multiply 2.5 by 3 -- 2.5 × 3 = 7.5

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4 years ago
Here is a linear equation in two variables: 2x+4y−31=123
Paha777 [63]

Answer:

y=−11x+77/2

Step-by-step explanation:

The procedure for solving simultaneous linear equations now called Gaussian elimination appears in the ancient Chinese mathematical text Chapter Eight: Rectangular Arrays of The Nine Chapters on the Mathematical Art. Its use is illustrated in eighteen problems, with two to five equations.[4]

Systems of linear equations arose in Europe with the introduction in 1637 by René Descartes of coordinates in geometry. In fact, in this new geometry, now called Cartesian geometry, lines and planes are represented by linear equations, and computing their intersections amounts to solving systems of linear equations.

The first systematic methods for solving linear systems used determinants, first considered by Leibniz in 1693. In 1750, Gabriel Cramer used them for giving explicit solutions of linear systems, now called Cramer's rule. Later, Gauss further described the method of elimination, which was initially listed as an advancement in geodesy.[5]

In 1844 Hermann Grassmann published his "Theory of Extension" which included foundational new topics of what is today called linear algebra. In 1848, James Joseph Sylvester introduced the term matrix, which is Latin for womb.

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Arthur Cayley introduced matrix multiplication and the inverse matrix in 1856, making possible the general linear group. The mechanism of group representation became available for describing complex and hypercomplex numbers. Crucially, Cayley used a single letter to denote a matrix, thus treating a matrix as an aggregate object. He also realized the connection between matrices and determinants, and wrote "There would be many things to say about this theory of matrices which should, it seems to me, precede the theory of determinants".[5]

Benjamin Peirce published his Linear Associative Algebra (1872), and his son Charles Sanders Peirce extended the work later.[6]

The telegraph required an explanatory system, and the 1873 publication of A Treatise on Electricity and Magnetism instituted a field theory of forces and required differential geometry for expression. Linear algebra is flat differential geometry and serves in tangent spaces to manifolds. Electromagnetic symmetries of spacetime are expressed by the Lorentz transformations, and much of the history of linear algebra is the history of Lorentz transformations.

The first modern and more precise definition of a vector space was introduced by Peano in 1888;[5] by 1900, a theory of linear transformations of finite-dimensional vector spaces had emerged. Linear algebra took its modern form in the first half of the twentieth century, when many ideas and methods of previous centuries were generalized as abstract algebra. The development of computers led to increased research in efficient algorithms for Gaussian elimination and matrix decompositions, and linear algebra became an essential tool for modelling and simulations.[5]

Vector spaces

Main article: Vector space

Until the 19th century, linear algebra was introduced through systems of linear equations and matrices. In modern mathematics, the presentation through vector spaces is generally preferred, since it is more synthetic, more general (not limited to the finite-dimensional case), and conceptually simpler, although more abstract.

A vector space over a field F (often the field of the real numbers) is a set V equipped with two binary operations satisfying the following axioms. Elements of V are called vectors, and elements of F are called scalars. The first operation, vector addition, takes any two vectors v and w and outputs a third vector v + w. The second operation, scalar multiplication, takes any scalar a and any vector v and outputs a new vector av. The axioms that addition and scalar multiplication must satisfy are the following. (In the list below, u, v and w are arbitrary elements of V, and a and b are arbitrary scalars in the field F.)[7]

8 0
3 years ago
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