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3 years ago

What is the solution to this system of equations? -5.9x-3.7y=-2.1 5.9x+3.7y=2.1

Mathematics

2 answers:

marin [14]3 years ago

8 0

Answer:

There are infinite solutions that solve the system $y=0.57-1.59x$

Step-by-step explanation:

The solution can be expressed as $y=0.57-1.59x$ (there are infinite solutions that solve the system). This results from rearranging terms in the first or second equation.

This is because the first equation is a linear combination of the second one (this is, the first equation equals (-1) times the second equation). This means that the second equation does not add any information about x and y: it says exactly the same as the first equation about the relationship between x and y.

Then, in terms of solving the system, you have one unique expression (not repeated) that shows the relationship between y and x. This means that, any pair of (x,y) that meet the requirements expressed by $y=0.57-1.59x$ , will solve the system. Then we have infinite solutions.

Novosadov [1.4K]3 years ago

7 0

Answer:

the answer is C

Step-by-step explanation:

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Okay now using distributive property write 2(4x-9) without parenthesis

riadik2000 [5.3K]

2(4x - 9) =
2 * 4x - 2 * 9 =
8x - 18

Hope this helps!!
~Kiwi

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2 years ago

Find the height of the building below by using your knowledge of trigonometry.

Lilit [14]

Answer:

50 $\sqrt{3}$

Step-by-step explanation:

The figure represents a 30-60-90 special triangle

in this special right triangle the measure of sides lengths is represented by

a, a $\sqrt{3}$ , and 2a

The side length that sees angle measure 30 is represented by a and here we see a = 50m

The height of the building, which sees angle measure 60, is equal to

50 $\sqrt{3}$

5 0

2 years ago

An instructor who taught two sections of Math 161A, the first with 20 students and the second with 30 students, gave a midterm e

uranmaximum [27]

Answer:

The answers are for option (a) 0.2070 (b)0.3798 (c) 0.3938

Step-by-step explanation:

Given:

Here Section 1 students = 20

Section 2 students = 30

Here there are 15 graded exam papers.

(a )Here Pr(10 are from second section) = ²⁰C₅ * ³⁰C₁₀/⁵⁰C₁₅= 0.2070

(b) Here if x is the number of students copies of section 2 out of 15 exam papers.

here the distribution is hyper-geometric one, where N = 50, K = 30 ; n = 15

Then,

Pr( x ≥ 10 ; 15; 30 ; 50) = 0.3798

(c) Here we have to find that at least 10 are from the same section that means if x ≥ 10 (at least 10 from section B) or x ≤ 5 (at least 10 from section 1)

so,

Pr(at least 10 of these are from the same section) = Pr(x ≤ 5 or x ≥ 10 ; 15 ; 30 ; 50) = Pr(x ≤ 5 ; 15 ; 30 ; 50) + Pr(x ≥ 10 ; 15 ; 30 ; 50) = 0.0140 + 0.3798 = 0.3938

Note : Here the given distribution is Hyper-geometric distribution

where f(x) = kCₓ)(N-K)C(n-x)/ NCK in that way all these above values can be calculated.

7 0

3 years ago

Solve for the equation for all x values by completing the square x^2 - 27= 67

navik [9.2K]

Answer:

The solutions for the equation $x^2\:-\:27=\:67$ are $x=\sqrt{94},\:x=-\sqrt{94}$ .