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Luden [163]
2 years ago
5

2x + 5y = 14 4x + 2y = -4 solving with elimination​

Mathematics
2 answers:
melamori03 [73]2 years ago
8 0

Answer:

(- 3, 4 )

Step-by-step explanation:

2x + 5y = 14 → (1)

4x + 2y = - 4 → (2)

multiplying (1) by - 2 and adding to (2) will eliminate the x- term

- 4x - 10y = - 28 → (3)

add (2) and (3) term by term to eliminate x

0 - 8y = - 32

- 8y = - 32 ( divide both sides by - 8 )

y = 4

substitute y = 4 into either of the 2 equations and solve for x

substituting into (1)

2x + 5(4) = 14

2x + 20 = 14 ( subtract 20 from both sides )

2x = - 6 ( divide both sides by 2 )

x = - 3

solution is (- 3, 4 )

m_a_m_a [10]2 years ago
6 0

Answer: x=-\frac{3}{179} =\\-0.016759777\\y=-\frac{142}{179} =\\-0.793296089

Step-by-step explanation: steps using substitution

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emmainna [20.7K]

Answer:

1+1=2

Step-by-step explanation:

3 0
2 years ago
Find the x- and y-intercepts of the graph of the equation y = x2 – 2.
sertanlavr [38]

Answer:

See below ~

Step-by-step explanation:

<u>X-intercept(s) : Take y = 0</u>

  • 0 = x² - 2
  • x = ±√2
  • X-intercepts are : (-√2, 0) and (√2, 0)

<u>Y-intercept : Take x = 0</u>

  • y = (0)² - 2
  • y = -2
  • Y-intercept is : (0, -2)
3 0
1 year ago
Rewrite the polynomial expression in factored form: 60x4−12x3−135x2+27x
Jet001 [13]

Answer:

3x(2x - 3)(2x + 3)(5x - 1).

Step-by-step explanation:

60x4 − 12x3 − 135x2 + 27x

= 3x(20x^3 - 4x^2 - 45x + 9)    We can factor what's in the parentheses by grouping:

=  3x(4x^2( 5x - 1) - 9(5x - 1))

= 3x (4x^2 - 9)(5x - 1)                 Now we factors  4x^2 - 9:

= 3x(2x - 3)(2x + 3)(5x - 1).

7 0
3 years ago
What is the distance between pints (-2 -5) and (3 -6)
sukhopar [10]

Answer:

\displaystyle d = \sqrt{26}

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

<u>Algebra II</u>

  • Distance Formula: \displaystyle d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

Step-by-step explanation:

<u>Step 1: Define</u>

Point (-2, -5)

Point (3, -6)

<u>Step 2: Find distance </u><em><u>d</u></em>

Simply plug in the 2 coordinates into the distance formula to find distance<em> d</em>

  1. Substitute in points [DF]:                     \displaystyle d = \sqrt{(3--2)^2+(-6--5)^2}
  2. (Parenthesis) Simplify:                         \displaystyle d = \sqrt{(3+2)^2+(-6+5)^2}
  3. (Parenthesis) Add:                               \displaystyle d = \sqrt{(5)^2+(-1)^2}
  4. [√Radical] Exponents:                         \displaystyle d = \sqrt{25+1}
  5. [√Radical] Add:                                    \displaystyle d = \sqrt{26}
4 0
3 years ago
Given is a finite set of spherical planets, all of the same radius and no two intersecting. On the surface of each planet consid
wariber [46]

Answer:

Step-by-step explanation:

To answer this question, First, we define a direction in space, which defines

the North Pole of each planet.

We also assume that the direction is not at 90° perpendicular to the axis of any two planet.

Assume we now define an order on the set of planet by saying that planet A ≥ B, when removing all the other planet from space, the north pole B, is visible from A.

If we refer to the direction of the planet in the diagram, we discover that,

1, A ≥ A

2, If A ≥ B and B≥ A,

Then A = B

3, If A ≥ B, and B ≥ C , then A ≥ C

4, Either A ≥ B or B ≥ A

It should also be noted that the array of order above has a unique maximal element  (M). This is the only  planet whose North Pole is not visible from another.

If we now consider a sphere of the same radius as the planets. Remove

from it all North Poles defined by directions that are perpendicular

to the axis of two of the planets. This is a set of area zero.

For every other point on this sphere, there exists a direction in

space that makes it the North Pole, and for that direction, there

exists a unique North Pole on one of the planets which is not visible

from the others. Hence the total area of invisible points is equal

to the area of this sphere, which in turn is the area of one of the  planets.

5 0
2 years ago
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