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2 years ago

Solve using system of equations by using elimination method for -5x+4y=2 ; 9x-4y=6

Mathematics

1 answer:

jek_recluse [69]2 years ago

6 0

Answer:

y = 3

x = 2

Step-by-step explanation:

-5x + 4y = 2

9x - 4y = 6

Sum both eq.

-5x + 9x = 4x

+4y - 4y = 0

2 + 6 = 8

then:

4x + 0 = 8

4x = 8

x = 8/4

x = 2

from the first eq.

-5x + 4y = 2

-5*2 + 4y = 2

-10 + 4y = 2

4y = 2 + 10

4y = 12

y = 12/4

y = 3

Check:

from the second eq.

9x - 4y = 6

9*2 - 4*3 = 6

18 - 12 = 6

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A coffee supply store waits until the orders for its special blend reach 100 pounds before making up a batch. Columbian coffee s

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X - Columbian coffee ( in pounds),
y - Brazilian coffee ( in pounds );
x + y = 100
8.85 x + 3.85 y = 6.55 ( x + y )
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x = 100 - y
8.85 ( 100 - y ) + 3.85 y = 6.55 * 100
855 - 8.55 y + 3.85 y = 655
- 5 y = - 230
y = ( - 230 ) : ( - 5 )
y = 46
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Answer: 54 pounds of Columbian coffee and 46 pounds of Brazilian coffee should be used.

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3 years ago

Point B has coordinates (1,2). The x-coordinate of point A is -8. The distance between point A and point B is 15 units. What

Xelga [282]

Answer:

The possible coordinates of point A are $A_{1} (x,y) = (-8, 14)$ and $A_{2} (x,y) = (-8, -10)$ , respectively.

Step-by-step explanation:

From Analytical Geometry, we have the Equation of the Distance of a Line Segment between two points:

$l_{AB} = \sqrt{(x_{B}-x_{A})^{2} + (y_{B}-y_{A})^{2}}$ (1)

Where:

$l_{AB}$ - Length of the line segment AB.

$x_{A}, x_{B}$ - x-coordinates of points A and B.

$y_{A}, y_{B}$ - y-coordinates of points A and B.

If we know that $l_{AB} = 15$ , $x_{A} = -8$ , $x_{B} = 1$ and $y_{B} = 2$ , then the possible coordinates of point A is:

$\sqrt{(1+8)^{2}+(2-y_{A})^{2}} = 15$

$81 + (2-y_{A})^{2} = 225$

$(2-y_{A})^{2} = 144$

$2-y_{A} = \pm 12$

There are two possible solutions:

1) $2-y_{A} = -12$

$y_{A} = 14$

2) $2 - y_{A} = 12$

$y_{A} = -10$

The possible coordinates of point A are $A_{1} (x,y) = (-8, 14)$ and $A_{2} (x,y) = (-8, -10)$ , respectively.

8 0

3 years ago

Expand $\left(x^2 \frac{1}{x}\right)^3$. (Write the terms with higher degree first, so for example an $x^2$ term would come befo

Ilya [14]

Answer:

$x^6+3x^3+3+\dfrac{1}{x^3}$

Step-by-step explanation:

We can use the pascal triangle to find the coefficients of the expasion of the cube. It holds that

$(x^2 + \dfrac{1}{x})^3=(x^2)^3+3(x^2)^2\dfrac{1}{x}+3x^2(\dfrac{1}{x})^2+(\dfrac{1}{x})^3\\\\=x^6+3x^4\dfrac{1}{x}+3x^2\dfrac{1}{x^2}+\dfrac{1}{x^3}\\\\=x^6+3x^3+3+\dfrac{1}{x^3}$