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algol [13]
3 years ago
14

A bag grass seed covers 3750 square feet and costs $27.50. If a gardener is reseeding a city park that has the shape of a right

triangle with a base of B feet and a height of H feet, write an equation for the cost, of reseeding the park.
Mathematics
2 answers:
solmaris [256]3 years ago
4 0

Answer:

C = (13.75 * B * H)/3750

Step-by-step explanation:

The area of a triangle is represented by the equation A = 1/2 * B * H

Every time another 3750 square feet is covered, $27.50 is spent. So, we can represent the cost using the equation C = A/3750 * 27.50

Plugging in the base and height, C = (1/2 * B * H * 27.50)/3750

C = (13.75 * B * H)/3750

Blizzard [7]3 years ago
3 0
The answer would be (base times height)/3750, and then take that answer and multiply by 27.50

[(BxH)/3759]x27.50
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<h2><u>1. Determining the value of x and y:</u></h2>

Given equation(s):

  • y = 3x - 1
  • 3x + y = -7

To determine the point of intersection given by the two equations, it is required to know the x-value and the y-value of both equations. We can solve for the x and y variables through two methods.

<h3 /><h3><u>Method-1: Substitution method</u></h3>

Given value of the y-variable: 3x - 1

Substitute the given value of the y-variable into the second equation to determine the value of the x-variable.

\implies 3x + y = -7

\implies3x + (3x - 1) = -7

\implies3x + 3x - 1 = -7

Combine like terms as needed;

\implies 3x + 3x - 1 = -7

\implies 6x - 1 = -7

Add 1 to both sides of the equation;

\implies 6x - 1 + 1 = -7 + 1

\implies 6x = -6

Divide 6 to both sides of the equation;

\implies \dfrac{6x}{6}  = \dfrac{-6}{6}

\implies x = -1

Now, substitute the value of the x-variable into the expression that is equivalent to the y-variable.

\implies y = 3(-1) - 1

\implies     \ \ = -3 - 1

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Therefore, the value(s) of the x-variable and the y-variable are;

\boxed{x = -1}   \boxed{y = -4}

<h3 /><h3><u>Method 2: System of equations</u></h3>

Convert the equations into slope intercept form;

\implies\left \{ {{y = 3x - 1} \atop {3x + y = -7}} \right.

\implies \left \{ {{y = 3x - 1} \atop {y = -3x - 7}} \right.

Clearly, we can see that "y" is isolated in both equations. Therefore, we can subtract the second equation from the first equation.

\implies \left \{ {{y = 3x - 1 } \atop {- (y = -3x - 7)}} \right.

\implies \left \{ {{y = 3x - 1} \atop {-y = 3x + 7}} \right.

Now, we can cancel the "y-variable" as y - y is 0 and combine the equations into one equation by adding 3x to 3x and 7 to -1.

\implies\left \{ {{y = 3x - 1} \atop {-y = 3x + 7}} \right.

\implies 0 = (6x) + (6)

\implies0 = 6x + 6

This problem is now an algebraic problem. Isolate "x" to determine its value.

\implies 0 - 6 = 6x + 6 - 6

\implies -6 = 6x

\implies -1 = x

Like done in method 1, substitute the value of x into the first equation to determine the value of y.

\implies y = 3(-1) - 1

\implies y = -3 - 1

\implies y = -4

Therefore, the value(s) of the x-variable and the y-variable are;

\boxed{x = -1}   \boxed{y = -4}

<h2><u>2. Determining the intersection point;</u></h2>

The point on a coordinate plane is expressed as (x, y). Simply substitute the values of x and y to determine the intersection point given by the equations.

⇒ (x, y) ⇒ (-1, -4)

Therefore, the point of intersection is (-1, -4).

<h3>Graph:</h3>

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