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3 years ago

A flower garden has been designed with an L-shaped walkway and viewing area around two sides as shown.

Mathematics

1 answer:

nikdorinn [45]3 years ago

8 0

Answer:

c. 2.4 yards

Step-by-step explanation:

If x is the width of the walkway, then the total landscape area is ...

(4.5 +x)(7.5 +x) = 68.31

This quadratic equation can be rearranged to standard form and solved in any of the usual ways.

(x +4.5)(x +7.5) -68.31 = 0

x² +12x -34.56 = 0

Using the quadratic formula, we have ...

x = (-12 ±√(12² -4(1)(-34.56)))/2 = -6 ± √70.56 = -6 ± 8.4

The positive solution is x = 2.4.

The width of the walkway is 2.4 yards.

_____

A graphing calculator or spreadsheet solver can also find the solution for you.

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Two trains, train a and train , weigh a total of 466 tons. Train A is heavier than train b. The difference of there weights is 3

AlexFokin [52]

Answer: A = 396 tons

B = 70 tons

Step-by-step explanation:

From the question,

Weight of A + Weight of B = 466

Weight of A - Weight of B = 326

A + B = 466 ......... i

A - B = 326 ..........ii

Subtract equation ii from I

2B = 140

B = 140/2

B = 70 tons

A = 466 - 70

A = 396 tons

5 0

3 years ago

What is the area of this figure?

Ugo [173]

Answer:

11610

Step-by-step explanation:

First you do the top rectangle which the area for that is 144 which is 4 times 4 times 3 times 3.

Then you do the middle rectangle which the area is 441 which is 7 times 7 times 3 times 3.

Then you do the last rectangle which the area is 11025 which is 15 times 15 times 7 times 7.

Then lastly you add all three areas which is 144+441+11025=11610. So therefore, the answer is 11610.

4 0

3 years ago

Solve using elimination<br> x+y-2z=8<br> 5x-3y+z=-6<br> -2x-y+4z=-13

Free_Kalibri [48]

So here is your answer with LaTeX issued format interpretation. Full process elucidated briefly, below:

$\begin{alignedat}{3}x + y - 2z = 8 \\ 5x - 3y + 2 = - 6 \\ - 2x - y + 4z = - 13 \end{alignedat}$

For this equation to get obtained under the impression of those variables we have to eliminate them individually for moving further and simplifying the linear equation with three variables along the axis.

Multiply the equation of x + y - 2z = 8 by a number with a value of 5; Here this becomes; 5x + 5y - 10z = 40; So:

$\begin{alignedat}{3}5x + 5y - 10z = 40 \\ 5x - 3y + z = - 6 \\ - 2x - y + 4z = - 13 \end{alignedat}$

Pair up the equations in a way to eliminate the provided variable on our side, that is; "x":

5x - 3y + z = - 6

-

5x + 5y - 10z = 40
______________

- 8y + 11z = - 46

Therefore, we are getting.

$\begin{alignedat}{3}5x + 5y - 10z = 40 \\ - 8y + 11z = - 46 \\ - 2x - y + 4z = - 13 \end{alignedat}$

Multiply the equation of 5x + 5y - 10z = - 40 by a number with a value of 2; Here this becomes; 10x + 10y - 20z = 80.

Multiply the equation of - 2x - y + 4z = - 13 by a number with a value of 5; Here this becomes; - 10x - 5y + 20z = - 65; So:

$\begin{alignedat}{3}10x + 10y - 20z = 80 \\ - 8y + 11z = - 46 \\ - 10x - 5y + 20z = - 65 \end{alignedat}$

Pair up the equations in a way to eliminate the provided variables on our side, that is; "x" and "z":

- 10x - 5y + 20z = - 65

+
10x + 10y - 20z = 80
__________________

5y = 15

$\begin{alignedat}{3}10x + 10y - 20z = 80 \\ - 8y + 11z = - 46 \\ 5y = 15 \end{alignedat}$

Multiply the equation of - 8y + 11z = - 46 by a number with a value of 5; Here this becomes; - 40y + 55z = - 230.

Multiply the equation of 5y = 15 by a number with a value of 8; Here this becomes; 40y = 120; So:

$\begin{alignedat}{3}10x + 10y - 20z = 80 \\ - 40y + 55z = - 690 \\ 40y = 120 \end{alignedat}$

Pair up the equations in a way to eliminate the provided variables on our side, that is; "y":

40y = 120

+

- 40y + 55z = - 230
_________________

55z = - 110

$\begin{alignedat}{3}10x + 10y - 20z = 80 \\ - 40y + 55z = - 230 \\ 55z = - 110 \end{alignedat}$

Solving for the variable of 'z':

$\mathsf{55z = - 110}$

$\bf{\dfrac{55z}{55} = \dfrac{-110}{55}}$

Cancel out the common factor acquired on the numerator and denominator, that is, "55":

$z = - \dfrac{\overbrace{\sout{110}}^{2}}{\underbrace{\sout{55}}_{1}}$

$\boxed{\mathbf{z = - 2}}$

Solving for variable "y":

$\mathbf{\therefore \quad - 40y - 55 \big(- 2 \big) = - 230}$

$\mathbf{- 40y - 55 \times 2 = - 230}$

$\mathbf{- 40y - 110 = - 230}$

$\mathbf{- 40y - 110 + 110 = - 230 + 110}$

Adding the numbered value as 110 into this equation (in previous step).

$\mathbf{- 40y = - 120}$

Divide by - 40.

$\mathbf{\dfrac{- 40y}{- 40} = \dfrac{- 120}{- 40}}$

$\mathbf{y = \dfrac{- 120}{- 40}}$

$\boxed{\mathbf{y = 3}}$

Solve for variable "x":

$\mathbf{10x + 10y - 20z = 80}$

$\mathbf{Since, \: z = - 2; \quad y = 3}$

$\mathbf{10x + 10 \times 3 - 20 \times (- 2) = 80}$

$\mathbf{10x + 10 \times 3 + 20 \times 2 = 80}$

$\mathbf{10x + 30 + 20 \times 2 = 80}$

$\mathbf{10x + 30 + 40 = 80}$

$\mathbf{10x + 70 = 80}$

$\mathbf{10x + 70 - 70 = 80 - 70}$

$\mathbf{10x = 10}$

Divide by this numbered value $\mathbf{10}$ to get the final value for the variable "x".

$\mathbf{\dfrac{10x}{10} = \dfrac{10}{10}}$

The numbered values in the numerator and the denominator are the same, on both the sides. This will mean the "x" variable will be left on the left hand side and numbered values "10" will give a product of "1" after the division is done. On the right hand side the numbered values get divided to obtain the final solution for final system of equation for variable "x" as "1".

$\boxed{\mathbf{x = 1}}$

Final solutions for the respective variables in the form of " (x, y, z) " is:

$\boxed{\mathbf{\underline{\Bigg(1, \: \: 3, \: \: - 2 \Bigg)}}}$

Hope it helps.

8 0

3 years ago

Read 2 more answers

Question 1 (33 points)

myrzilka [38]

Answer:

lol idk

Step-by-step explanation:

look it up

6 0

3 years ago

Please help to write the following

Vanyuwa [196]

Answer: option 2. x²-1

Step-by-step explanation:

concept to know: a²-b²=(a+b)(a-b)

------------

(a+b)(a-b)=a²+b²

(x+1)(x-1)=x²-1

Hope this helps!! :)

Please let me know if you have any question

4 0

3 years ago

Read 2 more answers