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3 years ago

PLEASE HELP! I’m stuck on number 7 and 8! I’ll mark brainliest for best answer

Mathematics

2 answers:

alexandr1967 [171]3 years ago

8 0

Question 7
-5/14

Question 8
17/3

kolbaska11 [484]3 years ago

7 0

Answer:

7.) -5/14

8.) 2 1/3

Step-by-step explanation:

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3 years ago

Graph the linear equation. Find three points that solve the equation, then plot on the graph. -3y=x-6

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Answer:

Step-by-step explanation: x - 6

The given equation can be re-written as y = ---------

-3

Arbitrarily choose x = 0. Then:

x - 6 0-6

y = --------- = ----------- = 2, so (0, 2) is a point on the graph which is also the

-3 -3 y-intercept

Arbitrarily choose x = 6. Then y = 0, and (6, 0) is another point on the graph

which happens to be the x-intercept

arbitrarily choose x = 12. Then y = (12 - 6) / (-3) = -2. Then (12, -2) is

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Plot (12, -2), (6, 0) and (0, 2). Draw a line through these three points.

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3 years ago

You have D dollars to buy fence to enclose a rectangular plot of land (see figure at right). The fence for the top and bottom co

alex41 [277]

The perimeter of the rectangular plot of land is given by the expression below

$P=2x+2y$

On the other hand, since the available money to buy fence is D dollars,

$\begin{gathered} D=4(2x)+3(2y) \\ \Rightarrow D=8x+6y \\ D\rightarrow\text{ constant} \end{gathered}$

Furthermore, the area of the enclosed land is given by

$A=xy$

Solving the second equation for x,

$\begin{gathered} D=8x+6y \\ \Rightarrow x=\frac{D-6y}{8} \end{gathered}$

Substituting into the equation for the area,

$\begin{gathered} A=(\frac{D-6y}{8})y \\ \Rightarrow A=\frac{D}{8}y-\frac{3}{4}y^2 \end{gathered}$

To find the maximum possible area, solve A'(y)=0, as shown below

$\begin{gathered} A^{\prime}(y)=0 \\ \Rightarrow\frac{D}{8}-\frac{3}{2}y=0 \\ \Rightarrow\frac{3}{2}y=\frac{D}{8} \\ \Rightarrow y=\frac{D}{12} \end{gathered}$

Therefore, the corresponding value of x is

$\begin{gathered} y=\frac{D}{12} \\ \Rightarrow x=\frac{D-6(\frac{D}{12})}{8}=\frac{D-\frac{D}{2}}{8}=\frac{D}{16} \end{gathered}$ <h2>Thus, the dimensions of the fence that maximize the area are x=D/16 and y=D/12.</h2><h2>As for the used money,</h2> $\begin{gathered} top,bottom:\frac{8D}{16}=\frac{D}{2} \\ Sides:\frac{6D}{12}=\frac{D}{2} \end{gathered}$ <h2>Half the money was used for the top and the bottom, while the other half was used for the sides.</h2>

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1 year ago