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3 years ago

2) A big league pitching coach tries to limit his pitchers to 110

1 answer:

8 0

P is less than or equal to (<) 110-52 you want to use the < sign with a line under it. The minus 52 is because he has already thrown 52 out of his 110 and P is how many pitches he is alloted

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viktelen [127]

Answer:

altemate interior angles

7 0

3 years ago

Slope of (-2,5)and(7,8)

grigory [225]

Answer:

The slope is 1/3

Step-by-step explanation:

8 – 5 = 3

_____________

7 – -2 = 9

3/9=1/3

4 0

3 years ago

Read 2 more answers

Help! Is it parallel, perpendicular, or neither? Thanks *ignore handwriting*

Digiron [165]

These lines are perpendicular!
y = x
y = -x–2
The slopes are reciprocals of each other, so they are perpendicular.

7 0

3 years ago

Read 2 more answers

Can someone help me with this please?

Karolina [17]

For the first image can you give us the choices for each one?

5 0

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>

7 0

1 year ago