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4vir4ik [10]
1 year ago
13

I've done the y = equations but I don't know how to do this one.

Mathematics
1 answer:
Tems11 [23]1 year ago
6 0

Consider the equation

-10x-y=-20

1) First row of the table

Set x=0 and solve as follows:

\begin{gathered} -10(0)-y=-20 \\ \Rightarrow-y=-20 \\ \Rightarrow y=20 \end{gathered}

The answer is y=20 and the pair x-y is (0,20)

2) Second row

Set x=1 and solve, as follows:

\begin{gathered} -10(1)-y=-20 \\ \Rightarrow-10-y=-20 \\ \Rightarrow-y=-10 \\ \Rightarrow y=10 \end{gathered}

The answers are y=10 and (1,10)

3) Third row.

Set y=0 and solve as follows:

\begin{gathered} -10x-(0)=-20 \\ \Rightarrow-10x=-20 \\ \Rightarrow x=\frac{20}{10}=2 \\ \Rightarrow x=2 \end{gathered}

The answers are x=2 and (2,0)

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2 years ago
You are a lifeguard and spot a drowning child 60 meters along the shore and 40 meters from the shore to the child. You run along
sukhopar [10]

Answer:

The lifeguard should run across the shore a distance of 48.074 m before jumpng into the water in order to minimize the time to reach the child.

Step-by-step explanation:

This is a problem of optimization.

We have to minimize the time it takes for the lifeguard to reach the child.

The time can be calculated by dividing the distance by the speed for each section.

The distance in the shore and in the water depends on when the lifeguard gets in the water. We use the variable x to model this, as seen in the picture attached.

Then, the distance in the shore is d_b=x and the distance swimming can be calculated using the Pithagorean theorem:

d_s^2=(60-x)^2+40^2=60^2-120x+x^2+40^2=x^2-120x+5200\\\\d_s=\sqrt{x^2-120x+5200}

Then, the time (speed divided by distance) is:

t=d_b/v_b+d_s/v_s\\\\t=x/4+\sqrt{x^2-120x+5200}/1.1

To optimize this function we have to derive and equal to zero:

\dfrac{dt}{dx}=\dfrac{1}{4}+\dfrac{1}{1.1}(\dfrac{1}{2})\dfrac{2x-120}{\sqrt{x^2-120x+5200}} \\\\\\\dfrac{dt}{dx}=\dfrac{1}{4} +\dfrac{1}{1.1} \dfrac{x-60}{\sqrt{x^2-120x+5200}} =0\\\\\\  \dfrac{x-60}{\sqrt{x^2-120x+5200}} =\dfrac{1.1}{4}=\dfrac{2}{7}\\\\\\ x-60=\dfrac{2}{7}\sqrt{x^2-120x+5200}\\\\\\(x-60)^2=\dfrac{2^2}{7^2}(x^2-120x+5200)\\\\\\(x-60)^2=\dfrac{4}{49}[(x-60)^2+40^2]\\\\\\(1-4/49)(x-60)^2=4*40^2/49=6400/49\\\\(45/49)(x-60)^2=6400/49\\\\45(x-60)^2=6400\\\\

x

As d_b=x, the lifeguard should run across the shore a distance of 48.074 m before jumpng into the water in order to minimize the time to reach the child.

7 0
3 years ago
2p-1<br>make p the subject​
jok3333 [9.3K]

Answer:

2p-1/p

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
More, less or equal to: <br> Absolute 4 or Absolute 1
Sloan [31]

|4| > |1|

Absolute value of 4 is just 4, and the abosulute valvue of 1 is just 1.

So , since 4 is greater than 1 , that would be your answer.

4 > 1


7 0
3 years ago
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