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3 years ago

X² + 2x - 3 = 0

Mathematics

1 answer:

tankabanditka [31]3 years ago

6 0

Step-by-step explanation:

x²+2x-3=0

(x²+3x)-(x-3)=0

x(x+3)-1(x+3)=0

(x-1)(x+3)=0

x=1 or x=-3

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A pole that is 3.2 m tall casts a shadow that is 1.29 m long. At the same time, a nearby tower casts a shadow that is 39.25 m lo

Debora [2.8K]

Answer: 97 meters

Step-by-step explanation:

Draw 2 right triangles KLM and ABC.

ML=3.2 m ( is a pole) KL =1.29 m( pole's shadow).

(The triangle KLM is right because we suppose that pole is staying vertically).

In triangle AB is tower's shadow= 39.25 m and BC is the height og tower=x meters?

Triangles KLM and ABC are similar .

So ML/KL=x/AB

3.2/1.29=x/39.25

x=3.2*39.25/1.29= approx 97.36 meters= 97 meters

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3 years ago

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NikAS [45]

Answer:

78.87

Step-by-step explanation:

I added 2.1 and 5.3 and subtracted by 90 and got the answer

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3 years ago

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On average, the moon is 384,400 kilometers from the Earth. How is this distance written in

KiRa [710]

Answer:

Step-by-step explanation:

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3 years ago

Read 2 more answers

Need help on this quadratic factoring problems

Effectus [21]

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3 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

4 years ago