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2 years ago

I’m confused on this one

Mathematics

1 answer:

tatyana61 [14]2 years ago

7 0

$6\sqrt2$

Why?

We have the isosceles right triangle. The length of legs are 6. Use the Pythagorean theorem:

$h^2=6^2+6^2\\\\h^2=36+36\\\\h^2=36\cdot2\to h=\sqrt{36\cdot2}\\\\h=\sqrt{36}\cdot\sqrt2\\\\h=6\sqrt2$

In general, the formula for calculating the hypotenuse in an isosceles right triangle is $h=a\sqrt2$

a - a leg, h - a hypotenuse

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Suppose a local park is rectangular in shape and has a length of 210 yards and a width of 105 yards. What is the length of the d

Otrada [13]

Answer:

234.784 ~ 235

Step-by-step explanation:

One side = 210

other side = 105

a2+b2 = c2

210 ^2 + 105 ^2 = c2

55125 = c ^2

c = $\sqrt{55125}$ = 234.784 ~ 235

hope its right!!!

7 0

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

2 years ago

Describe the relationship between two variables when the correlation coefficient r is one of the following. (a) near –1 strong

zepelin [54]

Answer:

a) strong negative linear correlation.

b) Weak or no linear correlation.

c) strong positive linear correlation.

Step-by-step explanation:

The correlation coefficient r measures the strength and direction (positive or negative) of two variables. The correlation coefficient r is always between -1 and 1. When the coefficient r is negative then the direction of the correlation is downhill (negative) and when it's positive then it's an uphill correlation (positive). Similarly, as the coefficient is closer to -1 or 1 the correlation is stronger, with zero being a non linear relationship.

Now back to the question:

a) Near -1: as we said before, this means an strong negative (-1) linear correlation.

b) Near 0: weak or no linear correlation (we cannot say if its positive or negative because we don't know it it's near zero from the right (positive numbers) or the left (negative numbers)

c) Near 1: strong positive (close to +1) linear correlation

3 0

2 years ago

Find the area of the composite shape below<br> help me

Ivenika [448]

Answer:

743.13

Step-by-step explanation:

area of the rectangle is 33x11 which is 363 and there are two equal half circles so we can just pretend they are connected so the radius is 11 A=pi times $radius^{2}$ which is 380.13 then add both of them and you get 743.13

8 0

2 years ago

Whats the radius of d = 29 m

VMariaS [17]

Answer:

6? .................

..............

5 0

2 years ago

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