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3 years ago

8

Design your own real-world scenario in which the solution requires the Pythagorean Theorem. Use complete sentences to describe t

he scenario.

1 answer:

Over [174]3 years ago

3 0

My house is located at (1,1) and the store is located at (4,3) Each square is a mile. How far do I walk to get to the store?

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A wheel has a diameter of 21cm calculate its circumference

notsponge [240]

21•3.14=65.94 the circumference is 65.94

8 0

4 years ago

Read 2 more answers

You start driving north for 9 miles, turn right, and drive east for another 40 miles. At the end of driving, what is your straig

Vsevolod [243]

The straight line distance from the starting point is 41 miles.

<u>Explanation:</u>

Given:

Distance covered towards north, n = 9 miles

Distance covered towards east, e = 40 miles

Distance from the origin to the end, x = ?

If we imagine this, then the route forms a right angle triangle

where,

n is the height

e is the base

x is the hypotenuse

Using pythagoras theorm:

(x)² = (n)² + (e)²

(x)² = (9)² + (40)²

(x)² = 1681

x = 41 miles

Therefore, the straight line distance from the starting point is 41 miles.

5 0

3 years ago

Explain please hurry ASAP

Nuetrik [128]

Answer:

x = 4

Step-by-step explanation:

we know that 72 and (2x + 10) add up to 90

90-72 = 18

(2x + 10) = 18

2x = 8

x = 4

6 0

3 years ago

Evaluate the integral. (sec2(t) i t(t2 1)8 j t7 ln(t) k) dt

polet [3.4K]

If you're just integrating a vector-valued function, you just integrate each component:

$\displaystyle\int(\sec^2t\,\hat\imath+t(t^2-1)^8\,\hat\jmath+t^7\ln t\,\hat k)\,\mathrm dt$

$=\displaystyle\left(\int\sec^2t\,\mathrm dt\right)\hat\imath+\left(\int t(t^2-1)^8\,\mathrm dt\right)\hat\jmath+\left(\int t^7\ln t\,\mathrm dt\right)\hat k$

The first integral is trivial since $(\tan t)'=\sec^2t$ .

The second can be done by substituting $u=t^2-1$ :

$u=t^2-1\implies\mathrm du=2t\,\mathrm dt\implies\displaystyle\frac12\int u^8\,\mathrm du=\frac1{18}(t^2-1)^9+C$

The third can be found by integrating by parts:

$u=\ln t\implies\mathrm du=\dfrac{\mathrm dt}t$

$\mathrm dv=t^7\,\mathrm dt\implies v=\dfrac18t^8$

$\displaystyle\int t^7\ln t\,\mathrm dt=\frac18t^8\ln t-\frac18\int t^7\,\mathrm dt=\frac18t^8\ln t-\frac1{64}t^8+C$

8 0

4 years ago

BabaBlast [244]

I think the answer is D but i could be wrong

4 0

3 years ago

Read 2 more answers