3 years ago

A triangle has sides with lengths of 11 miles, 15 miles, and 17 miles. Is it a right triangle?

Mathematics

1 answer:

dybincka [34]3 years ago

3 0

Answer:

Below.

Step-by-step explanation:

If its a right triangle it will obey the Pythagoras theorem.

17^2 = 289

11^2 + 15^2 = 121 + 225 = 346

- so NOT a right triangle.

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astra-53 [7]

$x+y=k\implies y=k-x$

Substitute this into the parabolic equation,

$k-x=6-(x-3)^2\implies x^2-7x+(k+3)=0$

We're told the line $x+y=k$ intersects $C$ twice, which means the quadratic above has two distinct real solutions. Its discriminant must then be positive, so we know

$(-7)^2-4(k+3)=49-4(k+3)>0\implies k$

We can tell from the quadratic equation that $C$ has its vertex at the point (3, 6). Also, note that

$-1\le\sin t\le1\implies3\le3+2\sin t\le5\implies3\le x\le5$

and

$-1\le\cos2t\le1\implies2\le4+2\cos2t\le6\implies2\le y\le6$

so the furthest to the right that $C$ extends is the point (5, 2). The line $x+y=k$ passes through this point for $2+5=k\implies k=7$ . For any value of $k$ , the line $x+y=k$ passes through $C$ either only once, or not at all.