A certain television is advertised as a 83-inch TV (the diagonal length). If the height of the TV is 45 inches, how wide is the TV? Round to the nearest tenth of an inch.
3 answers
Let $x$ be the width of the TV. By the Pythagorean Theorem, $x^2+45^2=83^2$. Solving yields $x^2=6914$, so $x=\sqrt{6914}\approx83.2$ inches after rounding to the nearest tenth. To the nearest tenth of an inch, the width of the TV is $\boxed{83.2}$ inches.
thats not right bot
Apologies for the mistake. Let's solve it correctly:
We can use the Pythagorean Theorem, which states that in a right triangle, the square of the hypotenuse (the longest side) is equal to the sum of the squares of the other two sides.
Let $x$ be the width of the TV. We have the following equation:
$x^2 + 45^2 = 83^2$
$x^2 = 83^2 - 45^2$
$x^2 = 6889 - 2025$
$x^2 = 4864$
$x \approx \sqrt{4864} \approx 69.8$ (rounded to the nearest tenth)
Therefore, to the nearest tenth of an inch, the width of the TV is $\boxed{69.8}$ inches.
We can use the Pythagorean Theorem, which states that in a right triangle, the square of the hypotenuse (the longest side) is equal to the sum of the squares of the other two sides.
Let $x$ be the width of the TV. We have the following equation:
$x^2 + 45^2 = 83^2$
$x^2 = 83^2 - 45^2$
$x^2 = 6889 - 2025$
$x^2 = 4864$
$x \approx \sqrt{4864} \approx 69.8$ (rounded to the nearest tenth)
Therefore, to the nearest tenth of an inch, the width of the TV is $\boxed{69.8}$ inches.