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3 years ago

Please help me with this ASAP ASAP please please help

Mathematics

1 answer:

hammer [34]3 years ago

4 0

9514 1404 393

Answer:

h = 7 in

LA = 42π in²

SA = 60π in²

Step-by-step explanation:

Use the appropriate formula with the numbers you know, and solve for the number you need to know.

V = πr²h

h = V/(πr²) = (63π in³)/(π(3 in)²)

h = 7 in

LA = 2πrh

LA = 2π(3 in)(7 in)

LA = 42π in²

SA = 2πr(r +h)

SA = 2π(3 in)(3 in +7 in)

SA = 60π in²

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The hypotenuse of a right triangle is 6 cm and one side is 2 cm longer than the other side. Find the length of each side to one

MA_775_DIABLO [31]

Answer:

The two legs of the right triangle are approximately 3.1 cm and 5.1 cm and the hypotenuse is 6 cm.

Step-by-step explanation:

Let x be one leg of the right triangle.

Since the other leg is two centimeters longer, it can be represented by the expression:

$x + 2 \text{ cm}$

According to the Pythagorean Theorem, for a right triangle:

$a^2 + b^2 = c^2$

Where c is the hypotenuse and a and b are the two legs.

The hypotenuse is given to be 6 cm and the two legs are x and (x + 2). Hence:

$(x)^2 + (x+2)^2 = 6^2$

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Subtract:

$2x^2 + 4x - 32 =0$

Divide:

$x^2 + 2x - 16 = 0$

The equation is not factorable, so we can consider using the quadratic formula:

$\displaystyle x = \frac{-b\pm\sqrt{b^2 -4ac}}{2a}$

In this case, a = 1, b = 2, and c = -16. Substitute:

$\displaystyle x= \frac{-(2)\pm\sqrt{(2)^2-4(1)(-16)}}{2(1)}$

And evaluate:

$\displaystyle \begin{aligned}x &= \frac{-2\pm\sqrt{68}}{2} \\ \\ &= \frac{-2\pm\sqrt{4\cdot 17}}{2} \\ \\ &= \frac{-2\pm2\sqrt{17}}{2} \\ \\ &= -1 \pm \sqrt{17} \end{aligned}$

Hence, our two solutions are:

$\displaystyle x = -1 + \sqrt{17} \approx 3.1\text{ or } x = -1 - \sqrt{17} \approx -5.1$

Lengths cannot be negative, so we can ignore the second solution.

Hence, the value of x or the first side length is about 3.1 centimeters.

Since the other side length is two centimeters longer, the other side is about 5.1 centimeters.

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3 years ago

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Answer:

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2 years ago

Number 3 and how you got the answer

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3 years ago

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