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Rufina [12.5K]
3 years ago
5

What is the area of a 6x6x6 equilateral triangle?

Mathematics
2 answers:
NARA [144]3 years ago
6 0
A 6x6x6 triangle
we cut it into half so consider one half
its 6x3
by Pythagoras theorem, we can find the length of the triangle sqrt (6^2 -3^2) = a
then we can find the area of the triangle by
1/2 x base x height
=
1/2 x 6 x a
maw [93]3 years ago
4 0
216 cubic units bc 6x6=36 and then 36x6=216
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What is m∠S? Enter your answer in the box. ° An isosceles triangle with vertices labeled R, S, and T. Side R T is the base. Side
yulyashka [42]

Answer:

m<S = 45°

Step-by-step explanation:

The sum of the measures of the angles of a triangle equals 180 deg.

m<R + m<S + m<T = 180

3x + 2x + 3x = 180

8x = 180

x = 22.5

m<S = 2x = 2(22.5) = 45

5 0
4 years ago
Which equation could represent p? Please please help
Andre45 [30]

Answer:

<h2>P(x) = (x+3)(x-2)^2</h2>

Step-by-step explanation:

Looking at the brackets you can see where the curve will intersect the x-axis.

The graph shows the curve intersecting at (0,-3) and (0,2).

This means:

x = -3

AND

x = 2

Rearrange the equations, equating them to 0.

x + 3 = 0

x - 2 = 0

This will be the values in the brackets.

Because the curve only touches 0,2 and DOES NOT cross it, we know that x - 2 is a repeated root, hence (x-2) is squared.

Therefore your brackets are: (x+3)(x-2)(x-2)

Which can be simplified:

(x+3)(x-2)^2

Where ^2 means squared.

7 0
3 years ago
How many tons are in 6,000 pounds ?
bagirrra123 [75]

Answer:

3

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
Solve for x. X/4+1=3?
Elan Coil [88]
Subtract 1: x/4=2 multiply by 4: x=8 which is the answer
8 0
3 years ago
The region bounded by y=x^2+1, y=x, x=-1, x=2 with square cross sections perpendicular to the x-axis.
VLD [36.1K]

Answer:

The bounded area is 5 + 5/6 square units. (or 35/6 square units)

Step-by-step explanation:

Suppose we want to find the area bounded by two functions f(x) and g(x) in a given interval (x1, x2)

Such that f(x) > g(x) in the given interval.

This area then can be calculated as the integral between x1 and x2 for f(x) - g(x).

We want to find the area bounded by:

f(x) = y = x^2 + 1

g(x) = y = x

x = -1

x = 2

To find this area, we need to f(x) - g(x) between x = -1 and x = 2

This is:

\int\limits^2_{-1} {(f(x) - g(x))} \, dx

\int\limits^2_{-1} {(x^2 + 1 - x)} \, dx

We know that:

\int\limits^{}_{} {x} \, dx = \frac{x^2}{2}

\int\limits^{}_{} {1} \, dx = x

\int\limits^{}_{} {x^2} \, dx = \frac{x^3}{3}

Then our integral is:

\int\limits^2_{-1} {(x^2 + 1 - x)} \, dx = (\frac{2^3}{2}  + 2 - \frac{2^2}{2}) - (\frac{(-1)^3}{3}  + (-1) - \frac{(-1)^2}{2}  )

The right side is equal to:

(4 + 2 - 2) - ( -1/3 - 1 - 1/2) = 4 + 1/3 + 1 + 1/2 = 5 + 2/6 + 3/6 = 5 + 5/6

The bounded area is 5 + 5/6 square units.

3 0
3 years ago
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