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Norma-Jean [14]
4 years ago
13

A triangle has a side lengths of 4cm,3cm, and 4cm. Classify its as acute, obtuse, or right.

Mathematics
1 answer:
DIA [1.3K]4 years ago
4 0

When given 3 triangle sides, to determine if the triangle is acute, right or obtuse:

1) Square all 3 sides.

4, 3, 4, 

16, 9, 16

2) Sum the squares of the 2 shortest sides.

16 + 9 = 25


3) Compare this sum to the square of the 3rd side.

25 > 16

if sum > 3rd side²   Acute Triangle

So, it is an acute triangle.

Source:
http://www.1728.org/triantest.htm


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Calculate the area of triangle ABC with altitude, CD, given
defon

Answer:

37.5 square units

Step-by-step explanation:

Given:

A(-6,-4), B(6,5), C(-1,6), and D(2, 2), where CD is the altitude if ∆ABC

Required:

Area of ∆ABC

SOLUTION:

Area of a ∆ABC =½*AB*CD

✍️AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{6 -(-6))^2 + (5 -(-4))^2

AB = \sqrt{12^2 + 9^2}

AB = \sqrt{144 + 81}

AB = \sqrt{225} = 15

✍️CD = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{2 -(-1))^2 + (2 - 6)^2

CD = \sqrt{(3)^2 + (-4)^2}

CD = \sqrt{9 + 16}

CD = \sqrt{25} = 5

✍️Area of a ∆ABC =½*AB*CD

= ½*15*5

✅ Area = 37.5 square units

5 0
3 years ago
Find the point P on the graph of the function y=√x closest to the point (9,0)
Sphinxa [80]

Answer:

\displaystyle \frac{17}{2}.

Step-by-step explanation:

Let the x-coordinate of P be t. For P\! to be on the graph of the function y = \sqrt{x}, the y-coordinate of \! P would need to be \sqrt{t}. Therefore, the coordinate of P \! would be \left(t,\, \sqrt{t}\right).

The Euclidean Distance between \left(t,\, \sqrt{t}\right) and (9,\, 0) is:

\begin{aligned} & d\left(\left(t,\, \sqrt{t}\right),\, (9,\, 0)\right) \\ &= \sqrt{(t - 9)^2 +\left(\sqrt{t}\right)^{2}} \\ &= \sqrt{t^2 - 18\, t + 81 + t} \\ &= \sqrt{t^2 - 17 \, t + 81}\end{aligned}.

The goal is to find the a t that minimizes this distance. However, \sqrt{t^2 - 17 \, t + 81} is non-negative for all real t\!. Hence, the \! t that minimizes the square of this expression, \left(t^2 - 17 \, t + 81\right), would also minimize \sqrt{t^2 - 17 \, t + 81}\!.

Differentiate \left(t^2 - 17 \, t + 81\right) with respect to t:

\displaystyle \frac{d}{dt}\left[t^2 - 17 \, t + 81\right] = 2\, t - 17.

\displaystyle \frac{d^{2}}{dt^{2}}\left[t^2 - 17 \, t + 81\right] = 2.

Set the first derivative, (2\, t - 17), to 0 and solve for t:

2\, t - 17 = 0.

\displaystyle t = \frac{17}{2}.

Notice that the second derivative is greater than 0 for this t. Hence, \displaystyle t = \frac{17}{2} would indeed minimize \left(t^2 - 17 \, t + 81\right). This t\! value would also minimize \sqrt{t^2 - 17 \, t + 81}\!, the distance between P \left(t,\, \sqrt{t}\right) and (9,\, 0).

Therefore, the point P would be closest to (9,\, 0) when the x-coordinate of P\! is \displaystyle \frac{17}{2}.

8 0
3 years ago
Which property of exponents must you apply to the expression p^1/2 to derive p as the result?
Elodia [21]

We apply the following property:

(x^n)^m = x^(n • m)

Let m = 2

(P^1/2)^2 = P^(1/2 • 2) = P

5 0
3 years ago
PLEASE HELP<br> I have no idea how to do this
tresset_1 [31]

Answer:

1st 2nd or both's answer you want

7 0
3 years ago
Please help me i will give brainliest..
Ghella [55]

Answer:

56

Step-by-step explanation:

5 0
3 years ago
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