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4 years ago

1. a triangle has the base of 5/3x and the height of x how do i work this equation

Mathematics

1 answer:

sashaice [31]4 years ago

3 0

Answer:

Area = 5/6 * x^2

Step-by-step explanation:

If you are given the base and height of a certain triangle, you will most likely be asked to find the area

The area of any triangle can be found by

Area = base*height * (1/2)

In your case

Area = (5/3)*x*x* (1/2) = 5/6 * x^2

In your case, x should be a known parameter of your triangle, so if you have the value of the height, you should be able to find the area. Or conversely, if you have the area, you can find the length and height of your triangle.

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Answer:

D (32, 10)

Step-by-step explanation:

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3 years ago

How is the pattern for a perfect square trinomial used to factor the trinomial?

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Answer:

Step-by-step explanation:

We're going to try two things here.

First an example

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3 years ago

A triangle has three sides 35cm 54 cm and 61 cm find its area Also find the smallest of its altitude

I am Lyosha [343]

Answer:

Area of given triangle is 939.15cm² and smallest altitude is 30.8cm

<h3>Solution:</h3>

We are given three sides of a triangle, Let the sides be :

( a ) = 35 cm

( b ) = 54 cm

( c ) = 61 cm

We can find the area of the triangle with its three sides using Heron's Formula

<u>Heron's </u><u>Formula</u>

Heron's formula was founded by hero of Alexandria, for finding the area of triangle in terms of the length of its sides. Heron's formula can be written as:

$\sf{ \pmb { \longrightarrow \: \sqrt{s(s - a)(s - b)(s - c)} }}$

where ( s ) :

$\sf \longrightarrow s = \dfrac{a + b + c}{2}$

Therefore, for the given triangle first we will calculate ( s )

$\begin {aligned}\quad & \quad \longmapsto \sf s = \dfrac{a + b + c}{2} \\ & \quad \longmapsto \sf s = \dfrac{35 + 54 + 61}{2} \\ & \quad \longmapsto \sf s = \dfrac{150}{2} \\ & \quad \longmapsto \sf s = 75cm \end{aligned}$

Now, Area of triangle will be:

$\begin{aligned}&:\implies \sf\quad \sf \: A = \sqrt{s(s - a)(s - b)(s - c)} \\ &:\implies \sf\quad \sf \: A = \sqrt{75(75 - 35)(75 - 54)(75 - 61)} \\&:\implies \sf\quad \sf \: A = \sqrt{75 \times 40 \times 21 \times 14} \\ &:\implies \sf\quad \sf \: A = \sqrt{5 \times 5 \times 3 \times 3 \times 2 \times 2 \times 7 \times 7 \times 2 \times 2 \times 5} \\ &:\implies \sf\quad \sf \: A =5 \times 3 \times 2 \times 7 \times 2 \sqrt{5} \\ &:\implies \sf\quad \sf \: A =420 \times 2.23 \\ &:\implies \sf\quad \sf \boxed{ \pmb{ \sf A =939.15 {cm}^{2} }} \end{aligned}$

Also, we have to find the smallest altitude, and the smallest altitude will be on the longest side. So,

$\begin{aligned}&:\implies \sf\quad \sf \: Area =939.15 \\ &:\implies \sf\quad \sf \: \dfrac{1}{2} \times b \times h =939.15 \\ &:\implies \sf\quad \sf \: \dfrac{1}{2} \times 61 \times h = 939.15 \\&:\implies \sf\quad \sf \: h =939.15 \times \dfrac{2}{61} \\&:\implies \sf\quad \sf \: h = \dfrac{1818.3}{61} \\ &:\implies \sf\quad \boxed{ \pmb{\sf \: h =30.79 \: (approx)}} \end{aligned}$

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[1, 2.5]

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For an interval (a, b) and a function f(x), the rate of change is defined as:

$r = \frac{f(b) - f(a)}{b - a}$

So the rate of change is only negative when the function evaluated on the first value of the interval is larger than the function evaluated in the second one.

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f(1) = 3

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If you want to learn more about rates of change, you can read:

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