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patriot [66]
3 years ago
8

GEOMETRY! PLEASE SHOW YOUR WORK. (updated picture)

Mathematics
2 answers:
Orlov [11]3 years ago
6 0

Given:

The base of the given triangle has two part:

Left( left side of the hypotenuse) = 25

Right ( right side of the hypotenuse) = x

The altitude (h) = 60

To find the value of x.

Formula:

By Altitude rule we know that, the altitude of a triangle is mean proportional between the right and left part of the hypotenuse,

\frac{left}{altitude} =\frac{altitude}{right}

Now,

Putting,

left = 25, altitude = 60 and right = x we get,

\frac{25}{60} =\frac{60}{x}

or, (25)(x )=(60)(60) [ by cross multiplication]

or, x = \frac{3600}{25}

or, x = 144

Hence,

The value of x is 144.

Eva8 [605]3 years ago
6 0

Answer:

<em>x = 144</em>

Step-by-step explanation:

<u><em>Height Theorem </em></u>

<em>60² = 25·x </em>

<em> </em>

<em>3600 = 25x </em>

<em> </em>

<em>x = 3600 : 25 </em>

<em> </em>

<em>x = 144</em>

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.Using a table of values, determine the solution to the equation below to the nearest fourth of a unit. x+4= -3^x+4
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Remark
The very first thing you should do with a question like this is get the graph. Then you will know what you are looking for. I have provided you with such a graph. Your table should center around -2 ≤ x ≤ -1

So let's set up a table and see what we get. Start with y = x + 5

x             x + 5
-1           4
-1.25      3.75
-1.5        3.5
-1.75      3.25
-2           3

Do the same thing for - (3)^x + 4 See below to see how this is entered your calculator

x            -(3)^x + 4
-1           3.66
-1.25      3.75
-1.5        3.8
-1.75      3.85
-2           3.89

Conclusion
When x = - 1.25 y = 3.75 for both graphs. <<<< Answer

Footnote
You may not be familiar with how to put this in your calculator. This is the way I would do it. I'm only doing it for y = -(3^x) + 4

Let x = - 1.25
3
^ Note your calculator might have x^y or y^x. You'll have 1 of the three.
1.25
+/- 
=
X
1
+/-
=     At this point you should have -0.25
+
4
= 
That gives you 3.75



5 0
3 years ago
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