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taurus [48]
3 years ago
14

The legs of a right triangle measure 11.4 meters and 15.1 meters. To the nearest tenth, what is the measure of the largest angle

?​
Mathematics
1 answer:
Korolek [52]3 years ago
6 0

Answer:

37.1 degrees

Step-by-step explanation:

tan x = 11.4/15.1

tan x = 0.7550

x = 37.05 degrees

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______________ conveys information about the relative scarcity of a good, such as whether a shortage or a surplus exists.
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I need this one too! Please help us!!
4 0
3 years ago
If x²-5x +1 =0, what is the value of -<br><img src="https://tex.z-dn.net/?f=%28x%20%2B%20%20%5Cfrac%7B1%7D%7Bx%7D%20%29%20%20" i
ra1l [238]
<h3>Answer:   5</h3>

=================================================

Work Shown:

x^2 - 5x + 1 = 0

x^2 + 1 - 5x = 0

x^2 + 1 = 5x

(x^2 + 1)/x = 5    .... where x is nonzero

(x^2)/x + (1/x) = 5

x + (1/x) = 5

---------------

An alternative method involves solving the original equation using the quadratic formula. After you get the two roots x = p and x = q, you should be able to find that p + 1/p = 5 and also q + 1/q = 5 as well.

In this case,

p = (5 + sqrt(21))/2

q = (5 - sqrt(21))/2

8 0
3 years ago
Read 2 more answers
Jenny climbed up 35 steps
frez [133]

Answer:

What’s the question

Step-by-step explanation:

3 0
3 years ago
Consider the equation below.
Korvikt [17]

Answer:

Equation in square form:

y=3(x+5)^2-4

Extreme value:

(h,k)=(-5,-4)

Step-by-step explanation:

We are given

y=3x^2+30x+71

we can complete square

y=3(x^2+10x)+71

we can use formula

a^2+2ab+b^2=(a+b)^2

y=3(x^2+2\times x\times 5)+71

now, we can add and subtract 5^2

y=3(x^2+2\times x\times 5+5^2-5^2)+71

y=3(x^2+2\times x\times 5+5^2)-3\times 5^2+71

y=3(x+5)^2-75+71

So, we get equation as

y=3(x+5)^2-4

Extreme values:

we know that this parabola

and vertex of parabola always at extreme values

so, we can compare it with

y=a(x-h)^2+k

where

vertex=(h,k)

now, we can compare and find h and k

y=3(x+5)^2-4

we get

h=-5

k=-4

so, extreme value of this equation is

(h,k)=(-5,-4)

6 0
4 years ago
Read 2 more answers
Find the absolute minimum and absolute maximum values of f on the given interval. f(x) = x − ln 8x, [1/2, 2]
insens350 [35]
The given function is 
f(x) =  x - ln(8x), on the interval [1/2, 2].

The derivative of f is
f'(x) = 1 - 1/x
The second derivative is 
f''(x) = 1/x²

A local maximum or minimum occurs when f'(x) = 0.
That is,
1 - 1/x = 0  => 1/x = 1  => x =1.
When x = 1, f'' = 1 (positive).
Therefore f(x) is minimum when x=1.
The minimum value is
f(1) = 1 - ln(8) = -1.079

The maximum value of f occurs either at x = 1/2 or at x = 2.
f(1/2) = 1/2 - ln(4) = -0.886
f(2)  = 2 - ln(16) = -0.773
The maximum value of f is
f(2) = 2 - ln(16) = -0.773
A graph of f(x) confirms the results.

Answer: 
Minimum value  = 1 - ln(8)
Maximum value = 2 - ln(16)


4 0
4 years ago
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