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3 years ago

If h(x)= 6x - 17, find h(8). *

Mathematics

2 answers:

Lena [83]3 years ago

3 0

The value of h(8) in h(x)= 6x - 17 is 31

What are functions:

Functions relate input and output. Therefore, a function takes elements from a set (the domain) and relates them to elements in a set (range).

h(x)= 6x - 17

Let's find h(8)

h(8) = 6(8) - 17

h(8) = 48 - 17

h(8) = 31

learn more on function here:brainly.com/question/15615479?referrer=searchResults

mixer [17]3 years ago

3 0

Answer:

h(8) = 31

Step-by-step explanation:

Substitute x = 8 into h(x) , that is

h(8) = 6(8) - 17 = 48 - 17 = 31

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3 years ago

Suppose that theta is an angle in standard position whose terminal side Intersects the unit circle at (-11/61, -60/61)

Blizzard [7]

Answer:

The exact values of the tangent, secant and cosine of angle theta are, respectively:

$\cos \theta = -\frac{11}{61}$

$\tan \theta = \frac{-\frac{60}{61} }{-\frac{11}{61} } = \frac{60}{11}$

$\sec \theta = \frac{1}{-\frac{11}{61} } = -\frac{61}{11}$

Step-by-step explanation:

The components of the unit vector are $x = -\frac{11}{61}$ and $y = -\frac{60}{61}$ . Since $r = 1$ , then $x = \cos \theta$ and $y = \sin \theta$ . By Trigonometry, tangent and secant can be calculated by the following expressions:

$\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{y}{x}$

$\sec \theta = \frac{1}{\cos \theta} = \frac{1}{x}$

Now, the exact values of the tangent, secant and cosine of angle theta are, respectively:

$\cos \theta = -\frac{11}{61}$

$\tan \theta = \frac{-\frac{60}{61} }{-\frac{11}{61} } = \frac{60}{11}$

$\sec \theta = \frac{1}{-\frac{11}{61} } = -\frac{61}{11}$

3 0

3 years ago

Bryan is buying notebooks and pens at an office supply store. Notebooks cost $3.59 and pens cost $1.49. He can spend up to $13 o

nordsb [41]

3.59x+1.49y=13.00
that should help

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3 years ago

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masha68 [24]

Answer:

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3 years ago

Given: M(-4,-1), B(-5,-3) find the midpoint

Dmitry [639]

$\bf \textit{middle point of 2 points }\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
M&({{ -4}}\quad ,&{{ -1}})\quad 
% (c,d)
B&({{ -5}}\quad ,&{{ -3}})
\end{array}\qquad
% coordinates of midpoint 
\left(\cfrac{{{ x_2}} + {{ x_1}}}{2}\quad ,\quad \cfrac{{{ y_2}} + {{ y_1}}}{2} \right)
\\\\\\
\left(\cfrac{{{ -5}}-4}{2}\quad ,\quad \cfrac{-3-1}{2} \right)$

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3 years ago

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