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3 years ago

Find the value of x. Write your answer in simplest form.

Mathematics

2 answers:

swat323 years ago

6 0

Answer:

Step-by-step explanation:

We can name the triangle as ABC

By using pythagoras theorem,

AC^2 = AB^2 + BC^2

(8√2)^2 = x^2 + x^2

128 = 2x^2

x^2 = 128/2

x^2 = 64

x = √64

x = 8

Hope it helps

Plz mark it as brainliest!!!!

givi [52]3 years ago

4 0

Answer:

X = 8

Step-by-step explanation:

Pythagorean triplate

8,8,8√2

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Given that f(x) = −x 4 and g(x) = −2x − 3, solve for f(g(x)) when x = 2. −14 −7 7 11

Rudiy27

F(x) = -x + 4
g(x) = -2x - 3

f(g(x)) = -(-2x - 3) + 4 = 2x + 3 + 4 = 2x + 7
f(g(2)) = 2(2) + 7 = 4 + 7 = 11

7 0

3 years ago

What is the answer to this

Setler [38]

The answer is 49.5 for the variable c

6 0

3 years ago

Read 2 more answers

1. Find the slope of the line (2,3) and (0,7)

aalyn [17]

Answer:

hope these helps you

Step-by-step explanation:

1. (2,3)&(0,7)

7-3/0-2

4/-2

=-2

2. (0,2)&(5,1)

1-2/5-0

=-1/5

3. (0,0)&(8,1)

1-0/8-0

=1/8

4.(0,2)&(6,4)

4-2/6-0

2/6

= 1/3

5.(0,1)&(4,2)

2-1/4-0

=1/4

6. (-1,6)&(4,2)

2-6/4-(-1)

=-4/5

5 0

4 years ago

How do I find the answer?

svlad2 [7]

Answer:

do you still need help

Step-by-step explanation:

4 0

4 years ago

Read 2 more answers

How do u graph this

iren [92.7K]

$\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\
% function transformations for trigonometric functions
\begin{array}{rllll}
% left side templates
f(x)=&{{ A}}sin({{ B}}x+{{ C}})+{{ D}}
\\\\
f(x)=&{{ A}}cos({{ B}}x+{{ C}})+{{ D}}\\\\
f(x)=&{{ A}}tan({{ B}}x+{{ C}})+{{ D}}
\end{array}
\\\\
-------------------\\\\$

$\bf \bullet \textit{ stretches or shrinks}\\
\left. \qquad \right. \textit{horizontally by amplitude } |{{ A}}|\\\\
\bullet \textit{ flips it upside-down if }{{ A}}\textit{ is negative}\\
\left. \qquad \right. \textit{reflection over the x-axis}
\\\\
\bullet \textit{ flips it sideways if }{{ B}}\textit{ is negative}\\
\left. \qquad \right. \textit{reflection over the y-axis}$

$\bf \bullet \textit{ horizontal shift by }\frac{{{ C}}}{{{ B}}}\\
\left. \qquad \right. if\ \frac{{{ C}}}{{{ B}}}\textit{ is negative, to the right}\\\\
\left. \qquad \right. if\ \frac{{{ C}}}{{{ B}}}\textit{ is positive, to the left}\\\\
\bullet \textit{vertical shift by }{{ D}}\\
\left. \qquad \right. if\ {{ D}}\textit{ is negative, downwards}\\\\
\left. \qquad \right. if\ {{ D}}\textit{ is positive, upwards}\\\\$

$\bf \bullet \textit{function period or frequency}\\
\left. \qquad \right. \frac{2\pi }{{{ B}}}\ for\ cos(\theta),\ sin(\theta),\ sec(\theta),\ csc(\theta)\\\\
\left. \qquad \right. \frac{\pi }{{{ B}}}\ for\ tan(\theta),\ cot(\theta)$

so hmmm $\bf \begin{array}{llcll}
f(x)=&0.5sin(x)&+2\\
&\uparrow &\uparrow \\
&A&D
\end{array}$

instead of going up/down by 1unit, it goes up/down by half
and then it's shifted upwards by 2, so the midline will end up at y = 2

but basically the same sin(x) graph, just a bit shorter due to the smaller amplitude.

5 0

3 years ago