Find the missing side lengths. Leave your answers as radicals in simplest form. Show your work to support your answer

Mathematics

1 answer:

kotegsom [21]1 year ago

3 0

Answer

x=2 and y=2√2

Step-by-step explanation:

$\sin( \alpha ) = \frac{2 \sqrt{2} }{x } \\ x = \frac{2 \sqrt{2} }{ \sin(45) } \\ x = 4 \\ {x}^{2} = {y}^{2} + 2 \sqrt{2} \\ {4}^{2} = {y}^{2} + (2 \sqrt{2 \: )}^{2} \\ 16 = {y}^{2} + 8 \\ 16 - 8 = {y}^{2} \\ y = \sqrt{8} \\ y = 2 \sqrt{2}$

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I will give u brainliest

goldfiish [28.3K]

Answer:

u = 3

Step-by-step explanation:

${u}^{3} = 27$

${3}^{3} = 27$

$3 \times 3 \times 3 = 27$

7 0

2 years ago

Read 2 more answers

This is a question I have been trying to do for quite a while. My additional mathematics exam is next week and I need help. Anyo

Radda [10]

First of all, let's consider the line, since it's simpler to graph: we draw the endpoints and connect them:

$f(x)=2-\dfrac{3x}{4\pi} \implies f(0)=2,\quad f(2\pi)=\dfrac{1}{2}$

So, you just need to draw the points

$(0,2),\quad \left(2\pi, \dfrac{1}{2}\right)$

As for the trigonometric function, we have to start from the parent function $y=\sin(x)$ and derive the graph of its child function via transformations:

When we multiply the whole function by 2, we stretch the graph vertically. So, the function has still period $2\pi$ , but now it ranges from -2 to 2 instead of from -1 to 1 (amplitude 2)
When we multiply the argument by 2, we compress the function horizontally. So, the new period becomes $\pi$ , and the function makes two complete oscillations from 0 to $2\pi$