Hai! What Is 4 Squared???

Mathematics

2 answers:

Nitella [24]2 years ago

4 0

Answer:

Step-by-step explanation:

$x^{2}$ = $x$ multiplied by $x$

so $4^{2}$ = 16

lakkis [162]2 years ago

4 0

Answer:

Step-by-step explanation:

4 squared should look like this:

4²

That means we just multiply 4 twice:

4 x 4

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The graph of a sinusoidal function has a minimum point at (0, - 10) and then has a maximum point at (2, - 4) Write the formula o

AleksAgata [21]

Answer: $\bold{y=3sin\bigg(\dfrac{\pi}{2}x-\dfrac{\pi}{2}\bigg)-7}$

<u>Step-by-step explanation:</u>

Minimum: (0, -10)

Maximum: (2, -4)

y = A sin (Bx - C) + D

Amplitude (A) = (Max - Min)/2
Period = 2π/B → B = 2π/Period
Phase Shift = C/B → C = B × Phase Shift
Midline (D) = (Max + Min)/2

$A=\dfrac{-4-(-10)}{2}\quad =\dfrac{-4+10}{2}\quad =\dfrac{6}{2}\quad =\large\boxed{3}$

$\text{x-value of Max minus x-value of Min}= \dfrac{1}{2}\text{Period}\\\\2 - 0 = \dfrac{1}{2}P\quad \rightarrow \quad P=4\\\\\\B=\dfrac{2\pi}{P}\quad =\dfrac{2\pi}{4}\quad =\large\boxed{\dfrac{\pi}{2}}\\$

$D = \dfrac{\text{Max + Min}}{2}\quad = \dfrac{-4+(-10)}{2}\quad =\dfrac{-14}{2}\quad =\large\boxed{-7}$

Sin usually starts at (0, 0). For this graph, the midline touches 0 when x = 1 so the Phase Shift = 1.

$C = B \times \text{Phase Shift}\quad = \dfrac{\pi}{2}\times 1\quad =\large\boxed{\dfrac{\pi}{2}}$

$A=3, \quad B=\dfrac{\pi}{2}, \quad C=\dfrac{\pi}{2},\quad D=-7\\\\\rightarrow \quad \large\boxed{y=3\sin \bigg(\dfrac{\pi}{2}x-\dfrac{\pi}{2}\bigg)-7}$

3 0

3 years ago

Prove the following

DerKrebs [107]

$\bf [cot(\theta )+csc(\theta )]^2=\cfrac{1+cos(\theta )}{1-cos(\theta )} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{doing the left-hand side}}{[cot(\theta )+csc(\theta )]^2}\implies cot^2(\theta )+2cot(\theta )csc(\theta )+csc^2(\theta ) \\\\\\ \cfrac{cos^2(\theta )}{sin^2(\theta )}+2\cdot \cfrac{cos(\theta )}{sin(\theta )}\cdot \cfrac{1}{sin(\theta )}+\cfrac{1}{sin^2(\theta )}\implies \cfrac{cos^2(\theta )}{sin^2(\theta )}+\cfrac{2cos(\theta )}{sin^2(\theta )}+\cfrac{1}{sin^2(\theta )}$

$\bf \cfrac{\stackrel{\textit{perfect square trinomial}}{cos^2(\theta )+2cos(\theta )+1}}{sin^2(\theta )}\implies \boxed{\cfrac{[cos(\theta )+1]^2}{sin^2(\theta )}} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{doing the right-hand-side}}{\cfrac{1+cos(\theta )}{1-cos(\theta )}}\implies \stackrel{\textit{multiplying by the denominator's conjugate}}{\cfrac{1+cos(\theta )}{1-cos(\theta )}\cdot \cfrac{1+cos(\theta )}{1+cos(\theta )}}$

$\bf \cfrac{[1+cos(\theta )]^2}{\underset{\textit{difference of squares}}{[1-cos(\theta )][1+cos(\theta )]}}\implies \cfrac{[cos(\theta )+1]^2}{1^2-cos^2(\theta )} \\\\\\ \cfrac{[cos(\theta )+1]^2}{1-cos^2(\theta )}\implies \boxed{\cfrac{[cos(\theta )+1]^2}{sin^2(\theta )}}$