Can anyone help me solve this.I need proper equation..Thanks

Mathematics

2 answers:

nikklg [1K]3 years ago

8 0

Answer: ?=9

√2

Step-by-step explanation:

x=9 =a side of squre

y=diagonal

use equation :

y=a√2

y=9√2

y=12.69

artcher [175]3 years ago

3 0

Answer:

y = 9 $\sqrt{2}$

Step-by-step explanation:

Using the sine ratio in the right triangle and the exact value

sin45° = $\frac{1}{\sqrt{2} }$

sin45° = $\frac{opposite}{hypotenuse}$ = $\frac{9}{y}$

Multiply both sides by y

y × sin45° = 9, that is

y × $\frac{1}{\sqrt{2} }$ = 9

Multiply both sides by $\sqrt{2}$

y = 9 $\sqrt{2}$

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3 years ago

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Plz help me with this

AnnZ [28]

Answer: $\bold{A)\quad y=5sin\bigg(\dfrac{6}{5}x-\pi\bigg)-4}$

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

The general form of a sin/cos function is: y = A sin/cos (Bx-C) + D where the period (P) = 2π ÷ B

In the given function, $B=\dfrac{6}{5}$ → $P=2\pi \cdot \dfrac{5}{6}=\dfrac{10\pi}{3}$

Half of that period is: $\dfrac{1}{2}\cdot \dfrac{10\pi}{3}=\large\boxed{\dfrac{5\pi}{3}}$

Calculate the period for each of the options to find a match:

$A)\quad B=\dfrac{6}{5}:\quad 2\pi \div \dfrac{6}{5}=2\pi \cdot \dfrac{5}{6}=\dfrac{5\pi}{3}\quad \leftarrow\text{THIS WORKS!}\\\\\\B)\quad B=\dfrac{6}{10}:\quad 2\pi \div \dfrac{6}{10}=2\pi \cdot \dfrac{10}{6}=\dfrac{10\pi}{3}\\\\\\C)\quad B=\dfrac{5}{6}:\quad 2\pi \div \dfrac{5}{6}=2\pi \cdot \dfrac{6}{5}=\dfrac{12\pi}{5}\\\\\\D)\quad B=\dfrac{3}{10}:\quad 2\pi \div \dfrac{3}{10}=2\pi \cdot \dfrac{10}{3}=\dfrac{20\pi}{3}$