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3 years ago

Simplify (2V5 - 4)(3 V5+2). a. 4 c. b. 8-4v5 d. 22-8v5

Mathematics

1 answer:

makkiz [27]3 years ago

5 0

Answer:

$(2\sqrt{5} - 4)(3\sqrt{5}+2)= 22 - 8\sqrt{5}$

Step-by-step explanation:

Given

$(2\sqrt{5} - 4)(3\sqrt{5}+2)$

Required

Simplify

$(2\sqrt{5} - 4)(3\sqrt{5}+2).$

Expand brackets

$2\sqrt{5}(3\sqrt{5}+2). - 4(3\sqrt{5}+2).$

Open each bracket

$2\sqrt{5}*3\sqrt{5}+2\sqrt{5}*2 - 4*3\sqrt{5}-4*2$

$6 * 5 + 4\sqrt{5} - 12\sqrt{5} - 8$

$30 + 4\sqrt{5} - 12\sqrt{5} - 8$

Collect Like Terms

$- 8+30 + 4\sqrt{5} - 12\sqrt{5}$

$22 - 8\sqrt{5}$

Hence:

$(2\sqrt{5} - 4)(3\sqrt{5}+2)= 22 - 8\sqrt{5}$

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3 years ago

Outside temperature over a day can be modeled as a sinusoidal function. Suppose you know the temperature varies between 73 and 9

Naddika [18.5K]

Answer: After about 9.03 hours the temperature first reach 82 degrees.

Step-by-step explanation:

The sinusoidal function is given by :

$y=A\sin[\omega(x-\alpha)]+C$

where, A = amplitude; $\omega=\dfrac{2\pi}{period}$ , α= phase shift on the Y-axis and C = midline.

As per given,

Average daily temperature= $C=\dfrac{73+97}{2}=85$ [midline is average of upper and lower limit.]

A= 97-85 = 12

Phase shift: $\alpha=10$

Period = 24 hours;

$\omega=\dfrac{2\pi}{24}=\dfrac{\pi}{12}$

Substitute all values in sinusoidal function, we get

$y=12\sin[\dfrac{\pi}{12}(x-10)]+85$

Put y= 82, we get

$82=12\sin[\dfrac{\pi}{12}(x-10)]+85\\\\\Rightarrow\ -3= 12\sin[\dfrac{\pi}{12}(x-10)]\\\\=\dfrac{-1}{4}= \sin[\dfrac{\pi}{12}(x-10)]\\\\\Rightarrow\ \dfrac{\pi}{12}(x-10)=\sin^{-1}(\dfrac{-1}{4})\\\\\Rightarrow\ x-10=\dfrac{12}{\pi}(\sin^{-1}(\dfrac{-1}{4}))\\\\\Rightarrow\ x=\dfrac{12}{\pi}(\sin^{-1}(\dfrac{-1}{4}))+10\\\Rightarrow\ x\approx9.03$