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3 years ago

PLEASE HELP ASAP!!!!I'll mark brainliest to the correct answers

Mathematics

1 answer:

Korvikt [17]3 years ago

7 0

Answer:

Option 2

Step-by-step explanation:

2 + ¾(t) > 10

¾(t) > 8

t > 32/3

t > 10 ⅔

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Evaluate the expression 1/4[x(2y+3z)] x=8 y=3z=5/3

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Answer:

Step-by-step explanation:

x = 8 ; y = 5/3

$3z = \dfrac{5}{3}$

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Which of the following functions are solutions of the differential equation y'' + y = 3 sin x? (select all that apply)

DiKsa [7]

Answer:

Step-by-step explanation:

We must compute the derivatives and check if the equation is satisfied.

A. $y'=(3\sin x)'=3\cos x$ . Differentiate again to get $y''=(3\cos x)'=-3\sin x$ , then $y''+y=-3\sin x+3\sin x=0\neq 3\sin x$ so this choice of y doesn't solve the equation.

B. $y'=3\sin x+3x\cos x-5\cosx +5x\sin x=(5x+3)\sin x+(3x-5)\cos x$ and $y''=5\sin x+(5x+3)\cos x+3\cos x-(3x-5)\sin x=(10-3x)\sin x+(5x+6)\cos x$ , then $y''+y=10\sin x+6\cos x\neq 3\sin x$ so y is not a solution

C. $y'=\frac{-3}{2}\cos x+\frac{3}{2}x\sin x$ hence $y''=\frac{3}{2}\sin x+\frac{3}{2}\sin x+\frac{3}{2}x\cos x=3\sin x+\frac{3}{2}x\cos x$ . Then $y''+y=3\sin x+\frac{3}{2}x\cos x-\frac{3}{2}x\cos x=3\sin x$ so y is a solution.

D. $y'=-3\sin x$ and $y''=-3\cos x$ , then $y''+y=0$ thus y isn't a solution

E. $y'=\frac{3}{2}\sin x+\frac{3}{2}x\cos x$ hence $y''=\frac{3}{2}\cos x+\frac{3}{2}\cos x-\frac{3}{2}x\sin x=3\cos x-\frac{3}{2}x\cos x$ . Then $y''+y=3\cos x-\frac{3}{2}x\cos x-\frac{3}{2}x\cos x=3\cos x\neq 3\sin x$ then y is not a solution.

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3 years ago

What is the height, x, of the equilateral triangle?

e-lub [12.9K]

Using the Pythagorean Theorem, the height of the equilateral triangle is given by:

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<h3>What is the Pythagorean Theorem?</h3>

The Pythagorean Theorem relates the length of the legs $l_1$ and $l_2$ of a right triangle with the length of the hypotenuse h, according to the following equation: