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3 years ago

Tara answered 68 questions correctly on an 80 questions test express this amount as a fraction percent and decimal

Mathematics

2 answers:

Mandarinka [93]3 years ago

4 0

Answer
85% and as a decimal it is 0.85

hope this helps and have a wonderful day :)

allsm [11]3 years ago

4 0

Fraction is 68/80
85%
decimal is .85

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The graphs of f(x) and g(x) are shown below:

rosijanka [135]

<h2>Answer:</h2>

Option: C is the correct answer.

C.) x = −3.8, 3

<h2>Step-by-step explanation:</h2>

The function f(x) is given by:

$f(x)=x^2-x-12$

and the function g(x) is given by:

$g(x)=-1.8x-0.6$

Now, we are asked to find the solution of the equation:

$f(x)=g(x)$

i.e. we have to find the value of x such that both the functions are equal i.e.

$x^2-x-12=-1.8x-0.6\\\\i.e.\\\\x^2-x+1.8x-12+0.6=0\\\\i.e.\\\\x^2+0.8x-11.4=0\\\\i.e.\\\\10x^2+8x-114=0\\\\i.e.\\\\5x^2+4x-57=0$

Now, on solving the equation using the quadratic formula i.e. the solution of the equation:

$ax^2+bx+c=0$

is given by:

$x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}$

Here we have:

$a=5,\ b=4\ and\ c=-57$

Hence, the solution is given by:

$x=\dfrac{-4\pm \sqrt{4^2-4\times 5\times (-57)}}{2\times 5}\\\\i.e.\\\\x=\dfrac{-4\pm \sqrt{16+1140}}{10}\\\\i.e.\\\\x=-3.8\ and\ x=3$

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3 years ago

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.