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mash [69]
3 years ago
14

Which expression represents the calculation “add 5 and 6, then multiply by 3"?

Mathematics
2 answers:
valina [46]3 years ago
8 0
(5+6)*6 Following the rules of PEMDAS you would have to put 5+6 in parentheses so multiplication wouldn't be first.  
Anarel [89]3 years ago
5 0
(5 + 6)3   would represent  what you want.
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Find the distance between the numbers 6.7 and 9.2 on the number line
igor_vitrenko [27]
That indeed issss 2.5
8 0
3 years ago
7) 28 is 35% of what number?​
Brums [2.3K]

Answer:

80

Step-by-step explanation:

28×100÷35=80

6 0
3 years ago
#7-9 please help with an explanation,, will mark brainliest
Mandarinka [93]

Answers:

<h2>7. 12</h2><h2>8. -√18</h2><h2>9. √220</h2>

Step-by-step explanation:

To find out which is greater, we must eliminate the radical symbol.

#7

\sqrt{88} = 9.38083151965

9.4 < 12

#8

-\sqrt{18} = -4.24264068712

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#9

\sqrt{220} = 14.8323969742

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I'm always happy to help :)

8 0
3 years ago
How do you simplify 33xy/3x?​
Llana [10]

Answer:

11y

Step-by-step explanation:

33xy/3x

33/3=11

x/x=1

so we have

33xy/3x=11y

5 0
3 years ago
Read 2 more answers
Which of the following functions are solutions of the differential equation y'' + y = 3 sin x? (select all that apply)
DiKsa [7]

Answer:

C

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.

3 0
3 years ago
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