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FrozenT [24]
3 years ago
8

HELP 7TH GRADE MATH 3 QUESTIONSSSS GIVING BRANLIEST

Mathematics
2 answers:
Elan Coil [88]3 years ago
5 0
The 3rd questions answer is: In order to solve this, you need to add 20% of the $7.50 TO the $7.50.
First, multiply 0.20 x $7.50 = $1.50
Then, add the $1.50 to the original $7.50 because the price is increasing by 20%, so the new price of the ticket is $9.00. (I got this answer because someone else asked this question on BRAINLY and this was an answer, it’s correct btw)
Rashid [163]3 years ago
4 0

Answer:

  1. 3x + 5
  2. x² + 3x + 2
  3. $9

Step-by-step explanation:

3. The price was $7.50

Increasing by 20%,

20% of $7.50 is 7.50 × 20/100

= 750/100 × 20/100

= 75/10 × 2/10

= 15/2 × 1/5

= 3/2 × 1

= 3/2

Increasing by 20% = $7.50 + 3/2

750/100 + 5/2

75/10 + 5/2

75/10 + 15/10 = (75+15)/10 = 90/10 = $9

HOPE THIS HELPS YOU

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I'm sorry but the solution you proposed was a bit hard to follow, so I'll just post my solution: we convert "2 and 1/2" to a single fraction:

2+\dfrac{1}{2} = \dfrac{4}{2}+\dfrac{1}{2}=\dfrac{5}{2}

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3 years ago
Solve: (3x^2-y)dx + (4y^3-x)dy =0 and find the solution passing through (1,1).
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Step-by-step explanation:

The given equation is

(3x^{2}-y)dx+(4y^{3}-x)dy=0\\M(x,y)dx+N(x,y)dy=0

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\frac{\partial N}{\partial x}=\frac{\partial M}{\partial y}\\\\\therefore \frac{\partial N}{\partial x}=\frac{\partial (4y^{3}-x)}{\partial x} =-1\\\\\frac{\partial M}{\partial y}=\frac{\partial (3x^{3}-y)}{\partial y} =-1\\\\\therefore \frac{\partial N}{\partial x}=\frac{\partial M}{\partial y}=-1

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