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borishaifa [10]
3 years ago
9

Which fraction below is equivalent to 10/12?

Mathematics
1 answer:
Effectus [21]3 years ago
5 0

Answer:

5/6 or 10/12 are the same 0.83

Step-by-step explanation:

6/6-5/6=1/6 left     12/12-10/12=2/12 left

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In a mathematics test, Azmah scored 17 marks more than Yazid while Suzana's score is twice of Yazid's score. If their total scor
natta225 [31]

Answer:

53 marks

Step-by-step explanation:

To sue a linear equation to solve the given problem, we must assign a function to the score of one of them. Let the score of Azmah be N. Then given that Azmah scored 17 marks more than Yazid, Yazid's score will be

= N - 17

Given that Suzana's score is twice of Yazid's score, then Suzana's score

= 2(N - 17)

If their total score is 161 then

N + N - 17 + 2(N - 17) = 161

4N = 161 + 51

4N = 212

N = 53 . This is Azmah's score

3 0
2 years ago
Stacey had 3 cakes for her party, She had 1/8 of a cake left after the party. How much cake was eaten at her party?​
hoa [83]

Answer: 3=24/8 24/8 - 1/8 = 23/8 or 2 7/8

Step-by-step explanation:

Exchange 3 into 24/8 which is still 3 but in 8th’s. Then subtract 1/8 from 24/8 which is 23/8 or 2 7/8

8 0
2 years ago
Read 2 more answers
I NEED THIS QUICK! What is the volume of the rectangular prism?
Anika [276]

Answer:

30 3/8

Step-by-step explanation:

3 0
2 years ago
Integration of ∫(cos3x+3sinx)dx ​
Murljashka [212]

Answer:

\boxed{\pink{\tt I =  \dfrac{1}{3}sin(3x)  - 3cos(x) + C}}

Step-by-step explanation:

We need to integrate the given expression. Let I be the answer .

\implies\displaystyle\sf I = \int (cos(3x) + 3sin(x) )dx \\\\\implies\displaystyle I = \int cos(3x) + \int sin(x)\  dx

  • Let u = 3x , then du = 3dx . Henceforth 1/3 du = dx .
  • Now , Rewrite using du and u .

\implies\displaystyle\sf I = \int cos\ u \dfrac{1}{3}du + \int 3sin \ x \ dx \\\\\implies\displaystyle \sf I = \int \dfrac{cos\ u}{3} du + \int 3sin\ x \ dx \\\\\implies\displaystyle\sf I = \dfrac{1}{3}\int \dfrac{cos(u)}{3} + \int 3sin(x) dx \\\\\implies\displaystyle\sf I = \dfrac{1}{3} sin(u) + C +\int 3sin(x) dx \\\\\implies\displaystyle \sf I = \dfrac{1}{3}sin(u) + C + 3\int sin(x) \ dx \\\\\implies\displaystyle\sf I =  \dfrac{1}{3}sin(u) + C + 3(-cos(x)+C) \\\\\implies \underset{\blue{\sf Required\ Answer }}{\underbrace{\boxed{\boxed{\displaystyle\red{\sf I =  \dfrac{1}{3}sin(3x)  - 3cos(x) + C }}}}}

6 0
3 years ago
I Need Assistance With This:
Tamiku [17]

Answer:

22.5 in^2

Step-by-step explanation:

h = 3 in

base 1 = 6 in

base 2 = 9 in

So plug in

A = 1/2 (3) (6 + 9)

A = 3/2 (15)

A = 22.5 in^2

5 0
3 years ago
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