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sukhopar [10]
3 years ago
6

Amy is 5 years older than Ben. Three times Amy's age added to six times Ben's age is 42. How old are Amy and Ben?

Mathematics
1 answer:
I am Lyosha [343]3 years ago
5 0

Answer:

Amy: 8, Ben: 3

Step-by-step explanation:

No explanation here

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Did I get this right?
natima [27]

Answer: Yes, the answer is 40 quarters.


Step-by-step explanation:

1. You need to remember that 4 quarters make a dollar.

2. Keeping this on mind, you can make a Rule of three, as following: If there are 4 quarters in 1 dollar, how many quarters are in 10 dollars?

Then:

4 quarters-------1 dollar

          x---10 dollars

x=\frac{(10dollars)(4quarters)}{1dollar}\\x=40quarters

3. Therefore, there are 40 quarters in one roll.


7 0
3 years ago
Find the solution to the initial value problem
Karo-lina-s [1.5K]

Answer:

y = (11x + 13)e^(-4x-4)

Step-by-step explanation:

Given y'' + 8y' + 16 = 0

The auxiliary equation to the differential equation is:

m² + 8m + 16 = 0

Factorizing this, we have

(m + 4)² = 0

m = -4 twice

The complimentary solution is

y_c = (C1 + C2x)e^(-4x)

Using the initial conditions

y(-1) = 2

2 = (C1 -C2) e^4

C1 - C2 = 2e^(-4).................................(1)

y'(-1) = 3

y'_c = -4(C1 + C2x)e^(-4x) + C2e^(-4x)

3 = -4(C1 - C2)e^4 + C2e^4

-4C1 + 5C2 = 3e^(-4)..............................(2)

Solving (1) and (2) simultaneously, we have

From (1)

C1 = 2e^(-4) + C2

Using this in (2)

-4[2e^(-4) + C2] + 5C2 = 3e^(-4)

C2 = 11e^(-4)

C1 = 2e^(-4) + 11e^(-4)

= 13e^(-4)

The general solution is now

y = [13e^(-4) + 11xe^(-4)]e^(-4x)

= (11x + 13)e^(-4x-4)

3 0
3 years ago
A rectangular plot of ground is 5 meters longer than it is wide, It’s area is 20,000 square meters
Artyom0805 [142]
I'm assuming you're looking for the dimensions of the plot.  I'm going with that.  ;)  If the length of the plot is 5 meters longer than the width, then L = w + 5.  The area for a rectangle is L*w, and we have an area value of 20,000 so our formula is 20000=(w+5)(w) and 20,000=w^2+5w.  We will bring the 20,000 over by subtraction and set the polynomial equal to 0 to factor and solve for w.  w^2+5w-20,000=0  Solving for w we get values of w=138.9 and -143.9.  Of course the 2 things in math that will never EVER be negative are time and distance/length, so -143.9 is out.  Our width is 138.9 and the length is 138.9 + 5 so the length is 143.9.  And there you go!  Hope that's what you needed!
8 0
3 years ago
Name two equivalent fractions for each <br> 2/4<br> 1/3<br> 1/4
Scrat [10]
2/4=1/2 and 4/8
1/3=2/6 and 3/9
1/4=2/8 and 3/12
8 0
3 years ago
Read 2 more answers
A shopkeeper allows 10% discount on the marked price of a bicycle. If a costomer pays Rs 4068 with 13% VAT find the marked price
Over [174]

\large{ \tt{❁ \: S \: O \: L \: U \: T \: I \: O \: N \: ❁}}

  • We're provided - Discount % = 10 % , Cost with VAT [ SP with VAT ] = Rs 4068 & VAT % = 13%. We're asked to find out the marked price of the bicycle. Let's start :

\large{ \tt{❇ \: FIND \: SP \: WITHOUT \: VAT \: / \: SP\: ❇}}

\large{ \tt{❃ \: SP  \: with \: VAT= SP + VAT\% \: of \: SP}}

\large{ \tt{⇢ \: 4068 = SP + 13\% \: of \: SP}}

\large{ \tt{⇢ \: 4068 = SP +  \frac{13}{100}  \: sp}}

\large{ \tt{⇢ \: 4068 =  \frac{100 \:  \: SP + 13 \: SP}{100} }}

\large{ \tt{⇢ \: 4068 =  \frac{113 \: SP}{100} }}

\large{ \tt{⇢ \: 113 \: SP = 406800}}

\large{ \tt{⇢ \: SP =  \frac{406800}{113} }}

\large{ \tt{⇢ \: SP = 3600}}

  • Hence , SP = Rs 3600

\large{ \tt{✽ \: NOW , \: FIND \: THE \: MP \:✽ }}

  • Let Marked Price [ MP ] be x.

\large {\tt{❃ \: SP = MP - dis\% \: of \: MP}}

\large{ \tt{⇾ \: 3600 = x - 10\% \: of \: x}}

\large{ \tt{⇾ \: 3600 = x -  \frac{10}{100} } \: x}

\large{ \tt{⇾ \: 3600 =  \frac{100x - 10x}{100} }}

\large{ \tt{⇾ \: 3600 =  \frac{90x}{100} }}

\large{ \tt{⇾ \: 90x = 360000}}

\large{ \tt{⇾ \: x =  \frac{360000}{90} }}

\large{ \tt{⇾ \: x = Rs \: 4000}}

\large{ \boxed{ \boxed{ \tt{☂ \: OUR \: FINAL \: ANSWER :  \boxed {\tt {\: Rs \: 4000}}}}}}

  • Hope I helped! Let me know if you have any questions regarding my answer and don't hesitate to reach out to me if you need any assistance! :)
6 0
3 years ago
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