MATHEMATICS MIDDLE SCHOOL

Solve the system using elimination.
3x + 5y = 55
5x + 4y = 57

Answers

Answer 1
Answer:

Answer:

x=5, y=8. (5, 8).

Step-by-step explanation:

3x+5y=55

5x+4y=57

---------------

5(3x+5y)=5(55)

-3(5x+4y)=-3(57)

-------------------------

15x+25y=275

-15x-12y=-171

---------------------

13y=104

y=104/13

y=8

3x+5(8)=55

3x+40=55

3x=55-40

3x=15

x=15/3

x=5

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Does anyone knows how to do this I don’t understand it.

Answers

Between the two, the difference in x is -1 and the difference in y is 3. 
Y's difference is rise, and x's difference is run. 
Divide rise over run to get the slope. 
3 / -1 = -3 
So the slope is -3, eliminating 2 and 4. 
The y-intercept that fits the graph is 7, not -7. 
That means the correct equation is C. 
Hope this helps!
The correct answer is c

Durning the month of January, temperatures can decrease by 19 per week . If this happens, the temperature on January 12 was 10 , what's is the temperature on January 19th

Answers

If it does drop it would be -9 by the end of the week.

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cn b CX b,hxbcxhcjdhx smuggling

Whats the answer for (2x+6)-(7x+3)

Answers

Well 2x plus 7x equals 9x
 6+3 equals 9
so 9x-9
The answer is -5x+3.
Hope this helps:)

Kareem rides her bike 25 2 over 5 miles Harry rides his bike 25 3 over 8 miles.Who rides farther

Answers

25 2/5 = 25.4 miles (Kareem)
25 3/8 = 25.375 miles (Harry)

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Help please! this is due tomorrow so please help out! will mark as brainiest!!

Answers

Answer:

\large\boxed{\bold{Q1}\ -(128)/(9)}\n\boxed{\bold{Q2}\ 2^(25)}

Step-by-step explanation:

\bold{Q1}\n\n(-2(3^2\cdot3^3)^2)/(3^(12)\cdot2^(-6))\qquad\text{use}\ a^n\cdot a^m=a^(n+m)\n\n=-(2(3^(2+3))^2)/(3^(12)\cdot2^(-6))=-(2(3^5)^2)/(3^(12)\cdot2^(-6))\qquad\text{use}\ (a^n)^m=a^(nm)\n\n=-(2\cdot3^(5\cdot2))/(3^(12)\cdot2^(-6))=-(2\cdot3^(10))/(3^(12)\cdot2^(-6))\qquad\text{use}\ (a^n)/(a^m)=a^(n-m)\n\n=-2^(1-(-6))\cdot3^(10-12)=-2^(7)\cdot3^(-2)\qquad\text{use}\ a^(-n)=(1)/(a^n)\n\n=-2^7\cdot(1)/(3^2)=-(2^7)/(3^2)=-(128)/(9)

\bold{Q2}\n\n(4^2)/(4^(-7))\cdot(2^4\cdot2^3)\cdot(3^(-2))^0\qquad\text{use}\ (a^n)/(a^m)=a^(n-m),\ a^n\cdot a^m=a^(n+m),\ a^0=1\n\n=4^(2-(-7))\cdot2^(4+3)\cdot1=4^(9)\cdot2^7=(2^2)^(9)\cdot2^7\qquad\text{use}\ (a^n)^m=a^(nm)\n\n=2^(2\cdot9)\cdot2^7=2^(18)\cdot2^7\qquad\text{use}\ a^n\cdot a^m=a^(n+m)\n\n=2^(18+7)=2^(25)
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