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nalin [4]
3 years ago
7

How many solutions does the system of equations have?

Mathematics
1 answer:
SIZIF [17.4K]3 years ago
8 0
The answer is d) none because substituting -1/4x + 5/3 for y in the first equation gives 3x - 3x + 5/3 = 20 so 5/3 = 20 which is not true. 
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If X and Y vary directly, as x decreases, what happens to the value of y?
diamong [38]
As x decreases y also decreased
5 0
3 years ago
Solve for m: <br> 8(m + 2) = 3(12 - 4m)
Tema [17]

Answer:

m=1

Step-by-step explanation:

step 1: 8(m + 2) = 3(12 - 4m) remove the parentheses

step 2: 8 m + 2 = 3 12 - 4m move the terms

step 3: 8m+ 12m= 36 - 16 collect like terms and calulate

step 4: 20m=20 divide both sides

solution: m=1

3 0
3 years ago
Read 2 more answers
Why is -9an acceptable answer to the equation x2=81 ?
Arte-miy333 [17]

Answer:

X=9 or x=-9

Step-by-step explanation:

X^2=81

Can be written as

x^2-81=0

Factorize by the property:

a^2-b^2=(a-b)(a+b)

So the expression becomes:

(x-9)(x+9)=0

Therefore, either x-9=0or x+9=0

Then the solutions are x=9 or x=-9

8 0
4 years ago
Read 2 more answers
Just once i would like math to solve it's own problems,<br> i ain't no therapist.
Artyom0805 [142]

Answer:

brainly.com/question/22779054 PLEASE HELP ME

Step-by-step explanation:

5 0
3 years ago
Ok whoever answers this correct will get 100 brainly points
Snezhnost [94]

Answer:

We are given an area and three different widths and we need to determine the corresponding length and perimeter.  

The first width that is provided is 4 yards and to get an area of 100 we need to multiply it by 25 yards.  This would mean that our length is 25 yards and our perimeter would be 2(l + w) which is 2(25 + 4) = 58 yards.

The second width that is given is 5 yards and in order to get an area of 100 yards we need to multiply by 20 yards.  This would mean that our length is 20 yards and our perimeter would be 2(l + w) which is 2(20 + 5) = 50 yards.

The final width that is given is 10 yards and in order to get an area of 100 yards we need to multiply by 10.  This would mean that our length is 10 yards and our perimeter would be 2(l + w) which is 2(10 + 10) = 40 yards.

Therefore the field that would require the least amount of fencing (the smallest perimeter) is option C, field #3.

<u><em>Hope this helps!</em></u>

3 0
2 years ago
Read 2 more answers
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