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Kisachek [45]
3 years ago
9

I NEED HELP ON THISSSSS

Mathematics
2 answers:
Dafna11 [192]3 years ago
7 0

Answer:

A is correct

Step-by-step explanation:

Lynna [10]3 years ago
3 0
The answer to that question is C
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I NEED HELP ASAP BEFORE IT TIMES OUT!
padilas [110]
Ahh ok so split it up into two rectangles:

Rectangle one:
A=lw
A=4(12)
A=48

Rectangle two:
A=lw
A=10(2)
A=20

Add both areas up to get 68 ft squared!! :)
7 0
3 years ago
Read 2 more answers
A. When solving the system of linear equations 6x+y = 1 and 7x+2y = -3 with substitution, it is easiest to first solve the equat
nalin [4]

Answer:

6x + y = 1

y = 1 - 6x

7x + 2(1 - 6x) = -3 substitute 1 - 6x in for y

7x + 2 - 12x = -3 combine like terms

2 - 5x = -3 subtract 2 from both sides

-5x = -5 divide by -5

x = 1

3 0
3 years ago
3(x -y) + 4 if X = 3 and y = 2
Alchen [17]

Answer:

7

Step-by-step explanation:

3(3-2)+4

3(1)+4

3+4

7

7 0
2 years ago
2) Find x if m2) = 6x + 1 and<br> mZSOR = 11x + 6.<br> s<br> P
Virty [35]
X = -1 11x+6=6x+1 which leads to 5x=-5
3 0
3 years ago
two positive numbers x and y, with the maximum value 4, add up to 5. what is the difference between the maximum and minimum valu
Svetach [21]

Answer:

  92

Step-by-step explanation:

Since the sum of the two numbers is 5, we can represent one of them by x and the other by 5-x. Then the desired product is ...

  x²(5-x)³

A graphing calculator can show the extreme values of this on the interval 1 ≤ x ≤ 4. The maximum is 108 at x=2; the minimum is 16 at x=4.

The difference between the maximum and minimum is 108-16 = 92.

_____

If you like, you can take the derivative and set it to zero.

  f(x) = x²(5 -x)³

  f'(x) = 2x(5 -x)³ +x²(-3)(5-x)² = x(5 -x)²(2(5-x) -3x)

  f'(x) = 5x(5-x)²(2-x)

This will be zero for x=0, x=5, and x=2. The points at x=0 and x=5 represent minima in the product. The values x=0 and x=5 are not in the domain of interest. The point at x=2 represents a maximum.

To find the function extremes on an interval, we need to evaluate the function where the derivative is zero, and also at the ends of the interval. So, the function values of interest are ...

  f(1) = 1²·4³ = 64

  f(2) = 2²·3³ = 108 . . . . product maximum

  f(4) = 4²·1³ = 16 . . . . . . product minimum

The difference between the maximum and minimum is 92.

5 0
3 years ago
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