I need help with 14 and 16 please!!!!
1 answer:
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Answer: x>_3.2 OR x<_ -0.75
Step-by-step explanation: first break down your compound inequality. 5x-4>_12
You first cancel out your constants by adding 4 to both sides. Now you’re left with 5x>_16 then to cancel five you have to divide on both sides by five which equals to 3.2. Then, x>_ 3.2.
Next you do your second part, 12x+5<_-4
So first cancel out the constant of 5 by subtracting 5 on both sides, making the equation 12x<_-9. Now, you divide by 12 on both sides, making it -9/12. Which effectively is -0.75. Therefor, the answer being x<_ -0.75. Add the two together x>_3.2 OR x<_0.75
Answer: The length is 13cm and the width is 7cm
Step-by-step explanation:
For a rectangle of length L and width W, the area is:
A = W*L
In this case we have:
L = (2*x + 9) cm
W=(3*x + 1) cm
Then the area of the rectangle is:
A = (2*x + 9)*(3*x + 1) cm^2
A = (6*x^2 + 2*x + 27*x + 9) cm^2
A = (6*x^2 + 29*x + 9) cm^2
now we remove two squares with sides of x cm
The area of each one of these squares is (x cm)*(x cm) = x^2 cm^2
Then the area of the figure will be:
area = (6*x^2 + 29*x + 9) cm^2 - (2*x^2 ) cm^2
area = (4*x^2 + 29*x + 9) cm^2
Now we know that the area of this shape is 83 cm^2, then we need to solve:
83 cm^2 = (4*x^2 + 29*x + 9) cm^2
0 = (4*x^2 + 29*x + 9) cm^2 - 83 cm^2
0 = (4*x^2 + 29*x - 74) cm^2
Then we need to solve:
0 = 4*x^2 + 29*x - 74
Here we can use Bhaskara's equation, the solutions of this equation are given by:
Then the two solutions are:
x = (-29 - 45)/8 = -9.25 (for how the length and width are defined, we can not have x as a negative number, then this solution can be discarded).
The other solution is:
x = (-29 + 45)/8 = 2
x = 2
Then the length and width of the rectangle are:
Length = (2*2 + 9)cm = 13 cm
Width = (3*2 + 1)cm = 7cm