Answer:
B.
Step-by-step explanation:
Here the best method to solve is by substituting the end values of the set in each option , otherwise it will a time consuming problem.
Now substitute x=-4 in all the options
A.
16+8-8=16>0
so out of option A and C option C is correct.
B.
16-8-8=0 which means for the values of x>-4 is less than 0
Now substitute x=2 in all the options
A.
4-4-8=-8<0 . so option A and C both are incorrect.
B.
4+4-8=0 which means for the values of x<2 is less than 0
Therefore the correct option is B
Answer:
20 in²
Step-by-step explanation:
Assuming that the width of the sign is x, then from the question, we're told that the length is 5 times it's width, so
b = x inches
l = 5x inches
Again, we're told that the perimeter of the sign is 24 inches, and we know already that the perimeter of a rectangle is given as
Perimeter = 2(l + b), substituting this, we have
24 = 2(5x + x)
24 = 10x + 2x
24 = 12x
x = 24 / 12
x = 2 inches.
Since x is the width of the rectangle, and it's 2 inches, we use it to find the length of the sign.
l = 5 * 2
l = 10 inches.
Then, we are asked to find the area of the sign. Area of a rectangle is given as
A = l * b, so if we substitute, we have
Area = 10 * 2
Area = 20 square inches
Answer:
The expression is already in decimal form.
1.2
Step-by-step explanation:
(づ￣ ³￣)づ
Answer:
Not similar
Step-by-step explanation:
Please refer to the figure attached for the diagram of this problem.
Steps needed to find the width of the field (CD):
First, we should note that angle d would be equal to 27 degrees because there are two parallel lines that are cut by a transversal. Furthermore, angle a would be equal to 5 degrees since we just need to subtract 27 from 32.
We then subtract 27 and 5 from 180 to get angle c.
. Angle c is therefore 148 degrees.
Next, we need to find angle e which is just the supplementary of angle c. Angle e therefore measures
degrees.
For the next step we use sine law to find the length of segment AC:
Lastly, we need to utilize the sine law again to find the length of segment CD or the width of the field:
ANSWER: The width of the field is 290 ft.
