Answer:
1/5... 1 on top 5 on bottom
Step-by-step explanation:
sub in the values:
will now be
now add:
simplify:
1/5
Find the y value for point F such that DF and EF form a 1:3 ratio. E(2,4) D(-3,-3)
2 answers:
I drew the segment and used Pythagorean theorem to solve for its measure. The line formed is the hypotenuse of the imaginary right triangle.
Among the choices only -1.33 and -1.25 is a feasible choice. But I am leaning towards -1.25 as the y-value of point F based on my diagram. Please see attachment
Among the choices only -1.33 and -1.25 is a feasible choice. But I am leaning towards -1.25 as the y-value of point F based on my diagram. Please see attachment
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volume of the box is 675 cubic inches
A machine produces open boxes using square sheets of plastic.
It is a square sheet so length and width are same
Lets assume length as x so width is also x
The machine cuts equal-sized squares measuring 3 inches on a side from each corner of the sheet.
After turning up the sides the height of the box becomes 3 inches
We know the volume of a box formula
Volume = Length * width * height
We know length is x , width is x and height = 3
So V = x * x * 3
Given volume = 675 cubic inches
Divide by 3 on both sides
Now we take square root on both sides
x = 15
the length of one side of the open box is 15 inches.
Answer:
see explanation
Step-by-step explanation:
Using the trigonometric identities
• 1 + cot² x = csc²x and csc x =
• sin²x + cos²x = 1 ⇒ sin²x = 1 - cos²x
Consider the left side
sin²Θ( 1 + cot²Θ )
= sin²Θ × csc²Θ
= sin²Θ × 1 / sin²Θ = 1 = right side ⇔ verified
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Consider the left side
cos²Θ - sin²Θ
= cos²Θ - (1 - cos²Θ)
= cos²Θ - 1 + cos²Θ
= 2cos²Θ - 1 = right side ⇒ verified