1/3 hour = 20 minutes
1 1/2 dozen = 12 + 6 = 18 cookies
18 cookies x 3 (3 sets of 20 minutes = 1 hour) = 54 cookies per hour
Answer:
sin (- 135°)= – sin 135°= – sin (1 × 90°+ 45°) = – cos 45° = – 1√2
cos (- 135°)= cos 135°= cos (1 × 90°+ 45°) = – sin 45°= – 1√2
tan (- 135°) = – tan 135° = – tan ( 1 × 90° + 45°) = – (- cot 45°) = 1
csc (- 135°)= – csc 135°= – csc (1 × 90°+ 45°)= – sec 45° = – √2
sec (- 135°)= sec 135°= sec (1 × 90°+ 45°)= – csc 45°= – √2
cot (- 135°) = – cot 135° = – cot ( 1 × 90° + 45°) = – (-tan 45°) = 1
Step-by-step explanation:
hope this helps
Ok don’t worry this is a very easy topic
It falls under “gradients of a straight line”
I’ll try my best to explain the question but u can find plenty of videos online (Fuse School or Cognito) explaining the topic in detail.
Gradient means slope or how steep a line is.
To find the gradient we use the formula
(Change in y) / (change in x)
*Step 1*
Choose any two coordinates (exact) of the line from the graph
For line 1 let’s take
(2,4) (-3,4)
Now from this
the change in y= (4-4) = 0
the change in x = (-3-2) = -5
So the gradient = 0/-5 = 0
*Now this makes sense because a straight horizontal line has no slope and thus has a gradient of zero. So even without all the calculation we can figure this out easily.
Now for line 2 let’s take the coordinates:
(-3,1) (-3,3)
The change in y= (3-1) = 2
The change in x = (-3- -3) = 0
So gradient = 2/0 = “Undefined”
*The gradient of vertical straight line is always undefined (no gradient)*
*The gradient of horizontal straight lines is always 0*
Hope this helped and best of luck for your exams!
Your answer would be 144.5 i hope this helps
Please help me out with my questions and I’ll help you as well
