A recipe for pie crust calls for 2/3 cup of water for every 3 cups of flour. If you doubled the recipe, you would need 1 1/3 cup
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The coordinates of point F are (15/4 , -5/2)
- The coordinates of point G = (-3/4 , -3)
From the image attached:
AB divided into 4 similar parts, hence
- E is the mid-point of AB
- D is the mid-point of AE
- F is the mid-point of EB
Note that the rule of the mid-point states that:
When M (x , y) is the mid-point of the segment of AB, where A (x1 , y1) and B (x2 , y2), Then x = (x1 + x2)/2 and that of y = (y1 + y2)/2
Then lets solve for points E, F, D
Since A (-3 , -1) and B (6 , -3),
E is the mid-point of AB
Then E =[(-3 + 6)/2, (-1 + -3)/2]
= (3/2 , -2)
Since F is the mid-point of EB,
E (3/2 , -2) , B (6 , -3)
Then F = [(3/2 + 6)/2 , (-2 + -3)/2]
= (15/4 , -5/2)
Since D is the mid-point of AE, A (-3 , -1), E (3/2 , -2)
Then D = [(-3 + 3/2)/2 , (-1 + -2)/2]
= (-3/4 , -3/2)
Note also that:
BC is said to be an horizontal segment due to the fact that B and C have similar y - coordinate and G can be found on BC.
Since the y-coordinate of G is similar to y-coordinate of B and C
Then y-coordinate of G is = -3
Since DG ⊥ BC
Then DG = vertical segment
So G has similar x-coordinate of D
Then The x-coordinate of G = -3/4
The G = (-3/4 , -3)
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Answer:
(t, t -1, t)
Step-by-step explanation:
You have three unknowns but only 2 equations, so you can't really SOLVE this...you can get a solution with a variable still in it (I forget what this is called. I think it refers to infinite many solutions). Here's how it works:
Set up your matrix:
You want to change the number in position 21 (the 2 in the scond row) to a 0 so you have y and z left. Do this by multiplying the top row by -2 then adding it to the second row to get that 2 to become a 0. Multiplying in a -2 to the top row gives you:
Then add, keeping the first row the same and changing the second to reflect the addition:
The second equation is this now:
3y - 3z = -3. Solving for y gives you y = z - 1. Let's let z = t (some random real number that will make the system true. Any number will work. I'll show you at the end. Just bear with me...)
lf z = t, and if y = z - 1, then y = t - 1. So far we have that y = t - 1 and z = t. Now we solve for x:
From the first equation in the original system,
x - 2y + z = 2. Subbing in t - 1 for y and t for z:
x - 2(t - 1) + t = 2. Simplify to get
x - 2t + 2 + t = 2 and x - t = 0, and x = t. So the solution set is (t, t - 1, t). Picking a random value for t of, let's say 2, sub that in and make sure it works. If:
x - 2y + z = 2, then t - 2(t - 1) + t = 2 becomes t - 2t + 2 + t = 2, and with t = 2, 2 - 2(2) + 2 + 2 = 2. Check it: 2 - 4 + 4 = 2 and 2 = 2. You could pick any value for t and it will work.