<h3>
Answer:</h3><h3>
The Direction of the rotation is counterclockwise around the origin.</h3>
Step-by-step explanation:
What was Given was that Point G is rotated 90°. The coordinate of the pre-image point G was (7, –5), and its image G’ is at the coordinate (5, 7).
What you need to find was : What is the direction of the rotation.
So coming to conclusions is we have given that Point G is rotated 90°.
The Pre-image of G = ( 7, -5).
Translated image G' = ( 5 ,7).
By the Rotation rule of 90° is : (-y, x) 90 degree rotation counterclockwise around the origin.
(y, -x) 90 degree rotation clockwise about the origin.
We can from G (7,-5) and G'( 5 ,7) it is rotation counterclockwise around the origin.
Therefore, Direction of rotation is counterclockwise around the origin.
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I believe the answer is 350,000.
Step-by-step explanation:
step 1. (1) 5a +8c +4p = 57
(2) 9a +8c + 7p = 89
(3) 8a + 4c + 5p = 68.
step 2. this is a 3 variable system. solve the system.
step 3. eliminate c from (1) and (2) and also from (2) and (3). so you now have 4a + 3p = 32 and 7a + 3p = 47. now you have a two variable system
step 4. eliminate p by subtracting: 3a = 15 so a = 5.
step 5. if a = 5 then p = 4 ( plug into step 3)
step 6. if a = 5, p = 4, then c = 2 (plug into step 1).
step 7. apples = 5, pears = 4, cherries = 2.
Even if a diagram is not given, if you visualize it, EDG forms a right triangles with a bisector line DF. FG is then the lower half of segment EFG. To determine FG, use trigonometric functions like sine, cosine or tangent to relate the sides. For FG, you could use
tan ∅/2 = FG/DG
FG = DG tan(∅/2)
The angle is denoted as ∅/2 because of the angle bisector. DG is the base of the right triangle.
9514 1404 393
Answer:
a) 5 cm
b) 8 cm
c) 9 m
Step-by-step explanation:
It can be worthwhile to work the last attachment (a) first, since these are all variations of the same triangle.
The Pythagorean theorem tells you the sum of the squares of the legs is the square of the hypotenuse.
<u>Problem 6a</u>:
3² +4² = x² . . . . fill in the given numbers; all measures in cm
9 + 16 = x² . . . . simplify exponents
25 = x² . . . . . . . simplify sum
x = √25 = 5 . . . take the square root.
x = 5 cm . . . . . . apply units
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Note that this triangle is a 3-4-5 right triangle. That is a set of side lengths (ratios) that is useful to remember. In this problem set, you see it again immediately.
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<u>Problem 6b</u>:
Shortest-to-longest, the side ratios of the given triangle are ...
6 : x : 10
For some x', this is ...
3 : x' : 5 . . . . . . . . . . . . . matches a 3-4-5 triangle with x' = 4
The scale factor is 6/3 = 10/5 = 2, so we have ...
x = 2·x' = 2·4 = 8
x = 8 cm . . . . . with units
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<u>Problem 6c</u>:
The side ratios are ...
x : 12 : 15 which reduces to x' : 4 : 5
This matches a 3-4-5 triangle with x' = 3, and a scale factor of 12/4 = 15/5 = 3.
Then ...
x = 3·x' = 3·3 = 9
x = 9 m . . . . . with units
