Answer:
160
Step-by-step explanation:
Is means equals and of means multiply
24 = 15% * N where N is the unknown number
Change to decimal form
24 = .15 N
Divide each side by .15
24/.15 = .15N/.15
160 = N
Answer:
The answer is 60cm. (b)
Step-by-step explanation:
For every 1.5cm on the map, it represents 3 ft on real-size scale.
In other words, for every 1cm on the map, it represents 3/1.5 = 2ft on real size scale.
If you want to find 120ft of real-size scale, you can use 120ft/2ft * 1cm which gives 60cm
This is a little long, but it gets you there.
- ΔEBH ≅ ΔEBC . . . . HA theorem
- EH ≅ EC . . . . . . . . . CPCTC
- ∠ECH ≅ ∠EHC . . . base angles of isosceles ΔEHC
- ΔAHE ~ ΔDGB ~ ΔACB . . . . AA similarity
- ∠AEH ≅ ∠ABC . . . corresponding angles of similar triangle
- ∠AEH = ∠ECH + ∠EHC = 2∠ECH . . . external angle is equal to the sum of opposite internal angles (of ΔECH)
- ΔDAC ≅ ΔDAG . . . HA theorem
- DC ≅ DG . . . . . . . . . CPCTC
- ∠DCG ≅ ∠DGC . . . base angles of isosceles ΔDGC
- ∠BDG ≅ ∠BAC . . . .corresponding angles of similar triangles
- ∠BDG = ∠DCG + ∠DGC = 2∠DCG . . . external angle is equal to the sum of opposite internal angles (of ΔDCG)
- ∠BAC + ∠ACB + ∠ABC = 180° . . . . sum of angles of a triangle
- (∠BAC)/2 + (∠ACB)/2 + (∠ABC)/2 = 90° . . . . division property of equality (divide equation of 12 by 2)
- ∠DCG + 45° + ∠ECH = 90° . . . . substitute (∠BAC)/2 = (∠BDG)/2 = ∠DCG (from 10 and 11); substitute (∠ABC)/2 = (∠AEH)/2 = ∠ECH (from 5 and 6)
- This equation represents the sum of angles at point C: ∠DCG + ∠HCG + ∠ECH = 90°, ∴ ∠HCG = 45° . . . . subtraction property of equality, transitive property of equality. (Subtract ∠DCG+∠ECH from both equations (14 and 15).)
The answer would be 8.5. 8.54 rounds down to 8.54.
Hope this helps!
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Plot the graph then you do the vertical line test. You draw a vertical line on the graph. If the line touches the graph at just one point,note just one point then it is a function