Answer:
C. 
Step-by-step explanation:
Let x represent number of miles.
We have been given that a taxi cab charges $1.50 for the flat fee and $0.50 for each mile.
The cost of riding x miles would be 
and the total cost of riding x miles would be 
.
Since Sharon has only $25 to spend on the taxi ride, so the cost of riding x miles on taxi must be less than or equal to 25.
We can represent this information in an inequality as:
Therefore, option C is the correct choice.
Answer:
3.5m is 500% of 70 cm and 70 cm is 20% of 3.5m.
Step-by-step explanation:
70 cm = 70cm/100= 0.7m
To solve the first problem use this equation:
3.5/0.7 *100 to get 500%
And to solve the second problem you use the same equation:
0.7/3.5*100 to get 20%
Hope this helps you.
Answer:
The required prove is shown below.
Step-by-step explanation:
Consider the provided statement.
The required figure is shown below:
Statement Reasons
PQ ≅ TQ; UQ ≅ QS Given
PQ = TQ, UQ = QS Definition of Congruence
PQ + QS = PS; TQ+QU = TU Segment Addition Postulate
TQ + QS = PS Substitution (PQ = TQ)
TQ + QS = TU Substitution (QU = QS)
PS = TU Transitive Property
Prove: PS ≅TU Definition of Congruence
Hence, the required prove is shown above.
Answer:
Let total pages be x
ATQ,
(x/5)+40=7x/10
(x+200)/5=7x/10
Cross multiply
10x+2000=35x
10x-35x=-2000
-25x=-2000
x=2000/25
x=80 pages
Therefore pages left= 3/10 of pages
=3*80/10=24 pages
Answer:
option 2 , 3 , 4 , 5
Step-by-step explanation:
Use Pythagorean theorem
hypotenuse² = base² + altitude².
If the lengths satisfies Pythagorean rule, then they form a right angle triangle.
2) 41² = 40² + 9²
1681 = 1600 + 81
1681 = 1681
41, 40 , 9 will form a right angle as it satisfies Pythagorean rule.
3) 15² = 12² + 9²
225 = 144 + 81
225 = 225
9, 12 , 15 will form a right angle as it satisfies Pythagorean rule.
4) 17² = 15² + 8 ²
289 = 225 + 64
289 = 289
17, 15 , 8 will form a right angle as it satisfies Pythagorean rule.
5)53² = 28² + 45²
2809 = 784 + 2025
2809 = 2809
53 , 28 , 45 will form a right angle as it satisfies Pythagorean rule.
