Step-by-step explanation:
I used Pythagoras Theorem to find the diagonal distance as placement of objects is right angles triangle.
a : Elm street
b : Birch street
c : Diagonal distance
c^2 = a^2 + b^2
c^2 = 15^2 + 8^2
c^2 = 225 + 64
c = √289
c = 17 blocks
I added the total distance of the original route Joline took
= 15 blocks + 8 blocks
= 23 blocks
How much less distance does she have to walk if she took the diagonal route
= Total distance of original route - distance of diagonal route
= 23 - 17
= 6 blocks
Joline have to walk less 6 blocks if she took the diagonal route compared to her original route
Answer:
Given the statement: The quotient of a number x and 4 is atmost 5.
Let x be a number.
"quotient of a number x and 4" represented as
"at most" means ''
Therefore, the inequality for the given statement is;
Solve the above inequality:
multiply both sides by 4 we get;
Simplify:
Therefore, the solutions for the given inequality is:
1. d. d = 66t
The maximum speed of the wildebeest is v=66 ft/s. Speed is defined as the ratio between the distance covered (d) and the time taken (t):
Re-arranging the formula, we get
And substituting v=66 ft/s, we find
2. c. y = 5 - n
In fact, we can check that this rule satisfies all the numbers of the sequence:
n=3 --> y = 5 - 3 = 2
n=4 --> y = 5 - 4 = 1
n=5 --> y = 5 - 5 = 0
n=6 --> y = 5 - 6 = -1
3. d. $5.50, $11.00, $16.50; E = 5.50h
In fact, if we use the rule E=5.50 h, we see that the table is completed as follows:
h=1 --> E=(5.50)(1 h)=5.50$
h=2 --> E=(5.50)(2 h)=11.00 $
h=3 --> E=(5.50)(3 h)=16.50$
h=4 --> E=(5.50)(4 h)=22.00$
h=5 --> E=(5.50)(5 h)=27.50$
4. c. if d ≥ 5 then C = 4.70d, if d < 5 then C = 4.70d - 1; $17.80, $32.90
If the number of days is more than 5, the cost is 4.70$ per day, therefore c=4.70d (where d is the number of days); if the number of days is less than 5, one dollar is returned, so the cost will be C=4.70d-1. By applying this rule, we can calculate the rental cost after d=4 days and d=7 days:
- d=4 --> c=4.70*4-1=17.80$
- d=7 --> c=4.70*7=32.90$
5. b. y = 3x + 3
In fact, we can check this is the correct rule by substituting the numbers:
x=0 --> y=3*0+3=3
x=1 --> y=3*1+3=6
x=2 --> y=3*2+3=9
x=3 --> y=3*3+3=12
x=4 --> y=3*4+3=15
7. x -2 -1 0 1 2
y (-16 ) (-6 ) (4 ) (14 ) (24 )
The rule we have to apply is:
-10x + y = 4
Re-arranging it, we can solve for y:
y = 10x + 4
So, let's substitute the different values of x:
x=-2 --> 10*(-2)+4=-20+4= -16
x=-1 --> 10*(-1)+4=-10+4= -6
x=0 --> 10*0+4= 4
x=1 --> 10*1+4=10+4= 14
x=2 --> 10*2 +4 =20+4 = 24
8. y = -5x - 1
In fact, if substitute the different values of x given by the problem, we find:
x = -2 --> y = -5*(-2) -1= 10 -1 = 9
x = -1 --> y = -5*(-1) -1 = 5 -1 = 4
x = 0 --> y = -5*0 -1 = -1
x = 1 --> y = -5*1 - 1 = -6
x = 2 --> y = -5*2 -1 = -11