Answer:
Douglas can race the go-karts at least 3 times.
Step-by-step explanation:
Given that:
Worth of game card = $20
Cost of go-kart = $3.50 each time
Amount Douglas wants to left = $7.75
Let,
x be the times Douglas can ride go-kart.
20 - 3.50x ≤ 7.75
-3.50x ≤ 7.75 - 20
-3.50x ≤ -12.25
3.50x ≤ 12.25
Dividing both sides by 3.50
Hence,
Douglas can race the go-karts at least 3 times.
The rules of significant figures are: 1. all non zeros are significant. 2. trailing zeros are significant. 3. zeros between non-zeros are signifincant. Hence, applying these to the given.
A. 4 sig figs
B. 1 sig fig
C.6 sig figs
D.1 sig fig
6. One variable only so pretty straightforward.
length-x+4
width-x
x+x+4=80
2x=76
x=38
x+4=42
answer: length 42cm and width 38cm
7. Another money problem!
n-# of nickels
q-# of quarters
n=3+q
0.05n+0.25q=2.85
Substitution works like a charm!
0.05(3+q)+0.25q=2.85
0.15+0.05q+0.25q=2.85
0.3q=2.7
q=9
n=3+q
n=3+9
n=12
answer: 9 nickels and 12 quarters
8. One variable situation again.
Ann's money-2b+9
Betty's money-b
b+2b+9=60
3b=51
b=17
2b+9=43
answer: Ann has $43 and Betty has $17.
9. # of red m&m's-x+1
# of blue m&m's-x
x+1+x=13
x=6
x+1=7
answer: 6 blue and 7 red m&m's
10. a-number of adult tickets
s-number of student tickets
a+s=785 ----> a=785-s
5a+2s=3280
5(785-s)+2s=3280
-3s=-645
s=215
a+s=785
a+215=785
a=570
answer: 215 children tickets and 570 adult tickets
Answer:
Yes they are similar because they are both right angles but the one on the right is bigger than the one on the left.
Step-by-step explanation:
