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zzz [600]
2 years ago
9

Austin goes to a trade show that sells reptiles. There is a $5.00 entrance fee. He visited a booth that sold geckos for $35 each

. How much did he spend if he bought 3 geckos?
A. 62.5%
B. 6 5/9
C. 22.5%
D. 110
Mathematics
1 answer:
vivado [14]2 years ago
7 0
The answer is D because 35x3 is 105 + the $5 fee is 110
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What is the opposite integer of 10
NeX [460]
Since 10 is a positive the opposite must be negative. Opposite integer of 10 is -10 ( negative 10).

Happy studying ^-^
4 0
3 years ago
Read 2 more answers
I'm so confused. Please help I tried the first time got 50 some thing then 33 then 20.25. How do I do this?
vazorg [7]

Answer:

Basically, you have to think: more bricks : more time

more workers : less time

2,400 bricks 6 workers takes 18 hours

You now have to solve for 4,500 blocks and 10 workers

4,500 / 2,400 = 1.875 (times greater)

10 / 6 = 1.666666666 (times less)

So, we get 18 hours, multiply it by 1.875 and divide it by 1.666666666

which equals 20.25 hours

So, it seems you were correct on your third try.


Step-by-step explanation:


5 0
2 years ago
(a) find the slope m of the tangent to the curve y = 5/ x at the point where x = a > 0.(b) find equations of the tangent line
Trava [24]
<span>(a) the slope of curve is calculated from the derivative of the curve expression y=5/x. In this problem, the slope m=-5/(a^2). (b) x=1, y=5,and m=-5, the tangent line is y-5=-5*(x-1). x=4, y=5/4, and m=-5/16, the tangent line is y-5/4=-5/16*(x-4)</span>
4 0
3 years ago
Yeah thanks if u help I love u no matter what..
skad [1K]

Answer:

please stop all this love comments

Step-by-step explanation:

okk? that will be better

3 0
2 years ago
How many different ways can you make 82 cents using current u.s. currency
andrew11 [14]
 <span>You can probably just work it out. 

You need non-negative integer solutions to p+5n+10d+25q = 82. 

If p = leftovers, then you simply need 5n + 10d + 25q ≤ 80. 


So this is the same as n + 2d + 5q ≤ 16 

So now you simply have to "crank out" the cases. 

Case q=0 [ n + 2d ≤ 16 ] 

Case (q=0,d=0) → n = 0 through 16 [17 possibilities] 
Case (q=0,d=1) → n = 0 through 14 [15 possibilities] 
... 
Case (q=0,d=7) → n = 0 through 2 [3 possibilities] 
Case (q=0,d=8) → n = 0 [1 possibility] 

Total from q=0 case: 1 + 3 + ... + 15 + 17 = 81 

Case q=1 [ n + 2d ≤ 11 ] 
Case (q=1,d=0) → n = 0 through 11 [12] 
Case (q=1,d=1) → n = 0 through 9 [10] 
... 
Case (q=1,d=5) → n = 0 through 1 [2] 

Total from q=1 case: 2 + 4 + ... + 10 + 12 = 42 

Case q=2 [ n + 2 ≤ 6 ] 
Case (q=2,d=0) → n = 0 through 6 [7] 
Case (q=2,d=1) → n = 0 through 4 [5] 
Case (q=2,d=2) → n = 0 through 2 [3] 
Case (q=2,d=3) → n = 0 [1] 

Total from case q=2: 1 + 3 + 5 + 7 = 16 

Case q=3 [ n + 2d ≤ 1 ] 
Here d must be 0, so there is only the case: 
Case (q=3,d=0) → n = 0 through 1 [2] 

So the case q=3 only has 2. 

Grand total: 2 + 16 + 42 + 81 = 141 </span>
3 0
3 years ago
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