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Klio2033 [76]
3 years ago
10

A man bought a bicycle for $400 and sold it at a loss of 35%. How much money did he lose?

Mathematics
1 answer:
DedPeter [7]3 years ago
8 0
If he bought the bike for $400 and lost 35% of his profit, you would divide 35 by 100. Next, you would multiply that number (0.35) by 400. When you do that, you get an answer of $140. In conclusion, if he bought the bike for $400 and sold it at a loss of 35%, he would lose $140 from that $400. 

Additionally, if you wanted to know what exactly he sold it for, you would simply subtract 140 from 400. Which would be $260.

So, he bought the bike for $400, but when he resold it, he lost $140 of that pay. Basically, he resold the bike for $260. 

I hope this helps! :) Let me know if you need help with anything else!
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What is the equation of the asymptote? f (x) = 3^x-5+1<br> y=?
Natasha_Volkova [10]

The equation of the asymptote is y=1

Explanation:

The given equation is f(x)=3^{x-5}+1

We need to determine the horizontal asymptote of the equation.

The given equation is of the exponential function of the form f(x)=c \cdot n^{a x+b}+k and has a horizontal asymptote y=k

Hence, from the above equation, the value of k is 1.

Therefore, we have,

k=1

Thus, the horizontal asymptote is y=1

Therefore, the asymptote of the equation is y=1

4 0
3 years ago
Hi lots of work today ​
dimaraw [331]

Answer:

x-axis

Step-by-step explanation:

The lateral line on a graph is known as the x-axis

5 0
2 years ago
Let V denote the set of ordered triples (x, y, z) and define addition in V as in
icang [17]

Answer:

a) No

b) No

c) No

d) No

Step-by-step explanation:

Remember, a set V wit the operations addition and scalar product is a vector space if the following conditions are valid for all u, v, w∈V and for all scalars c and d:

1. u+v∈V

2. u+v=v+u

3. (u+v)+w=u+(v+w).

4. Exist 0∈V such that u+0=u

5. For each u∈V exist −u∈V such that u+(−u)=0.

6. if c is an escalar and u∈V, then cu∈V

7. c(u+v)=cu+cv

8. (c+d)u=cu+du

9. c(du)=(cd)u

10. 1u=u

let's check each of the properties for the respective operations:

Let u=(u_1,u_2,u_3), v=(v_1,v_2,v_3)

Observe that  

1. u+v∈V

2. u+v=v+u, because the adittion of reals is conmutative

3. (u+v)+w=u+(v+w). because the adittion of reals is associative

4. (u_1,u_2,u_3)+(0,0,0)=(u_1+0,u_2+0,u_3+0)=(u_1,u_2,u_3)

5. (u_1,u_2,u_3)+(-u_1,-u_2,-u_3)=(0,0,0)

then regardless of the escalar product, the first five properties are met for a), b), c) and d). Now let's verify that properties 6-10 are met.

a)

6. c(u_1,u_2,u_3)=(cu_1,u_2,cu_3)\in V

7.

c(u+v)=c(u_1+v_1,u_2+v_2,u_3+v_3)=(c(u_1+v_1),u_2+v_2,c(u_3+v_3))\\=(cu_1+cv_1,u_2+v_2,cu_3+cv_3)=c(u_1,u_2,u_3)+c(v_1,v_2,v_3)=cu+cv

8.

(c+d)u=(c+d)(u_1,u_2,u_3)=((c+d)u_1,u_2,(c+d)u_3)=\\=(cu_1+du_1,u_2,cu_3+du_3)\neq (cu_1+du_1,2u_2,cu_3+du_3)=cu+du

Since 8 isn't satify then V is not a vector space with the addition as in R^3 and the scalar product a(x,y,z)=(ax,y,az)

b)  6. c(u_1,u_2,u_3)=(cu_1,0,cu_3)\in V

7.

c(u+v)=c(u_1+v_1,u_2+v_2,u_3+v_3)=(c(u_1+v_1),0,c(u_3+v_3))\\=(cu_1+cv_1,0,cu_3+cv_3)=c(u_1,u_2,u_3)+c(v_1,v_2,v_3)=cu+cv

8.

(c+d)u=(c+d)(u_1,u_2,u_3)=((c+d)u_1,0,(c+d)u_3)=\\=(cu_1+du_1,0,cu_3+du_3)=(cu_1,0,cu_3)+(du_1,0,du_3) =cu+du

9.

c(du)=c(d(u_,u_2,u_3))=c(du_1,0,du_3)=(cdu_1,0,cdu_3)=(cd)u

10

1u=1(u_1,u_2,u3)=(1u_1,0,1u_3)=(u_1,0,u_3)\neq(u_1,u_2,u_3)

Since 10 isn't satify then V is not a vector space with the addition as in R^3 and the scalar product a(x,y,z)=(ax,0,az)

c) Observe that 1u=1(u_1,u_2,u3)=(0,0,0)\neq(u_1,u_2,u_3)

Since 10 isn't satify then V is not a vector space with the addition as in R^3 and the scalar product a(x,y,z)=(0,0,0).

d)  Observe that 1u=1(u_1,u_2,u3)=(2*1u_1,2*1u_2,2*1u_3)=(2u_1,2u_2,2u_3)\neq(u_1,u_2,u_3)=u

Since 10 isn't satify then V is not a vector space with the addition as in R^3 and the scalar product a(x,y,z)=(2ax,2ay,2az).

8 0
3 years ago
Help (4/d)^3 anyway help
Lelu [443]

Answer:

(4√d)³

d^3/4

(d^1/4)³

required answer is=A.

7 0
3 years ago
Can someone explain how to work this out please ?
cupoosta [38]

Answer:

x=4

Step-by-step explanation:

f(x) = (x-2)^5  +3

Let f(x) = 35

35 = (x-2)^5  +3

Subtract 3 from each side

35-3 = (x-2)^5  +3-3

32 = (x-2)^5

Take the fifth root of each side

32 ^ (1/5) =  (x-2)^5^ 1/5

2 = x-2

Add 2 to each side

2 +2 = x-2+2

4 =x

6 0
2 years ago
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