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2 years ago

1+4 =5, 2+5=12, 3+6=21 then 8+11 ...

2 answers:

5 0

You take the 5 from the first part and add it to the 2+5. This gives you 12. Then take the 12+(3+6) and this gives you 21. 21+(8+11) and you get 40.

Katena32 [7]2 years ago

4 0

The answer to 8 + 11 is 19

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Find four consecutive even integers such that seven times the first exceeds their sum by 18.

Ganezh [65]

Answer:

10, 12, 14, 16

Step-by-step explanation:

Given: Four consecutive even integers such that seven times the first exceeds their sum by 18.

Lets assume the first number be "x".

As it is even integers

∴ Second consecutive number will be $(x+2)$

Third consecutive number will be $(x+4)$

Fourth consecutive number will be $(x+6)$

Now, as given seven times the first exceeds their sum by 18.

∴ $[x+(x+2)+(x+4)+(x+6)]+18 = 7x$

solving the equation to find the number.

⇒ $[x+(x+2)+(x+4)+(x+6)]+18 = 7x$

Opening parenthesis

$x+x+2+x+4+x+6+18 = 7x$

⇒ $4x+30= 7x$

subtracting both side by 4x

⇒ $30= 3x$

Dividing both side by 3

∴ x= 10.

Hence, subtituting the value x to find four consecutive even integers.

First number is 10

Second consecutive number will be $(10+2)$ = 12

Third consecutive number will be $(10+4)$ = 14

Fourth consecutive number will be $(10+6)$ = 16

∴ Four consecutive even integers are 10,12,14,16.

8 0

2 years ago

How would I do this problem? In trapezoid PQRS, PQ is parallel to SR. Find the area of PQRS. Leave your answer is simplest radic

Mrac [35]

Can u take its pictures and post it bcuz it might be easer to answer

7 0

2 years ago

Read 2 more answers

The taxi costs $2.25 for the first mile plus 15cents for each additional tenth of a mile. For a 5.2 mile trip, Eric payed $10 an

Verizon [17]

2.25 for the first mile. That leaves 4.2 miles.

4 * 0.1 = 0.4

0.15 * 4 = $6.00

0.15 * 2 = 0.3

$6.30 + $2.25 = $8.55

$10.00 - $8.55 = $1.45

The driver's tip was $1.45.

7 0

2 years ago

Suppose a randomly generated list of numbers from 0 to 9 that has a probability success of 80%. which of these groups of numbers

maria [59]

A is the answer because it is the widest variety

3 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.

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

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

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

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

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

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

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

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

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

2 years ago