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Ronch [10]
3 years ago
6

-7 (-6) + 17 help plz

Mathematics
1 answer:
Elden [556K]3 years ago
3 0

Answer: 59

Step-by-step explanation:

-7 (-6) + 17  (multiply -7 and -6)  

= 42+17      (simplify)

=59

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Find the nth term of this quadratic sequence<br> 3, 11, 25, 45, .
Hoochie [10]

The term of the given quadratic sequence is found to be 3n² - n + 1 using the principle of mathematical induction.

Given,

In the question:

The quadratic sequence is :

3, 11, 25, 45, ...

To find the nth term of the quadratic sequence.

Now, According to the question;

The first term of the sequence is 3, the second term is 11, the third term is 25, and the fourth term is 45.

The difference between the first and second terms can be calculated as follows:

11-3 = 8

The difference between the second and third terms can be calculated as follows:

25-11 = 14

The difference between the third and fourth terms can be calculated as follows:

45-25 = 20

The sequence is expressed as follows:

3,3+8,11+11,25+20,...

The difference between consecutive terms expands by 6.

Use the principle of mathematical induction.

6(\frac{n(n+1)}{2} )

= 3n(n+1)

The sequence's nth term can be calculated as follows:

term = 3n(n+1) - 4n + 1

             = 3n² - n + 1

Hence, the term of the given quadratic sequence is found to be 3n² - n + 1 using the principle of mathematical induction.

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6 0
1 year ago
Consider F and C below. F(x, y, z) = yz i + xz j + (xy + 6z) k C is the line segment from (2, 0, −2) to (5, 4, 2) (a) Find a fun
WITCHER [35]

Answer:

Required solution (a) f(x,y,z)=xyz+3z^2+C (b) 40.

Step-by-step explanation:

Given,

F(x,y,z)=yz \uvec i +xz\uvec j+(xy+6z)\uvex k

(a) Let,

F(x,y,z)=yz \uvec i +xz\uvec j+(xy+6z)\uvex k=f_x \uvec{i} +f_y \uvex{j}+f_z\uvec{k}

Then,

f_x=yz,f_y=xz,f_z=xy+6z

Integrating f_x we get,

f(x,y,z)=xyz+g(y,z)

Differentiate this with respect to y we get,

f_y=xz+g'(y,z)

compairinfg with f_y=xz of the given function we get,

g'(y,z)=0\implies g(y,z)=0+h(z)\implies g(y,z)=h(z)

Then,

f(x,y,z)=xyz+h(z)

Again differentiate with respect to z we get,

f_z=xy+h'(z)=xy+6z

on compairing we get,

h'(z)=6z\implies h(z)=3z^2+C   (By integrating h'(z))  where C is integration constant. Hence,

f(x,y,z)=xyz+3z^2+C

(b) Next, to find the itegration,

\int_C \vec{F}.dr=\int_C \nabla f. d\vec{r}=f(5,4,2)-f(2,0,-2)=(52+C)-(12+C)=40

3 0
3 years ago
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