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2 years ago

6

Which equation shows the variable terms isolated on one side and the constant terms isolated on the other side for the equation

-3x+3=4-4x?

1 answer:

olchik [2.2K]2 years ago

7 0

Answer: x= 1

Step-by-step explanation: solve for x by simplifying both sides of the equation, then isolating the variable.

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Can someone help me I’m stuck on how to do this

Aleksandr [31]

Answer:

Step-by-step explanation:

7 0

2 years ago

30 sweets are shared between 2 groups in the ratio 3:7. How many sweets are in each group?

jolli1 [7]

Answer:

See explanation below

Step-by-step explanation:

Given the ratio between two groups to be 3:7

Total ratio = 3+7 = 10

Total sweet shared = 30sweets

For the group with ratio of 3;

Share = 3/10×30

Share = 3×3

Share = 9 sweet

For the group with ratio of 5

This share = 7/10×30

This share = 7×3

This share = 21sweet

4 0

2 years ago

One-third of a number t is equal to 7.<br> Write the word sentence as an equation then solve

Artemon [7]

Answer:

21

Step-by-step explanation:

One third of the number can be represented by 1/3t.

To make an equation, set this equal to 7.

1/3t = 7

Then, solve for t by dividing each side by 1/3:

1/3t = 7

t = 21

So, the number is 21.

6 0

2 years 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

2 years ago

Find the Maclaurin series for f(x) using the definition of a Maclaurin series. [Assume that f has a power series expansion. Do n

Mazyrski [523]

We can use the fact that, for $|x|$ ,

$\displaystyle\frac1{1-x}=\sum_{n=0}^\infty x^n$

Notice that

$\dfrac{\mathrm d}{\mathrm dx}\left[\dfrac1{1-x}\right]=\dfrac1{(1-x)^2}$

so that

$f(x)=\displaystyle\frac5{(1-x)^2}=5\frac{\mathrm d}{\mathrm dx}\left[\sum_{n=0}^\infty x^n\right]$

$f(x)=\displaystyle5\sum_{n=0}^\infty nx^{n-1}$

$f(x)=\displaystyle5\sum_{n=1}^\infty nx^{n-1}$

$f(x)=\displaystyle5\sum_{n=0}^\infty(n+1)x^n$

By the ratio test, this series converges if

$\displaystyle\lim_{n\to\infty}\left|\frac{(n+2)x^{n+1}}{(n+1)x^n}\right|=|x|\lim_{n\to\infty}\frac{n+2}{n+1}=|x|$

so the series has radius of convergence $R=1$ .

5 0

2 years ago