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3 years ago

14

What is the solution to the equation 3(2x+5) = 3x+4x?

1 answer:

n200080 [17]3 years ago

4 0

X=15
expand the brackets and collect like terms so 6x+15=7x
then -6x to get x=15

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Miguel wants to wallpaper the back wall of his bedroom. the wall is in the shape of a rectangle. its length is 13 feet and its w

Bingel [31]

First of all lets find the area of the rectangle.
13 ft × 12 ft = 156 square feet.
Now lets find the price which is $8 / square foot. Using proportions we get:
$8/sq ft × 156 sq ft. Simplifying we get
$8 × 156 = $1248.
The wallpaper will cost $1248

3 0

3 years ago

A rectangle width is one fourth it’s length, outs are is 9 square units. The equation I(1/4I)=9 can be used to find I, the lengt

OlgaM077 [116]

Answer:

The length of rectangle is 6 units and the width is 1.5 units

Step-by-step explanation:

Let

L -----> the length of rectangle

W ----> the width of rectangle

we know that

The area of rectangle is equal to

$A=LW$

we have

$A=9\ units^{2}$

so

$9=LW$ -----> equation A

A rectangle width is one fourth it’s length

so

$W=\frac{1}{4}L$ ----> equation B

substitute equation B in equation A and solve for L

$9=L(\frac{1}{4}L)$

$L^{2}=9*4$

$L^{2}=36$

take square root both sides

$L=6\ units$

Find the value of W

$W=\frac{1}{4}(6)$

$W=1.5\ units$

5 0

3 years ago

5+b=25<br> b=20<br> 5+20=25<br> =?

Arisa [49]

The numbers are equal. 25=25

4 0

3 years ago

Read 2 more answers

Really need help with Dilations in the Coordinate Plane

zimovet [89]

Answer:

I'd say (x,y) --> (2x,2y)

Step-by-step explanation:

Please understand if I'm wrong because I couldn't enhance the image. The quality made it a bit hard for me to accurately tell.

Sorry if I'm wrong.

4 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

3 years ago