Ask question

Ask question

All categories

uranmaximum [27]

3 years ago

15

Justin is redoing his bathroom floor with tiles measuring 6in by 13 in. the floor has an area of 8,500 in². what is the least nu

mber of tiles he will need?

2 answers:

enyata [817]3 years ago

7 0

He will need 109 tiles

natulia [17]3 years ago

7 0

Each tile is 6 in. by 13 in.....so the area of each tile is L * W = 6 * 13 = 78 in^2

the entire floor has an area of 8500 in^2

so the least number of tiles needed is : 8500/78 = 108.97.....109 tiles are needed

You might be interested in

3y''-6y'+6y=e*x sexcx

Simora [160]

From the homogeneous part of the ODE, we can get two fundamental solutions. The characteristic equation is

$3r^2-6r+6=0\iff r^2-2r+2=0$

which has roots at $r=1\pm i$ . This admits the two fundamental solutions

$y_1=e^x\cos x$
$y_2=e^x\sin x$

The particular solution is easiest to obtain via variation of parameters. We're looking for a solution of the form

$y_p=u_1y_1+u_2y_2$

where

$u_1=-\displaystyle\frac13\int\frac{y_2e^x\sec x}{W(y_1,y_2)}\,\mathrm dx$
$u_2=\displaystyle\frac13\int\frac{y_1e^x\sec x}{W(y_1,y_2)}\,\mathrm dx$

and $W(y_1,y_2)$ is the Wronskian of the fundamental solutions. We have

$W(e^x\cos x,e^x\sin x)=\begin{vmatrix}e^x\cos x&e^x\sin x\\e^x(\cos x-\sin x)&e^x(\cos x+\sin x)\end{vmatrix}=e^{2x}$

and so

$u_1=-\displaystyle\frac13\int\frac{e^{2x}\sin x\sec x}{e^{2x}}\,\mathrm dx=-\int\tan x\,\mathrm dx$
$u_1=\dfrac13\ln|\cos x|$

$u_2=\displaystyle\frac13\int\frac{e^{2x}\cos x\sec x}{e^{2x}}\,\mathrm dx=\int\mathrm dx$
$u_2=\dfrac13x$

Therefore the particular solution is

$y_p=\dfrac13e^x\cos x\ln|\cos x|+\dfrac13xe^x\sin x$

so that the general solution to the ODE is

$y=C_1e^x\cos x+C_2e^x\sin x+\dfrac13e^x\cos x\ln|\cos x|+\dfrac13xe^x\sin x$

7 0

3 years ago

What is the name of the figure?

Serhud [2]

That is a Triangular Prism.

4 0

3 years ago

Read 2 more answers

I NEED NUMBER 24 ONLY PLEASEEEEEEEEEEEE I NEED HELP I DON’T GET IT!!!!!!

siniylev [52]

Answer:

$325,120.00

Step-by-step explanation:

This is my first answer so hopefully I do this correctly. In essence it's exactly the same as number 22!

The realtor earns $10,566.40 off of a comission of 3.25%. A comission is pretty much just a cut of however much the house cost; which a realtor gets for helping to sell the property. So we could say that the comission is equal to 3.25% of however much the house cost.

Say the house price = $x, so our equation would be:

$10,566.40 = (3.25%)•($x) or

10,566.40 = (.0325)•($x)

By solving for x we get:

$x = (10,566.40)÷(.0325) = $325,120

4 0

3 years ago

Read 2 more answers

34. For the following exercises, given each set of information, find a linear equation satisfying the conditions, if possible.

skad [1K]

Answer:

The linear equation for the line which passes through the points given as (-1,4) and (5,2), is written in the point-slope form as $y=\frac{1}{3} x-\frac{13}{3}$ .

Step-by-step explanation:

A condition is given that a line passes through the points whose coordinates are (-1,4) and (5,2).

It is asked to find the linear equation which satisfies the given condition.

Step 1 of 3

Determine the slope of the line.

The points through which the line passes are given as (-1,4) and (5,2). Next, the formula for the slope is given as,

$m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

Substitute 2&4 for $y_{2}$ and $y_{1}$ respectively, and $5 \&-1$ for $x_{2}$ and $x_{1}$ respectively in the above formula. Then simplify to get the slope as follows,

$m=\frac{2-4}{5-(-1)}$\\ $m=\frac{-2}{6}$\\ $m=-\frac{1}{3}$$

Step 2 of 3

Write the linear equation in point-slope form.

A linear equation in point slope form is given as,

$y-y_{1}=m\left(x-x_{1}\right)$

Substitute $-\frac{1}{3}$ for m,-1 for $x_{1}$ , and 4 for $y_{1}$ in the above equation and simplify using the distributive property as follows,

$y-4=-\frac{1}{3}(x-(-1))$\\ $y-4=-\frac{1}{3}(x+1)$\\ $y-4=-\frac{1}{3} x-\frac{1}{3}$$

Step 3 of 3

Simplify the equation further.

Add 4 on each side of the equation $y-4=\frac{1}{3} x-\frac{1}{3}$ , and simplify as follows,

$y-4+4=\frac{1}{3} x-\frac{1}{3}+4$\\ $y=\frac{1}{3} x-\frac{1+12}{3}$\\ $y=\frac{1}{3} x-\frac{13}{3}$$

This is the required linear equation.

5 0

1 year ago

ATRI has vertices T(-2,4), R(2,4), and 1(0,0). IS ATRI scalene, isosceles, or equilateral?

klasskru [66]

The answer is quilateral

3 0

3 years ago