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Ksenya-84 [330]

3 years ago

6

You want to place a towel bar that is 10 1⁄4 inches long in the center of a door that is 26 1⁄3 inches wide. How far should you

place the bar from each edge of the door? (Write the answer as a mixed number.)

1 answer:

djverab [1.8K]3 years ago

8 0

First we want to find the part of the door not covered by the towel bar.

So subtract 10.25 from 26.33

In order to do this we need common denominators

So we multiply the fractions to get

(26 & 4/12) - (10 & 3/12)

Now we subtract to get

15 & 1/12

Since the 15 is odd, add 12/12 to the fraction

14 & 13/12

Now divide that in half to get the distance from each edge

7 & 13/24

You might be interested in

The longest side of a triangle is 4 more than twice the shortest. The third side is 3 more than the shortest. The perimeter is 3

kondaur [170]

Answer:

The shortest side = 6

The third side = 9

The hypotenuse (the longest side) = 16

Step-by-step explanation:

First, let's establish the following based on the information given:

The shortest side = x

The third side = (x + 3)

The hypotenuse (the longest side) = (2x + 4)

The perimeter = 31

Since the perimeter is the total of all 3 sides, we are left with this equation:

(x) + (x + 3) + (2x + 4) = 31

From here, combine like-terms and solve for x.

(x) + (x + 3) + (2x + 4) = 31

(4x + 7) = 31

4x = 24

x = 6

Now that we know the value of x, we can apply this to the predetermined formulas to find the measurements of the remaining two sides.

The shortest side = 6

The third side = (x + 3) = 9

The hypotenuse (the longest side) = (2x + 4) = (2(6) + 4) = (12 + 4) = 16

To check, add all of the sides together to make sure they equal 31.

6 + 9 + 16 = 31

~Hope this Helps!~

7 0

3 years ago

Peter is a beekeeper who sells jars of honey and combs at the market. For each 10 jars of honey that he takes to the market carr

dsp73

Can you list the options? so i can help, theres not enough info for me to answer

5 0

3 years ago

The base of a solid in the region bounded by the two parabolas y2 = 8x and x2 = 8y. Cross sections of the solid perpendicular to

Usimov [2.4K]

The two curves intersect at two points, (0, 0) and (8, 8):

$x^2=8y\implies y=\dfrac{x^2}8$

$y^2=\dfrac{x^4}{64}=8x\implies\dfrac{x^4}{64}-8x=0\implies\dfrac{x(x-8)(x^2+8x+64)}{64}=0$

$\implies x=0,x=8\implies y=0,y=8$

The area of a semicircle with diameter $d$ is $\dfrac{\pi d^2}8$ . The diameter of each cross-section is determined by the vertical distance between the two curves for any given value of $x$ between 0 and 8. Over this interval, $y^2=8x\implies y=\sqrt{8x}$ and $\sqrt{8x}>\dfrac{x^2}8$ , so the volume of this solid is given by the integral

$\displaystyle\frac\pi8\int_0^8\left(\sqrt{8x}-\dfrac{x^2}8\right)^2\,\mathrm dx=\frac{288\pi}{35}$

6 0

3 years ago

Solution for dy/dx+xsin 2 y=x^3 cos^2y

vichka [17]

Rearrange the ODE as

$\dfrac{\mathrm dy}{\mathrm dx}+x\sin2y=x^3\cos^2y$
$\sec^2y\dfrac{\mathrm dy}{\mathrm dx}+x\sin2y\sec^2y=x^3$

Take $u=\tan y$ , so that $\dfrac{\mathrm du}{\mathrm dx}=\sec^2y\dfrac{\mathrm dy}{\mathrm dx}$ .

Supposing that $|y|$ , we have $\tan^{-1}u=y$ , from which it follows that

$\sin2y=2\sin y\cos y=2\dfrac u{\sqrt{u^2+1}}\dfrac1{\sqrt{u^2+1}}=\dfrac{2u}{u^2+1}$
$\sec^2y=1+\tan^2y=1+u^2$

So we can write the ODE as

$\dfrac{\mathrm du}{\mathrm dx}+2xu=x^3$

which is linear in $u$ . Multiplying both sides by $e^{x^2}$ , we have

$e^{x^2}\dfrac{\mathrm du}{\mathrm dx}+2xe^{x^2}u=x^3e^{x^2}$
$\dfrac{\mathrm d}{\mathrm dx}\bigg[e^{x^2}u\bigg]=x^3e^{x^2}$

Integrate both sides with respect to $x$ :

$\displaystyle\int\frac{\mathrm d}{\mathrm dx}\bigg[e^{x^2}u\bigg]\,\mathrm dx=\int x^3e^{x^2}\,\mathrm dx$
$e^{x^2}u=\displaystyle\int x^3e^{x^2}\,\mathrm dx$

Substitute $t=x^2$ , so that $\mathrm dt=2x\,\mathrm dx$ . Then

$\displaystyle\int x^3e^{x^2}\,\mathrm dx=\frac12\int 2xx^2e^{x^2}\,\mathrm dx=\frac12\int te^t\,\mathrm dt$

Integrate the right hand side by parts using

$f=t\implies\mathrm df=\mathrm dt$
$\mathrm dg=e^t\,\mathrm dt\implies g=e^t$
$\displaystyle\frac12\int te^t\,\mathrm dt=\frac12\left(te^t-\int e^t\,\mathrm dt\right)$

You should end up with

$e^{x^2}u=\dfrac12e^{x^2}(x^2-1)+C$
$u=\dfrac{x^2-1}2+Ce^{-x^2}$
$\tan y=\dfrac{x^2-1}2+Ce^{-x^2}$

and provided that we restrict $|y|$ , we can write

$y=\tan^{-1}\left(\dfrac{x^2-1}2+Ce^{-x^2}\right)$

5 0

3 years ago

Add the group of numbers below. Make sure<br> your answer is fully reduced.<br> 1/3+1/5+2/5

RideAnS [48]

Answer:

14/15

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers