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3 years ago

LMNP is a rectangle. If LN = 9x – 14 and MP = 7x + 4, find the values of x, LN, and MP.

1 answer:

7 0

I am thinking of a rectangle that has the two sides parallel to one another. Set the two functions equal to one another, so

9x-14=7x+4

After a bunch of algebra and math magic, 2x=18 => x=9

So if you just insert 9 into both equations, both will end up with a value of 67, so it ends up looking like a right triangle. Don't do that. Instead, to find the rectangle widths, use 9x-4 (+10 added to intercept) instead, while keeping 7x+4, so that the intercepts match.

LN) 9x-4 = 9(9)-4 = 81-4 = 77
MP) 7(9)+4 = 63+4 = 67

*If you are also looking for the diagonals, use Pythagorean Theorem 77^2+67^2=(hypotenuse)^2*

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ArbitrLikvidat [17]

Answer:

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Step-by-step explanation:

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3 years ago

Read 2 more answers

What is the common difference for this arithmetic sequence?

OleMash [197]

Answer:

Concept: Series

Start by taking the first and second An and finding the difference
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3 years ago

The United States five-dollar bill are 155.956 mm long and 66.294 wide. A stack of these bills fits inside a 156 x 66.3 x 66.3 m

maxonik [38]

Answer:

A) $5 × 25 = $75

B) 37.69%

Step-by-step explanation:

The dimension of five dollar bill:

Length = 155.956mm

Width = 66.294

Volume of box being occupied by 5 dollar bill = 258,473.68mm^3

Dimension of box :

156 x 66.3 x 66.3 mm

If volume of the 5—dollar bill is 258,473.68mm^3

Then, the Thickness of the five dollar stack equals :

Volume = Length × width × height

258,473.68mm^3 = 155.956mm × 66.294 × height

258,473.68 = 10338.947064 × h

h = (258473.68/10338.947064)

h = 25.000000003288 = 25

Therefore, the amount of money in the box is:

$5 × 25 = $75

Percentage of box volume taken up by $5 bill

Volume of the box:

156 x 66.3 x 66.3 mm = 685727.6399mm^3

Volume of box being occupied by 5 dollar bill = 258,473.68mm^3

(258,473.68 / 685727.6399) × 100

=37.69%

4 0

3 years ago

Find a linear second-order differential equation f(x, y, y', y'') = 0 for which y = c1x + c2x3 is a two-parameter family of solu

Alisiya [41]

Let $y=C_1x+C_2x^3=C_1y_1+C_2y_2$ . Then $y_1$ and $y_2$ are two fundamental, linearly independent solution that satisfy

$f(x,y_1,{y_1}',{y_1}'')=0$
$f(x,y_2,{y_2}',{y_2}'')=0$

Note that ${y_1}'=1$ , so that $x{y_1}'-y_1=0$ . Adding $y''$ doesn't change this, since ${y_1}''=0$ .

So if we suppose

$f(x,y,y',y'')=y''+xy'-y=0$

then substituting $y=y_2$ would give

$6x+x(3x^2)-x^3=6x+2x^3\neq0$

To make sure everything cancels out, multiply the second degree term by $-\dfrac{x^2}3$ , so that

$f(x,y,y',y'')=-\dfrac{x^2}3y''+xy'-y$

Then if $y=y_1+y_2$ , we get

$-\dfrac{x^2}3(0+6x)+x(1+3x^2)-(x+x^3)=-2x^3+x+3x^3-x-x^3=0$

as desired. So one possible ODE would be

$-\dfrac{x^2}3y''+xy'-y=0\iff x^2y''-3xy'+3y=0$

(See "Euler-Cauchy equation" for more info)

6 0

3 years ago

What are the common factors of 24 64 88

8090 [49]

Answer is 8

8 times 3 is 24

8 times 8 is 64

8 times 11 is 88

7 0

3 years ago