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3 years ago

A construction worker took 1over8

Mathematics

1 answer:

astraxan [27]3 years ago

8 0

Answer:

1/32

Step-by-step explanation:

1/8 divided by 4 = .03125

.03125= 1/32

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Help now!!!!!!!!!!!!!!!

uysha [10]

Answer:

Step-by-step explanation:

1) 165

2) 108

3)24

4) 75

5)2

6)7

3 0

3 years ago

2a + 3b = 5<br> b = a - 5

Lunna [17]

<h2>1st Equation:</h2>

Answer for A:

$a = \frac{5 - 3b}{2}$

Answer for B:

$b = \frac{5 - 2b}{3}$

<h2>2nd Equation:</h2>

Answer:

$a = b + 5$

<u>hope it helps</u><u>.</u>

5 0

2 years ago

Find all the solutions for the equation:

Contact [7]

$2y^2\,\mathrm dx-(x+y)^2\,\mathrm dy=0$

Divide both sides by $x^2\,\mathrm dx$ to get

$2\left(\dfrac yx\right)^2-\left(1+\dfrac yx\right)^2\dfrac{\mathrm dy}{\mathrm dx}=0$

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{2\left(\frac yx\right)^2}{\left(1+\frac yx\right)^2}$

Substitute $v(x)=\dfrac{y(x)}x$ , so that $\dfrac{\mathrm dv(x)}{\mathrm dx}=\dfrac{x\frac{\mathrm dy(x)}{\mathrm dx}-y(x)}{x^2}$ . Then

$x\dfrac{\mathrm dv}{\mathrm dx}+v=\dfrac{2v^2}{(1+v)^2}$

$x\dfrac{\mathrm dv}{\mathrm dx}=\dfrac{2v^2-v(1+v)^2}{(1+v)^2}$

$x\dfrac{\mathrm dv}{\mathrm dx}=-\dfrac{v(1+v^2)}{(1+v)^2}$

The remaining ODE is separable. Separating the variables gives

$\dfrac{(1+v)^2}{v(1+v^2)}\,\mathrm dv=-\dfrac{\mathrm dx}x$

Integrate both sides. On the left, split up the integrand into partial fractions.

$\dfrac{(1+v)^2}{v(1+v^2)}=\dfrac{v^2+2v+1}{v(v^2+1)}=\dfrac av+\dfrac{bv+c}{v^2+1}$

$\implies v^2+2v+1=a(v^2+1)+(bv+c)v$

$\implies v^2+2v+1=(a+b)v^2+cv+a$

$\implies a=1,b=0,c=2$

Then

$\displaystyle\int\frac{(1+v)^2}{v(1+v^2)}\,\mathrm dv=\int\left(\frac1v+\frac2{v^2+1}\right)\,\mathrm dv=\ln|v|+2\tan^{-1}v$

On the right, we have

$\displaystyle-\int\frac{\mathrm dx}x=-\ln|x|+C$

Solving for $v(x)$ explicitly is unlikely to succeed, so we leave the solution in implicit form,

$\ln|v(x)|+2\tan^{-1}v(x)=-\ln|x|+C$

and finally solve in terms of $y(x)$ by replacing $v(x)=\dfrac{y(x)}x$ :

$\ln\left|\frac{y(x)}x\right|+2\tan^{-1}\dfrac{y(x)}x=-\ln|x|+C$

$\ln|y(x)|-\ln|x|+2\tan^{-1}\dfrac{y(x)}x=-\ln|x|+C$

$\boxed{\ln|y(x)|+2\tan^{-1}\dfrac{y(x)}x=C}$

7 0

3 years ago

The length of a rectangle is 4 units less than the width. The area of the rectangle is 21 units. What is the length, in units, o

bazaltina [42]

The formula of an area of a rectangle:

A = wl

We have l = w - 4 and A = 21.

Substitute:

w(w - 4) = 21 <em>use distributive property</em>

(w)(w) + (w)(-4) = 21

w² - 4w = 21 <em>subtract 21 from both sides</em>

w² - 4w - 21 = 0

w² - 7w + 3w - 21 = 0

w(w - 7) + 3(w - 7) = 0

(w - 7)(w + 3) = 0 ↔ w - 7 = 0 ∨ w + 3 = 0

w = 7 ∨ w = -3 < 0

l = w - 4 → l = 7 - 4 = 3

<h3>Answer: the length = 3 u.</h3>

4 0

3 years ago

Find the common ratio of the sequence. 2, –10, 50, –250, –5 1/-5 –12 12

Sholpan [36]

-5
This is just to fill out the space of twenty characters

6 0

3 years ago

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