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3 years ago

-4|x-11|=-16 solve for x in the equation

Mathematics

1 answer:

BigorU [14]3 years ago

5 0

Answer:

x = 7 and 15

Step-by-step explanation:

Divide both sides by -4...

| x - 11 | = 4

because

| -4 | = 4 and

| 4 | = 4

x - 11 = -4 and x - 11 = 4

solve for x in both equations.

x = 7 and x = 15

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3 years ago

If Wally plans on using 30 sections of fence (one section is pictured and described above), about how many total feet of wood wi

d1i1m1o1n [39]

Answer:

Step-by-step explanation:

7 0

3 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

I will give brainliest fast please I will give points please explain well

ANEK [815]

If the equal is 55x + 35y = 200

x: time-driven at 55 miles/hour

y: time-driven at 35 miles/hour

If the car spends 2.5 hours at 35 miles per hour, then it drove:

55x + 35(2.5) = 200

55x =35

y = 112.5

There the drive spent about 2.5-hour driving at 55 mph.

Hope it helps!

4 0

2 years ago

PLEASE SHOW FULL SOLUTIONS !!!!! WILL MARK BRAINLIEST FOR THE BEST ANSWER. THANK YOU AND GOD BLESS

yanalaym [24]

Answer:

System is given below

Step-by-step explanation:

Let (applicable to all three lines below)
Hard candy = x kg with price $1.60/kg
Gummy worms = y kg with price $2.20/kg
Total weight = 50 kg with mixed price $1.75/kg

Required equations:

x + y = 50 total weight
1.60x + 2.20y = 50*1.75 total price

=========================================

Note. It says don't solve but the solution below for those who is interested to know the answer.

Simplify the second equation and solve by substitution x = 50 - y:

1.6(50 - y) + 2.2y = 87.5
80 - 1.6y +2.2y = 87.5
0.6y = 7.5
y = 7.5/0.6
y = 12.5

Find the value of x:

x = 50 - 12.5 = 37.5

Hard candy = 37.5 kg and gummy worms = 12.5 kg

7 0

3 years ago