Solve for X<br><br> PLEASE HELP!!

8x+4(x-3)=4(6x+4)-4 please help

kvasek [131]

Answer:

x = -2

Step-by-step explanation:

Step 1: Simplify both sides of the equation.

8x + 4(x - 3) = 4(6x + ) -4

8x +(4)(x) + (4)(-3) = (4)(6x) + (4)(4) + -4 (distribute)

8x + 4x + -12 = 24x + 16 + -4

(8x) + (4x) +(-12) = (24x) + (16 + -4) (combine like terms)

12x + -12 = 24x + 12

12x -12 = 24x + 12

Step 2: Subtract 24x from both sides.

12x -12 -24x = 24x + 12 -24x

-12x -12 = 12

Step 3: Add 12 to both sides.

-12x - 12 + 12 = 12 + 12

-12x = 24

Step 4: Divide both sides by -12.

-12x/-12 = 24/-12 <u>x = -2</u>

HOPE THIS HELPED!!!

5 0

3 years ago

A boat leaves a.M. Marina at 7 AM traveling at 3.4 knots bearing is N38°E the boat docks on an island at 2pm how far north of th

Jlenok [28]

Answer:

The distance between the docks & island is 44.07 km.

Step-by-step explanation:

Speed of the boat S = 3.4 knot = 6.3 $\frac{km}{hr}$

Time taken by bot to reach the island = 7 hr

We know that the distance is given by

Distance = Speed × Time

Put all the values in above equation we get

D = 6.3 × 7

D = 44.07 km

Therefore the distance between the docks & island is 44.07 km.

6 0

2 years ago

Use Simpson's Rule with n = 10 to approximate the area of the surface obtained by rotating the curve about the x-axis. Compare y

DiKsa [7]

The area of the surface is given exactly by the integral,

$\displaystyle\pi\int_0^5\sqrt{1+(y'(x))^2}\,\mathrm dx$

We have

$y(x)=\dfrac15x^5\implies y'(x)=x^4$

so the area is

$\displaystyle\pi\int_0^5\sqrt{1+x^8}\,\mathrm dx$

We split up the domain of integration into 10 subintervals,

[0, 1/2], [1/2, 1], [1, 3/2], ..., [4, 9/2], [9/2, 5]

where the left and right endpoints for the $i$ -th subinterval are, respectively,

$\ell_i=\dfrac{5-0}{10}(i-1)=\dfrac{i-1}2$

$r_i=\dfrac{5-0}{10}i=\dfrac i2$

with midpoint

$m_i=\dfrac{\ell_i+r_i}2=\dfrac{2i-1}4$

with $1\le i\le10$ .

Over each subinterval, we interpolate $f(x)=\sqrt{1+x^8}$ with the quadratic polynomial,

$p_i(x)=f(\ell_i)\dfrac{(x-m_i)(x-r_i)}{(\ell_i-m_i)(\ell_i-r_i)}+f(m_i)\dfrac{(x-\ell_i)(x-r_i)}{(m_i-\ell_i)(m_i-r_i)}+f(r_i)\dfrac{(x-\ell_i)(x-m_i)}{(r_i-\ell_i)(r_i-m_i)}$

Then

$\displaystyle\int_0^5f(x)\,\mathrm dx\approx\sum_{i=1}^{10}\int_{\ell_i}^{r_i}p_i(x)\,\mathrm dx$

It turns out that the latter integral reduces significantly to

$\displaystyle\int_0^5f(x)\,\mathrm dx\approx\frac56\left(f(0)+4f\left(\frac{0+5}2\right)+f(5)\right)=\frac56\left(1+\sqrt{390,626}+\dfrac{\sqrt{390,881}}4\right)$