All categories

3 years ago

Solving Rational equations. LCD method. Show work. Image attached.

Mathematics

1 answer:

Dmitry [639]3 years ago

3 0

Answer:

$b=1$

Step-by-step explanation:

The given rational equation is

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

The Least Common Denominator is $(b+2)(b-2)$ .

Multiply each term in the equation by the LCD.

$(b+2)(b-2)\times \frac{2b-5}{b-2} -(b+2)(b-2)\times2=(b+2)(b-2)\times\frac{3}{b+2}$

Simplify;

$(b+2)\times \frac{2b-5}{1} -2(b+2)(b-2)=(b-2)\times\frac{3}{1}$

$(b+2)(2b-5) -2(b+2)(b-2)=3(b-2)$

Expand and group similar terms

$2b^2-5b+4b-10 -2(b^2-4)=3b-6$

$2b^2-5b+4b-10 -2b^2+8=3b-6$

$-b-2=3b-6$

$3b+b=-2+6$

$4b=4$

$b=1$

You might be interested in

Simplify 4 (x3 – 3y2) – 2 (3x3 – 5y2)

Phoenix [80]

$\huge \bold{answer}$

5 0

3 years ago

A wide receiver catches a ball and begins to run for the endzone following a path defined by . A defensive player chases the rec

postnew [5]

Answer:

Receiver path:

x-5 = t(0)

y-50 = -10t

defender path:

x-10 = -0.9t

y-54 =-10.72t

Yes, he will reach the goal line (y = 0) before the defensive player catches him.

Step-by-step explanation:

Vector equation: (x-x1, y-y1) = t(a1, a2)

Horizontal component : x-x1= ta1

Vertical component: y-y1= ta2

(x-5, y-50) = t(0, -10)

x-5 = t(0)

y-50 =t(-10)

Parametric equation for the receiver path:

x-5 = t(0)

y-50 = -10t

(x - 10, y - 54) = t(-0.9, -10.72)

x - 10 = t(-0.9)

y - 54 = t(-10.72)

Parametric equation for the defender path:

x-10 = -0.9t

y-54 =-10.72t

At the 50yard line, using the receiver parametric equation for vertical component:

y-50 = -10t

At y= 50

50-50 = -10t

0= -10t

t= 0/-10 = 0

At y= 0

0-50 = -10t

t = -50/-10 = 5

Defender at y = 50

y-54 =-10.72t

50-54 =-10.72t

t = -4/-10.72 = 0.37

at y = 0

0-54 =-10.72t

t = -54/-10.72 = 5.04

t of receiver < t of defender

Since time of receiver at y=0 is less than time of defender, he will reach the goal line (y = 0) before the defensive player catches him.

3 0

3 years ago

F(prime)(t) = t^2 (1+f(t))<br><br> f(0) = 3<br><br> f(t) = ?

Andreyy89

Answer: $F(t) = 4 e ^ {\frac{t^{3} }{3}}-1$

Step-by-step explanation:

$F'(t) = t^{2} (1+ F(t))$

$\frac{dF}{dt} = t^{2} (1+F)$

First, we separated the variables:

$\frac{dF}{1+F} = t^{2} dt$

We integrate:

$\int\limits^{F(t)}_{F(0)} {\frac{1}{1+F'} } \, dF' = \int\limits^t_0 {t'^{3}} \, dt$

We change the variable for make more easy the integral:

$u = 1+F', du = dF'$

$\int\limits^{}_{} {\frac{1}{u} } \, du = \int\limits^0_t {t'^{3}} \, dt$

$ln(u)= \frac{t^{3}}{3}$

Now replace with $u = 1+F'$ and evaluate the limit:

$ln(1+F(t)) - ln (1+F(0)) = \frac{t^{3}}{3}$

Using properties of natural logarithm

$ln(1+F(t))= \frac{t^{3}}{3} + ln(4)$

Taking exponential of both sides

$e^{ln(1+F(t))} = e ^ {\frac{t^{3} }{3} + ln4}$

$e^{ln(1+F(t))} = e ^ {\frac{t^{3} }{3}}e^{ln4} =e ^ {\frac{t^{3} }{3}} * 4$

$1+F(t) = 4 e ^ {\frac{t^{3} }{3}}$

$F(t) = 4 e ^ {\frac{t^{3} }{3}}-1$

7 0

3 years ago

Determine whether the reflection is ovee the x-axis le the y-axis (3,0)(-3,0

d1i1m1o1n [39]

From (x,y) to (-x,y) would be a reflection across the y axis

so (3,0) to (-3,0) would be reflection across y axis

8 0

3 years ago

Draw 8 lines that are between 1 inch and 3 inches long. Measure each line to the nearest fourth inch, and make a line plot.

IRISSAK [1]

The length of the line is the difference between the endpoints of the line

The length of each line to the nearest fourth inch is 0.25 inch

<h3>How to measure the length of each line</h3>

The length of the horizontal line is given as:

Length = 2 inches

This is calculated as:

Length = 3 inches - 1 inch

Length = 2 inches

8 lines are to be drawn between the 1 inch and 3 inches point.

So, the length (l) of each line is:

$l = \frac{2\ inches}{8}$

Simplify the fraction

$l = \frac{1\ inches}{4}$

Divide

$l = 0.25\ inch$

Hence, the length of each line to the nearest fourth inch is 0.25 inch

Read more about line measurements at:

brainly.com/question/14366932

6 0

2 years ago