All categories

3 years ago

Simplify the rational expressions. State any excluded values. Show your work. (x-2)/(x^2+3x-10)

Mathematics

1 answer:

Elan Coil [88]3 years ago

7 0

1/(x+5) because the (x-2) is canceled from both the numerator and denominator of the equation.

You might be interested in

Due later pls help

Andre45 [30]

Answer:

1. 4z=-8

z= -8/4= -2

2. 2/3t=8/1

2= 24t

t= 0.08

3. 15-1=-8d

14=-8d

d= -1.75

4. 5r-2r=18

3r= 18

r= 6

5. 24= 8y

y=3

6. 3t-3= -t+5

3t+t= 5+3

4t=8

t=2

7. 2x-x= 3+6

x=9

5 0

2 years ago

Debbie is making an abstract painting with two triangles. The dimensions of the painting are shown below: 6x8 and 10x8 The total

Fiesta28 [93]

128 inches bevy se you have to find the areas of each and then add so it is 48 + 80 =128

5 0

3 years ago

There are 156 sixth graders and 7 sixth grade teachers.There are 120 fifth graders and 5 fifth grade teachers.Which grade has lo

Katen [24]

Fifth grade has 24 students per each teacher.
Sixth grade has about 23 students per teacher.

Sixth grade is lower.

4 0

3 years ago

Read 2 more answers

Simplify the following expression.<br> log(9x^5) + 5 log (1/x)

Darya [45]

Answer:

log9

Step-by-step explanation:

hello :

note : a>0 and b>0

1) log(ab) = loga+logb

2) loga^n = nloga.....n in N

log(9x^5) + 5 log (1/x) = log(9x^5) + log (1/x)^5 = log((9x^5) (1/x)^5)

= log(9x^5)/(x^5)) = log9 because : (1/x)^5 = 1/x^5

6 0

3 years ago

A rectangular box which is open at the top can be made from an12-by-18-inch piece of metal by cutting a square from each corner

algol [13]

Answer:

h=2.35in, l=13.3in and w=7.3in

Step-by-step explanation:

In order to solve this problem, we must start by drawing a diagram that will represent the prolem. (Look at attached picture)

From the diagram we can see that the length ad the width of the box are found by using the following expressions:

L=18-2x

W= 12-2x

H=x

next, we can make use of the volume formula of a rectangular prism, which is the following:

V=WLH

so we can substitute the expressions we got before on the formula:

V=x(12-2x)(18-2x)

and for ease of calculation, we can multiply the terms, so we get:

$V=x(216-24x-36x+4x^{2})$

$V=x(4x^{2}-60x+216)$

$V=4x^{3}-60x^{2}+216x$

So now we can find the derivative of the volume:

$V'=12x^{2}-120x+216$

and set it equal to zero, so we get:

$12x^{2}-120x+216=0$

and now we can solve. We can start by factoring the left side of the equation, so we get:

$12(x^{2}-10x+18)=0$

which yields:

x^{2}-10x+18=0

This one can only be solved by using the quadratic formula:

$x=\frac{-b\pm \sqrt{-b^{2}-4ac}}{2a}$

so we substitute the corresponding values:

$x=\frac{-(-10)\pm \sqrt{-(-10)^{2}-4(1)(18)}}{2(1)}$

which yields:

x=7.65in or x=2.35in

we keep the 2.35in only, because the 7.65in returns an impossible box, since that would make the width a negative width.

So the 2.35in height would give us the greatest box. Now we can use it to find the height, the width and the length of the box.

Height=x=2.35

W=12-2x=12-2*2.35=7.3in

L=18-2x=18-2(2.35)=13.3in

5 0

3 years ago