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3 years ago

Simplify the sum. State any restrictions on the variables. (x-2)/(x+3) + (10x)/(x^(2)-9)

Mathematics

2 answers:

Trava [24]3 years ago

7 0

Answer:

(x+2)/(x-3)

Step-by-step explanation:

(x-2)/(x+3) + (10x)/(x²+9)

(x-2)(x-3)+10x/(x+3)(x-3)

(x²-3x-2x+6+10x)/(x+3)(x-3)

(x²+5x+6)/(x+3)(x-3)

(x+3)(x+2)/(x+3)(x-3)

Ans: (x+2)/(x-3)

nlexa [21]3 years ago

6 0

$\frac{x-2}{x+3} + \frac{10x}{ x^{2} -9}= \\ = \frac{x-2}{x+3}+ \frac{10x}{(x-3)(x+3)}=$
Restrictions are: x ≠ - 3, x ≠ 3.
$= \frac{(x+2)(x-3)+10x}{(x+3)(x-3)}= \\ \frac{ x^{2} -3x+2x-6+10x}{(x+3)(x-3)}= \\ = \frac{ x^{2} +9x-6}{(x+3)(x-3)}$

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A right rectangular prism's edge lengths are 14 inches, 6 inches, and 3 1/3 inches. How many unit cubes with edge lengths of 2/5

DochEvi [55]

ANSWER

4375 unit cubes

EXPLANATION

The volume of a right rectangular prism is given by;

$V=lbh$

We substitute the given dimension to obtain,

$V=14 \times 6 \times \frac{10}{3} = 280i {n}^{3}$

The volume of the unit cube with edge lengths

$\frac{2}{5}$

$(\frac{2}{5})^{3} = \frac{8}{125} {in}^{3}$

To find the number of unit cubes that can fit inside the prism, we divide the volume of the rectangular prism by the volume of the unit cube.

$= \frac{280}{ \frac{8}{125} }$

$= 4375$

7 0

3 years ago

Gus is making a Chilli recipe that calls 3 parts beans to five parts ground beef .if he is using 8 cups of ground beef for a big

masya89 [10]

First you need to find the percent equivalent to that of the ratio. The ratio is 3:5. From there you would take the 3, and divide it by 5. 3/5=0.6. Apply that to the information that you are given: 8(0.6)=4.8.

4 0

3 years ago

A study of long-distance phone calls made from General Electric Corporate Headquarters in Fairfield, Connecticut, revealed the l

Katena32 [7]

Answer:

(a) The fraction of the calls last between 4.50 and 5.30 minutes is 0.3729.

(b) The fraction of the calls last more than 5.30 minutes is 0.1271.

(c) The fraction of the calls last between 5.30 and 6.00 minutes is 0.1109.

(d) The fraction of the calls last between 4.00 and 6.00 minutes is 0.745.

(e) The time is 5.65 minutes.

Step-by-step explanation:

We are given that the mean length of time per call was 4.5 minutes and the standard deviation was 0.70 minutes.

Let X = the length of the calls, in minutes.

So, X ~ Normal( $\mu=4.5,\sigma^{2} =0.70^{2}$ )

The z-score probability distribution for the normal distribution is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

where, $\mu$ = population mean time = 4.5 minutes

$\sigma$ = standard deviation = 0.7 minutes

(a) The fraction of the calls last between 4.50 and 5.30 minutes is given by = P(4.50 min < X < 5.30 min) = P(X < 5.30 min) - P(X $\leq$ 4.50 min)

P(X < 5.30 min) = P( $\frac{X-\mu}{\sigma}$ < $\frac{5.30-4.5}{0.7}$ ) = P(Z < 1.14) = 0.8729

P(X $\leq$ 4.50 min) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{4.5-4.5}{0.7}$ ) = P(Z $\leq$ 0) = 0.50

The above probability is calculated by looking at the value of x = 1.14 and x = 0 in the z table which has an area of 0.8729 and 0.50 respectively.

Therefore, P(4.50 min < X < 5.30 min) = 0.8729 - 0.50 = 0.3729.

(b) The fraction of the calls last more than 5.30 minutes is given by = P(X > 5.30 minutes)

P(X > 5.30 min) = P( $\frac{X-\mu}{\sigma}$ > $\frac{5.30-4.5}{0.7}$ ) = P(Z > 1.14) = 1 - P(Z $\leq$ 1.14)

= 1 - 0.8729 = 0.1271

The above probability is calculated by looking at the value of x = 1.14 in the z table which has an area of 0.8729.

(c) The fraction of the calls last between 5.30 and 6.00 minutes is given by = P(5.30 min < X < 6.00 min) = P(X < 6.00 min) - P(X $\leq$ 5.30 min)

P(X < 6.00 min) = P( $\frac{X-\mu}{\sigma}$ < $\frac{6-4.5}{0.7}$ ) = P(Z < 2.14) = 0.9838

P(X $\leq$ 5.30 min) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{5.30-4.5}{0.7}$ ) = P(Z $\leq$ 1.14) = 0.8729

The above probability is calculated by looking at the value of x = 2.14 and x = 1.14 in the z table which has an area of 0.9838 and 0.8729 respectively.

Therefore, P(4.50 min < X < 5.30 min) = 0.9838 - 0.8729 = 0.1109.

(d) The fraction of the calls last between 4.00 and 6.00 minutes is given by = P(4.00 min < X < 6.00 min) = P(X < 6.00 min) - P(X $\leq$ 4.00 min)

P(X < 6.00 min) = P( $\frac{X-\mu}{\sigma}$ < $\frac{6-4.5}{0.7}$ ) = P(Z < 2.14) = 0.9838

P(X $\leq$ 4.00 min) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{4.0-4.5}{0.7}$ ) = P(Z $\leq$ -0.71) = 1 - P(Z < 0.71)

= 1 - 0.7612 = 0.2388

The above probability is calculated by looking at the value of x = 2.14 and x = 0.71 in the z table which has an area of 0.9838 and 0.7612 respectively.

Therefore, P(4.50 min < X < 5.30 min) = 0.9838 - 0.2388 = 0.745.

(e) We have to find the time that represents the length of the longest (in duration) 5 percent of the calls, that means;

P(X > x) = 0.05 {where x is the required time}

P( $\frac{X-\mu}{\sigma}$ > $\frac{x-4.5}{0.7}$ ) = 0.05

P(Z > $\frac{x-4.5}{0.7}$ ) = 0.05

Now, in the z table the critical value of x which represents the top 5% of the area is given as 1.645, that is;

$\frac{x-4.5}{0.7}=1.645$

${x-4.5}{}=1.645 \times 0.7$

x = 4.5 + 1.15 = 5.65 minutes.

SO, the time is 5.65 minutes.

7 0

3 years ago

1. The sides of a triangle have lengths of 5, 8, and 6.(1/ 2). An angle bisector meets the side of length 6.(1/ 2). Find the len

Advocard [28]

Answer:

Step-by-step explanation:

i assume x+y=6(1/2)=\frac{13}{2}

$\frac{5}{8}=\frac{x}{y}=\frac{x}{\frac{13}{2}-x}=\frac{2x}{13-2x}\\65-10x=16x \\26 x=65\\2x=5\\x=\frac{5}{2}\\y=\frac{13}{2}-\frac{5}{2}=\frac{8}{2}=4$

6 0

3 years ago

Pls answer and thanks!!

Keith_Richards [23]

Answer:

Problem:

Solve 3x−y=6;y=x−4

Steps:

I will try to solve your system of equations.

y=x−4;3x−y=6

Step: Solvey=x−4for y:

Step: Substitutex−4foryin3x−y=6:

3x−y=6

3x−(x−4)=6

2x+4=6(Simplify both sides of the equation)

2x+4+−4=6+−4(Add -4 to both sides)

2x=2

(Divide both sides by 2)

x=1

Step: Substitute1forxiny=x−4:

y=x−4

y=1−4

y=−3(Simplify both sides of the equation)

Answer:

y=−3 and x=1

Step-by-step explanation:

Hope this helped you

3 0

2 years ago

Read 2 more answers