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4 years ago

What is the product x^2-16 over 2x+8 • x^3-2x^2+x over x^2+3x-4

Mathematics

1 answer:

Oxana [17]4 years ago

7 0

Answer:

$\dfrac{x^3-5x^2 +4x}{2x+8}$

Step-by-step explanation:

The basic idea is to factor each expression and cancel common factors from numerator and denominator.

$\dfrac{x^2-16}{2x+8}\cdot\dfrac{x^3-2x^2 +x}{x^2+3x-4}=\dfrac{(x-4)(x+4)}{2(x+4)}\cdot\dfrac{x(x-1)^2}{(x+4)(x-1)}\\\\=\dfrac{(x-4)x(x-1)}{2(x+4)}\cdot\dfrac{(x+4)(x-1)}{(x+4)(x-1)}\\\\=\boxed{\dfrac{x^3-5x^2 +4x}{2x+8}}$

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Sinplify the expression

Sunny_sXe [5.5K]

To simplify a division of exponents, you can subtract the powers. 2-(1/4)=7/4

Now, the expression is 3^7/4, which is equal to 6.8385.

Hope this helps!!

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4 years ago

Franco's Pizza is selling their pizzas 35% cheaper than usual if a pizza normally cost $12 how much is it now

avanturin [10]

Answer:

New or current cost = 12 - 4.2 = $7.8

Step-by-step explanation:

Franco's Pizza is selling their pizzas 35% cheaper than usual. This is a discount on the normal cost. To get the amount it costs now, we will find the percentage of the discount on the original cost and subtract what we get from the original cost.

Discount = 35℅

Original cost of pizza is $12 (if a pizza normally cost $12).

So the discount in terms of dollars

= ℅ discount / 100 × normal cost of pizza

= 35 / 100 × 12 =0.35 × 12 = $4.2

New or current cost of pizza would be normal or original cost - the discount in dollars.

New or current cost = 12 - 4.2 = $7.8

4 0

3 years ago

The number of defective circuit boards coming off a soldering machine follows a Poisson distribution. During a specific ten-hour

Alexus [3.1K]

Answer:

a) the probability that the defective board was produced during the first hour of operation is $\frac{1}{10}$ or 0.1000

b) the probability that the defective board was produced during the last hour of operation is $\frac{1}{10}$ or 0.1000

c) the required probability is 0.2000

Step-by-step explanation:

Given the data in the question;

During a specific ten-hour period, one defective circuit board was found.

Lets X represent the number of defective circuit boards coming out of the machine , following Poisson distribution on a particular 10-hours workday which one defective board was found.

Also let Y represent the event of producing one defective circuit board, Y is uniformly distributed over ( 0, 10 ) intervals.

f(y) = $\left \{ {{\frac{1}{b-a} }\\\ }} \right _0$ ; ( a ≤ y ≤ b ) $_{elsewhere$

= $\left \{ {{\frac{1}{10-0} }\\\ }} \right _0$ ; ( 0 ≤ y ≤ 10 ) $_{elsewhere$

f(y) = $\left \{ {{\frac{1}{10} }\\\ }} \right _0$ ; ( 0 ≤ y ≤ 10 ) $_{elsewhere$

Now,

a) the probability that it was produced during the first hour of operation during that period;

P( Y < 1 ) = $\int\limits^1_0 {f(y)} \, dy$

we substitute

= $\int\limits^1_0 {\frac{1}{10} } \, dy$

= $\frac{1}{10} [y]^1_0$

= $\frac{1}{10} [ 1 - 0 ]$

= $\frac{1}{10}$ or 0.1000

Therefore, the probability that the defective board was produced during the first hour of operation is $\frac{1}{10}$ or 0.1000

b) The probability that it was produced during the last hour of operation during that period.

P( Y > 9 ) = $\int\limits^{10}_9 {f(y)} \, dy$

we substitute

= $\int\limits^{10}_9 {\frac{1}{10} } \, dy$

= $\frac{1}{10} [y]^{10}_9$

= $\frac{1}{10} [ 10 - 9 ]$

= $\frac{1}{10}$ or 0.1000

Therefore, the probability that the defective board was produced during the last hour of operation is $\frac{1}{10}$ or 0.1000

no defective circuit boards were produced during the first five hours of operation.

probability that the defective board was manufactured during the sixth hour will be;

P( 5 < Y < 6 | Y > 5 ) = P[ ( 5 < Y < 6 ) ∩ ( Y > 5 ) ] / P( Y > 5 )

= P( 5 < Y < 6 ) / P( Y > 5 )

we substitute

$= (\int\limits^{6}_5 {\frac{1}{10} } \, dy) / (\int\limits^{10}_5 {\frac{1}{10} } \, dy)$

$= (\frac{1}{10} [y]^{6}_5) / (\frac{1}{10} [y]^{10}_5)$

= ( 6-5 ) / ( 10 - 5 )

= 0.2000

Therefore, the required probability is 0.2000

4 0

3 years ago

How many times greater is the value of 5 in 3,590 than the value of 5 in 359 ?

Tresset [83]

Answer:

10 times greater is the value of 5 in 3,590 than the value of 5 in 359.

Step-by-step explanation:

To find : How many times greater is the value of 5 in 3,590 than the value of 5 in 359 ?

Solution :

The place value stem is

Thousand Hundred Tens Ones

1000 100 10 1

The value of 5 in 3,590 is at hundred place

So, $5\times 100=500$

The value of 5 in 359 is at tens place

So, $5\times 10=50$

Number of time greater is the value of 5 in 3,590 than the value of 5 in 359 is given by,

$n=\frac{500}{50}$

$n=10$

Therefore, 10 times greater is the value of 5 in 3,590 than the value of 5 in 359.

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3 years ago

PLS HELP ME PLS SOVLE THIS WITH THE LIKE TERMS I ONLY HAVE 25 MINS TO FINISH THIS

Basile [38]

Answer:

Step-by-step explanation:

Here are the 3 answers you needed :)!!!

8 0

3 years ago

Read 2 more answers

What is the product x^2-16 over 2x+8 • x^3-2x^2+x over x^2+3x-4￼￼

What is the product x^2-16 over 2x+8 • x^3-2x^2+x over x^2+3x-4