All categories

3 years ago

An expression is shown below:

Mathematics

1 answer:

aleksandr82 [10.1K]3 years ago

5 0

Answer:

(a) $3y(x^3+4x-3x^2-12)$

(b) $3y(x^2+4)(x-3)$

Step-by-step explanation:

Given the expression: $3x^3y + 12xy - 9x^2y -36y$

If we consider the terms: $3x^3y ,12xy,9x^2y,36y$ in the expression. The common terms are 3 and y.

Therefore the greatest common factor of the terms is 3y.

Factoring this out, we obtain:

$3x^3y + 12xy - 9x^2y -36y=3y(x^3+4x-3x^2-12)$

Next, we factor the expression completely

$3x^3y + 12xy - 9x^2y -36y\\=3y(x^3+4x-3x^2-12)\\$Rearranging we obtain:\\=3y(x^3-3x^2+4x-12)\\=3y[x^2(x-3)+4(x-3)]\\=3y(x^2+4)(x-3)$

You might be interested in

-2x+y=1 and y=-x+4 What is x?

Vikentia [17]

When -2x+y=1 and y=-x+4, x = 1

4 0

2 years ago

Huilan's age is three times Thomas's age. The sum of their ages is 24. What Is Thomas's age?

mafiozo [28]

Step-by-step explanation:

8 because 24 ÷3=8 so yeah

6 0

2 years ago

Let D be the event that a randomly chosen person has seen a dermatologist. Let S be the event that a randomly chosen person has

SSSSS [86.1K]

Given:

D be the event that a randomly chosen person has seen a dermatologist.

S be the event that a randomly chosen person has had surgery for skin cancer.

To find:

The correct notation for the probability that a randomly chosen person has had surgery for skin cancer, given that the person has seen a dermatologist.

Solution:

Conditional probability: Probability of A given B is:

$P(A|B)=\dfrac{P(A\cap B)}{P(B)}$

Let D be the event that a randomly chosen person has seen a dermatologist.

Let S be the event that a randomly chosen person has had surgery for skin cancer.

Using the conditional probability, the correct notation for the probability that a randomly chosen person has had surgery for skin cancer, given that the person has seen a dermatologist is P(S|D).

Therefore, the correct option is D.

8 0

2 years ago

Which ordered pair is a solution to - 8x + 6y <3x - 30?

Ivenika [448]

Answer:

B ( 6,-6) ..

I think I'm correct

4 0

2 years ago

Read 2 more answers

A country's population in 1994 was 182 million. In 2002 it was 186 million. Estimate the population in 2004 using the exponentia

iogann1982 [59]

$\bf =ae^{kt}\qquad 
\begin{cases}
1994\impliedby \textit{year 0, starting point}\\
t=0\qquad P=182
\end{cases}\implies 182=ae^{k0}
\\\\\\
182=a\cdot e^0\implies 182=a\cdot 1\implies 182=a
\\\\\\
thus\qquad P=182e^{kt}\\\\
-------------------------------\\\\$

$\bf P=182e^{kt}\qquad 
\begin{cases}
2002\impliedby \textit{8 years later}\\
t=8\qquad P=186
\end{cases}\implies 186=182e^{k8}
\\\\\\
\cfrac{186}{182}=e^{8k}\implies ln\left( \frac{93}{91} \right)=ln(e^{8k})\implies ln\left( \frac{93}{91} \right)=8k
\\\\\\
\cfrac{ln\left( \frac{93}{91} \right)}{8}=k\implies 0.0027\approx k\implies \boxed{P=182e^{0.0027t}}$

what's the population in 2004? well, from 1994 to 2004 is 10 years later, so t = 10

plug that in, to get P for 2004

3 0

3 years ago