All categories

2 years ago

Simplify (3a^3 b^3)(4ab^2)^2

1 answer:

7 0

$\bf (3a^3b^3)(4ab^2)^2\implies (3a^3b^3)(4^2a^2b^{2\cdot 2})\implies (3a^3b^3)(4^2a^2b^4)
\\\\\\
3a^3b^34^2a^2b^4\implies 3\cdot 4^2a^3a^2b^3b^4\implies 3\cdot 16a^{3+2}b^{3+4}\implies 48a^5b^7$

You might be interested in

Find the value of y.

mixas84 [53]

Answer:

y=9

Step-by-step explanation:

3 0

2 years ago

Solve -3/5+1/5>7/20 Drag and drop a number or symbol into each box to show the solution.

Korolek [52]

Not sure what you mean by drag and drop but Im going to attempt this.

-3/5 = -12/20
1/5 = 4/20

That makes no sense. It should be -8/20 < 7/20?

3 0

2 years ago

For the school fair, 33 English teachers, 72 math teachers, 55 science teachers, and 20 social studies teachers volunteered for

IrinaVladis [17]

Answer:

0.4

Step-by-step explanation:

33+72+55+20=180

probability of choosing of choosing a maths teacher=72/180

=0.4

3 0

2 years ago

A Pizza place offers 4 different styles: Cheese, Pepperoni, Meat Lovers, and Hawaiian. They offer 3 sizes: Medium, Large, and X-

blondinia [14]

Answer:

36 different pizzas

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Select the two values of x that are roots of this equation x^2-5x+2=0

Nataly [62]

The roots of the equation is x = 4.56 OR x = 0.44

<h3>Quadratic equation</h3>

From the question, we are to determine the roots of the given equation

The given equation is

x² -5x +2 = 0

Using the formula method,

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

In the given equation,

a = 1, b = -5, c = 2

Putting the values into the formula,

$x =\frac{-(-5) \pm \sqrt{(-5)^{2}-4(1)(2) } }{2(1)}$

$x =\frac{5 \pm \sqrt{25-8} }{2}$

$x =\frac{5 \pm \sqrt{17} }{2}$

$x =\frac{5 + \sqrt{17} }{2}$ OR $x =\frac{5 - \sqrt{17} }{2}$

$x =\frac{5 + 4.12}{2}$ OR $x =\frac{5 - 4.12}{2}$

$x =\frac{9.12}{2}$ OR $x =\frac{0.88}{2}$

x = 4.56 OR x = 0.44

Hence, the roots of the equation is x = 4.56 OR x = 0.44

Learn more on Quadratic equation here: brainly.com/question/8649555

#SPJ1

3 0

2 years ago