All categories

2 years ago

Simplify 4a³b² × 3a⁶b⁵

Mathematics

2 answers:

Rasek [7]2 years ago

4 0

Answer:

12a^9b^7

Step-by-step explanation:

Multiplying variables of the same base, will require you to add the exponents.

4a^3b^2 * 3a^6b^5

4*3 = 12

a^3 * a^6 = a^9

b^2 * b^5 = b^7

12a^9b^7

Amanda [17]2 years ago

4 0

Answer:

$12a^{9}b^{7}$

Step-by-step explanation:

$4a^{3} b^{2} \times 3a^{6} b^{5}$

We'll use rule of exponents to simplify the expression:

$4^{1}a^{3}b^{2}\times 3^{1}a^{6}b^{5}$

Use the commutative property of Multiplication:

$4^{1}\times 3^{1}a^{3}a^{6}b^{2}b^{5}$

In order to multiply power of the same base, add their exponents:

$4^{1}\times 3^{1}a^{3+6}b^{2+5}$

Add exponents 2 and 5:

$4^{1}\times 3^{1}a^{9}b^{7}$

Multiply 4 times 3:

$12a^{9}b^{7}$

OAmalOHopeO

You might be interested in

Louis started a table showing a multiplication patterns complete the table describe a pattern you see in the products

kvasek [131]

One pattern that you can see in a multiplication table is the perfect square numbers. It runs from the top left hand corner directly through the middle to the bottom right hand corner. A perfect square is a number that is multiplied by itself. The perfect square numbers are 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144. They keep going forever on but those are the main ones from 1x1 to 12x12.

8 0

2 years ago

Sketch the region of integration for the following integral. ∫π/40∫6/cos(θ)0f(r,θ)rdrdθ

solong [7]

Answer:

The graph is sketched by considering the integral. The graph is the region bounded by the origin, the line x = 6, the line y = x/6 and the x-axis.

Step-by-step explanation:

We sketch the integral ∫π/40∫6/cos(θ)0f(r,θ)rdrdθ. We consider the inner integral which ranges from r = 0 to r = 6/cosθ. r = 0 is located at the origin and r = 6/cosθ is located on the line x = 6 (since x = rcosθ here x= 6)extends radially outward from the origin. The outer integral ranges from θ = 0 to θ = π/4. This is a line from the origin that intersects the line x = 6 ( r = 6/cosθ) at y = 1 when θ = π/2 . The graph is the region bounded by the origin, the line x = 6, the line y = x/6 and the x-axis.

3 0

2 years ago

A bin of 5 transistors is known to contain 2 that are defective. The transistors are to be tested, one at a time, until the defe

xxMikexx [17]

Answer:

$P(N_1 = a , N_2 = b)= \frac{1}{5-a C 1} * \frac{5-a C 1}{5C2} = \frac{1}{5C2}=\frac{1}{10}$

Step-by-step explanation:

For the random variable $N_1$ we define the possible values for this variable on this case $[1,2,3,4,5]$ . We know that we have 2 defective transistors so then we have 5C2 (where C means combinatory) ways to select or permute the transistors in order to detect the first defective:

$5C2 = \frac{5!}{2! (5-2)!}= \frac{5*4*3!}{2! 3!}= \frac{5*4}{2*1}=10$

We want the first detective transistor on the ath place, so then the first a-1 places are non defective transistors, so then we can define the probability for the random variable $N_1$ like this:

$P(N_1 = a) = \frac{5-a C 1}{5C2}$

For the distribution of $N_2$ we need to take in count that we are finding a conditional distribution. $N_2$ given $N_1 =a$ , for this case we see that $N_2 \in [1,2,...,5-a]$ , so then exist $5-a C 1$ ways to reorder the remaining transistors. And if we want b additional steps to obtain a second defective transistor we have the following probability defined: