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3 years ago

Simply (2x^-3 y^5) (-4x^8 y^2)

Mathematics

1 answer:

In-s [12.5K]3 years ago

5 0

The ans is B

(2x^-3 y^5) (-4x^8 y^2)
= (-2x^-3 y^5) (4x^8 y^2)
= -8x^5 y^7

Hope can help u

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One counting number is 4 times great as a counting number.The product of the two number is 36.What is the sum of the two number

VARVARA [1.3K]

Answer:

The sum of the two numbers is 15

Step-by-step explanation:

Let the counting numbers be x and y.

Then we can write the following equations

x=4y.......1 (one is four times greater )

x(y)=36.....2 (the product of the two)

Putting equation 1 into equation 2, we get

4y(y)=36

${4y}^{2} = 36$

Dividing through by 4,we get

${y}^{2} = \frac{36}{4}$

${y}^{2} = 9$

Taking square root of both side, we obtain

y=3

Putting the value of y into equation , we get

x=4(3)=12

Hence the numbers are 3 and 12

Therefore the sum of the two numbers is 3+12=15

3 0

3 years ago

Help!!!!! Solve the distance??

hram777 [196]

Answer:

531.4

Step-by-step explanation:

ED/AE=BC/AB

ED=AE*BC/AB

ED=240*620/280

ED=531.4

6 0

2 years ago

Two children own two-way radios that have a maximum range of 3 miles. One leaves a certain point at 1:00 P.M., walking due north

Grace [21]

Answer:

The answer is 21 minutes

Step-by-step explanation:

We use the equation Xf = Xo + vt

1) At 1:00 PM, child one leaves the starting point heading north at a constant velocity of 6 mi/hr or .1 [mi/min] (divide by 60 to convert from [mi/hr] to [mi/min])

2) He walks for 15 minutes before kid 2 starts walking. In 15 minutes he is able to cover 1.5 [mi]

$x_{1f1} =x_{o} +v_{1} t_{1} \\x_{1f1} =0+.1(15)\\x_{1f1} =1.5 [mi]$

3) Now, child 2 starts walking and we know that when the range reaches 3 miles, they won´t be able to communicate. So the sum of the final position of child 1 and child 2 must be 3[mi]