(−14x^3y^3+4x^5 y6)÷(2x^4y^5)

1 answer:

7 0

( - 14x^3)(y^3) + 4x^5y^6)/(2x^4y^5)

= 4x^5y^6 - 14x^3y^3/2x^4y^5

= 4x^2y^3 - 14/2xy^2

= 2x^2y^3 - 7/xy^2

Answer: 2x^2y^3 - 7/xy^2

Hope that helps!!!!

You might be interested in

I don't remember how to do this

meriva

B. Fractions have to have the same denominator so $\frac{9}{10} - \frac{2}{5} = \frac{9}{10} - \frac{4}{10}$ which = $\frac{5}{10}$ then you simplfy it to $\frac{1}{2}$ so it would be B

4 0

2 years ago

Solve the equation show steps -2m+4m+5=3

Dafna11 [192]

-2m+4m+5=3
(-2m+4m)+(5)=3
2m+5=3
2m+5-5=3-5
2m=-2
2m/2=-2/2
m=-1

5 0

2 years ago

Read 2 more answers

#15 I don’t know how to do the proof

Mila [183]

What's the question.....

3 0

3 years ago

The low temperature on Monday was 16°F. On Tuesday, the low was 18°F cooler. On Wednesday, the low temperature was –4 times Tues

drek231 [11]

Answer:

$-4(16-18)$ , $[16 + (-18)](-4)$

Step-by-step explanation:

Given: The low temperature on Monday was 16°F. The low temperature on Tuesday was 18°F cooler. The low temperature on Wednesday was –4 times Tuesday’s temperature.

To find: expression that can be used to describe the low temperature on Wednesday

Solution:

Temperature on Monday = 16°F

So,

Temperature on Tuesday = $(16-18)\°F$