Replace ∗ with a monomial so that the result is an identity:
1. (2a+*)(2a-*)=4a^2-b^2
2. (5x+*)(5x-*)=25x^2-0.16y^4
3. (*-b^4)(b^4+*)=121a^10-b^8
4. (*-3x)(*+3x)=16y^2-9x^2
5. 100m^4-4n^6=(10m^2-*)(*+10m^2)
6. m^4–225c^10 = (m^2 − *)(* +m^2)

QUESTION 1

Taking the right hand side we have;

We can rewrite this as;

We apply difference of two squares.

QUESTION 2

Taking the right hand side we have;

We can rewrite this as;

We apply difference of two squares.

QUESTION 3

Taking the right hand side of the given equation, we have;

We rewrite this as;

QUESTION 4

From the RHS;

This implies that;

QUESTION 5

From the left hand side, we have

This implies that;

Using difference of two squares, we have;

QUESTION 6

From the LHS;

We rewrite to obtain;

Using difference of two squares, we obtain;

2 & 7

step-by-step explanation:

he climbs 300 ft/hr at a constant rate. the peak is at 1,500 ft.

when going up, he reaches 600 ft in 2 hours. remember, that he is going in a constant rate. divide 600 with 300:

600/300 = 2

assuming that he takes the same route back (meaning that it's a 1500 ft route back, you can solve.

note also that you are solving when he gets back to 600 ft after he reaches the peak.

first, find the amount of hours it takes to get up the peak. divide the total amount (1500), with the ft/her (300) to get the amount of hours (x):

1500/300 = x

x = 5

it takes jamie 5 hours to get up. now, he is returning down, and reaches 600 (his current destination). remember that he climbs at a constant 300 ft/hr. divide 600 with 300:

600/300 = x

x = 2

now add the two numbers together.

x = 5 + 2 = 7

jamie takes 7 hours to go up the mountain and reach his current position when descending.

(total amount/constant rate) = (time).

remember that it is a constant rate, so in total, when he gets back down the mountain, it will take the same amount of time as when he goes up. the total amount for the round trip is: 5 + 5 = 10.

~

but which class are u may know that first