All categories

3 years ago

(a + b - c )(a + b + c )

Mathematics

2 answers:

ss7ja [257]3 years ago

8 0

Answer:

$(a+b-c)(a+b+c)=a^2+2ab+b^2-c^2$ .

Step-by-step explanation:

We want to expand: $(a+b-c)(a+b+c)$ .

We use the distributive property to obtain:

$(a+b-c)(a+b+c)=a(a+b+c)+b(a+b+c)-c(a+b+c)$ .

We now expand to get:

$(a+b-c)(a+b+c)=a^2+ab+ac+ab+b^2+bc-ac-bc-c^2$ .

This finally simplifies to

$(a+b-c)(a+b+c)=a^2+2ab+b^2-c^2$ .

True [87]3 years ago

6 0

If you are looking for the expanded version, this is it. It is also possible to group either an a from (a^2 + 2b) or a b from (2ab + b^2). Hope this helps!

You might be interested in

If a tank holds 5000 gallons of water, which drains from the bottom of the tank in 40 minutes, then Torricelli's Law gives the v

pochemuha

Answer:

V'(t) = $-250(1 - \frac{1}{40}t)$

If we know the time, we can plug in the value for "t" in the above derivative and find how much water drained for the given point of t.

Step-by-step explanation:

Given:

V = $5000(1 - \frac{1}{40}t )^2$ , where 0≤t≤40.

Here we have to find the derivative with respect to "t"

We have to use the chain rule to find the derivative.

V'(t) = $2(5000)(1 - \frac{1}{40} t)d/dt (1 - \frac{1}{40}t )$

V'(t) = $2(5000)(1 - \frac{1}{40} t)(-\frac{1}{40} )$

When we simplify the above, we get

V'(t) = $-250(1 - \frac{1}{40}t)$

If we know the time, we can plug in the value for "t" and find how much water drained for the given point of t.

4 0

4 years ago

5x=y-4 write it in standard form

Tems11 [23]

5x - y = -4 is the standard form

7 0

4 years ago

Gina looked at the architectural plan of a room with four walls in which the walls meet each other at right angles. The length o

Lerok [7]

For a better understanding of the solution provided here, please find the diagram attached.

In the diagram, ABCD is the room.

AC is the diagonal whose length is 18.79 inches.

The length of wall AB is 17 inches.

From the given information, we have to determine the length of the BC, which is depicted a $x$ , because for the room to be a square, the length of the wall AB must be equal to the length of the wall BC.