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Alja [10]
3 years ago
6

How would you combine like terms with exponents? Do you add the exponents?

Mathematics
2 answers:
choli [55]3 years ago
6 0

Answer: When you combine like terms in a algebraic expression, you do not add the exponents! When you combine like terms in an equation, such as a^2 and 4a^2, you only combine the numbers, not the exponents. The exponent attached to the number stays the same. a^2 and 4a^2 would be 5a^2.

matrenka [14]3 years ago
5 0

Answer: When combining like terms, add or subtract the coefficients. Keep the exponents as they are.

Step-by-step explanation:

You can combine 2x^2 + 3x^2 to get 5x^2. If x =3, 3^2=9 So 2(3)^2 is 18 and 3(3)^2 is 27. 18+27=45 And 5(3)^2= 45. Same result!

You can not combine 2x^2 and 2x^3. If the value of x is 3, 2(3)^2 this term works out to =18 and 2(3)^3 =54

If you add the exponents x^5 becomes 2(3)^5 or 2×243=486. Vastly different values! Don't add exponents unless you are multiplying terms.

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8 0
3 years ago
Evaluate 6x + 11 when x = 7
nlexa [21]

Answer:

53

Step-by-step explanation:

To evaluate 6x + 11, we must substitute the value of x.

Since we know that x = 7, it is easier to evaluate the expression;

6x + 11

6(7) + 11

Since 6 is outside the parenthesis, we must multiply everything inside the parenthesis by 6;

6(7) + 11

42 + 11

When you add the two numbers you get:

53

4 0
3 years ago
30 1/4 % in simplest form
11111nata11111 [884]
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5 0
3 years ago
Can someone please help me with this question?!? I am so confused and I don't know how to answer it.
lbvjy [14]

9514 1404 393

Answer:

  a. f(0) = 1

  b. DNE (does not exist)

  c. DNE

  d. lim = 3

Step-by-step explanation:

The function exists at a point if it is defined there. The function is defined anywhere on the solid line and at solid dots. It is not defined at open circles. So, the function is defined everywhere except (2, 3), which has an open circle.

The open circle at (0, 4) prevents the function from being doubly-defined at x=0, since it is already defined to be 1 at x=0.

This discussion tells you ...

  f(0) = 1

 f(2) does not exist. There is a "hole" in the function definition there.

__

The function has a limit at a point if approaching from the left and approaching from the right have you approaching that same point.

Consider the point (1, 2). The graph is a solid line through that point. Approaching from values less than x=1, we get to the same point (1, 2) as when we approach from values greater than x=1.

Similarly, consider the point (2, 3). Approaching from values of x less than 2, we get to the same point (2, 3) as when we approach from x-values greater than 2. The limit at x=2 is 3. The only difference from the previous case is that the function is not actually defined to be that value there.

__

Now consider what happens at x=0. When we approach from the left, we approach the point (0, 4). When we approach from the right, we approach the point (0, 1). These are different points. Because they are different coming from the left and from the right, we say "the limit as x→0 does not exist."

__

In summary, ...

  a) f(0) = 1

  b) lim x → 0 does not exist

  c) f(2) does not exist

  d) lim x → 2 = 3

_____

<em>Additional comment</em>

The significance of the function not being defined at a point where the limit exists, (2, 3), is that <em>the function is not continuous there</em>. This kind of discontinuity is called "removable", because we could make the function continuous at x=2 by defining f(2) = 3 (that is, "filling the hole").

6 0
2 years ago
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Your bakery is 450 square ft/////this is because you have to do 4,500•1/10
6 0
3 years ago
Read 2 more answers
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