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

2 answers:

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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According to the fundamental theorem of algebra how many roots exist for the polynomial function f(x)=8x^7-x^5+x^3+6

mojhsa [17]

Answer:

Step-by-step explanation:

This is a 7th degree polynomial. There should be 7 roots. Note how degree of poly = number of roots.

5 0

3 years ago

Read 2 more answers

PLEASE HELP

Sholpan [36]

1) We have that the equation is $x^2=20y$ , hence y=x^2/20. The standard equation of such an equation is y= $\frac{1}{4p} x^2$ . Hence, p=5 in this case. The focus is at (0,5) and the directrix is at y=-5 (a tip is that the directrix is always "opposite" the focus point of a parabola; if the directrix is at x=-7 for example, the focus is at (7,0)).
2) Similarly, we have that the equation is $x=3y^2 \\ \frac{1}{4p} =3$ . Thus, p=1/12. In this case, the parabola opens along the x-axis and the focus is at (1/12, 0). Also, the directrix is at x=-1/12. Hence the correct answer is B.
3) We are given that the parabola has a p of 9. Also, the focus lies along the y-axis, hence the parabola is opening along the y-axis. Finally, the focus is on the positive half, so the parabola is opening upwards. The equation for this case is y= $y=\frac{1}{4p} x^2= \frac{1}{36 } x^2$ .
4) Similarly as above. The directrix is superfluous, we only need the p-value. THe same comments about the parabola apply and if we substitute p=8 in the formula: $y= \frac{1}{4p} x^2$ we get y= $\frac{1}{32} x^2$ .
5) This is somewhat different, even though we do not need the directrix again. The focus lies on the x-axis, thus the parabola opens in this direction. The focus lies on the positive part of the axis, thus the parabola opens to the right. We also are given p=7. Hence, the equation we need is of the form $x= \frac{1}{4p} y^2$ . Substituting p=7, we get $x= \frac{1}{28} y^2$ .
6) The equation of a prabola with a vertex at (0,0) is of the form y=-ax^2. The minus sign is needed since the parabola is downwards. Since we are given anothe point, we can calculate a. We have to take y=-74 and x=14 feet (since left to right is 28, we need to take half). $-a= \frac{y}{x^2} = \frac{-74}{14^2} =-0.378$ . Thus a=0.378. Hence the correct expressions is y=-0.378* $x^2$