Answer:
(d) f(x) = -x²
Step-by-step explanation:
For the vertex of the quadratic function to be at the origin, both the x-term and the constant must be zero. That is, the function must be of the form ...
f(x) = a(x -h)² +k . . . . . . . . . . vertex form; vertex at (h, k)
f(x) = a(x -0)² +0 = ax² . . . . . vertex at the origin, (h, k) = (0, 0)
Of the offered answer choices, the only one with a vertex at the origin is ...
f(x) = -x² . . . . . a=-1
Answer:
V= 14.13
Step-by-step explanation:
Information needed:
V= 4/3
r^3
= 3.14
r= 3/2
Solve:
V= 4/3
r^3
V= 4/3(3.14)(3/2)^3
V= 4/3(3.14)(27/8)
V= 14.13
Normally, we could add exponents.
however, that only is possible when the bases are the same
recall what exponents mean
12³=12*12*12
so we cannot add exponents for 12³*11³ because that means 12*12*12*11*11*11
it would not equal 12⁶ or 11⁶
or you could refer to the rule

notice when x=x then we can add the bases
fun fact below
we can reverse a previous exponential rule like this
since

then

therefor

we can't add the exponents because the bases are not the same
Answer:
place the decimal point in the product so that the number of decimal places in product is the sum of decilal places in the factors. keep all zeros in the product when you place the decimal point. you can drop the zeros on the right once the decimal point has been placed in the product.