If the integer $152AB1$ is a perfect square, what is the sum of the digits of its square root?

9514 1404 393

  13

Step-by-step explanation:

152AB1 is not a square in hexadecimal, so we assume A and B are supposed to represent single digits in decimal.

If A=B=0, √152001 ≈ 389.9

If A=B=9, √152991 ≈ 391.1

The least significant digit of 152AB1 being non-zero, we know it is not the square of 390. Hence, it must be the square of 391.

For 152AB1 to be a perfect square, we must have ...

  152AB1 = 391² = 152881

The sum of the digits of the square root is 3+9+1 = 13.

the answer is the option b

the range is all real numbers

step-by-step explanation:

we know that

the absolute value parent function is equal to

see the attached figure to better understand the problem

so

a. it is v-shaped > is true

b. the range is all real numbers > is not true

because the range is the > [0,∞)

all real numbers greater than or equal to zero

c. it goes through the origin > is true

d. the domain is all real numbers > is true

step-by-step explanation:

1/3

step-by-step explanation:

to find this, you can divide 12 by 4 which is 3 or you can multiply 4 until you reach 12 (4x3=12). both should give you 3, meaning theres three times less bananas than there are oranges.