Which graph represents a function? On a coordinate plane, a line with 2 angles crosses the x-axis at (negative .5, 0), the y-axis at (0, 1), turns at (1, 3), crosses the x-axis at (1, 0), turns at (1, negative 2), and crosses the x-axis at (2, 0). On a coordinate plane, a curved line crosses the y-axis at (0, 1.5), the x-axis at (negative 2, 0), and the y-axis at (0, negative 1.5). On a coordinate plane, a line enters the plane at point (negative 5, 4), makes a 90-degree turn at (negative 1, 0), and leaves the plane at point (negative 5, negative 4).

i agree because i see the other persons thinking is clear

16.80 times .20= 3.36

step-by-step explanation:

hello!

to begin with, we're given that on earth, a rock weighs 20 pounds, and while on jupiter, that same rock weighs 42 pounds. now, we're trying to find the weight of a rock that weighs 10.5 pounds on earth, on jupiter.

to solve this, first, set up a proportion, cross-multiply, and solve it algebraically.

setting up the proportion is the hardest and the most confusing part of the problem, but it's simple.

if on earth, the rock weighs 20 lbs, and on jupiter it weighs 42 lbs, our first part of our proportion is 20/42 lbs. the second part has to be equal to each other. this means that 10.5 has to be on the numerator of the fraction, and x, the value we're trying to find, has to be on the denominator. remember that the rock weighs 10.5 lbs on earth, and we're trying to find the weight of it on jupiter (which is x).

20/42 = 10.5/x

next, you would cross-multiply.

20x = 42(10.5)

20x = 441

now, solve it algebraically by using division.

x = 22.05 lbs

this can be rounded to 22.1 lbs.

therefore, the weight of the rock on jupiter is 22.1 lbs, the fourth choice.

the answer is:

the lateral area is equal to

to find the lateral area of the square pyramid, first we need to find the perimeter of the base, and then, find the lateral area using the following formulas:

where,

s, is the base length

slant height, is the height of a the side of the pyramid

we are given:

now, calculating the perimeter of the base, we have:

then, calculating the lateral area:

hence, we have that the lateral area is equal to

have a nice day!