Using the table below, what was Oscar's distance at the beginning of the season (at month O)? ​

2 & 7

step-by-step explanation:

he climbs 300 ft/hr at a constant rate. the peak is at 1,500 ft.

when going up, he reaches 600 ft in 2 hours. remember, that he is going in a constant rate. divide 600 with 300:

600/300 = 2

assuming that he takes the same route back (meaning that it's a 1500 ft route back, you can solve.

note also that you are solving when he gets back to 600 ft after he reaches the peak.

first, find the amount of hours it takes to get up the peak. divide the total amount (1500), with the ft/her (300) to get the amount of hours (x):

1500/300 = x

x = 5

it takes jamie 5 hours to get up. now, he is returning down, and reaches 600 (his current destination). remember that he climbs at a constant 300 ft/hr. divide 600 with 300:

600/300 = x

x = 2

now add the two numbers together.

x = 5 + 2 = 7

jamie takes 7 hours to go up the mountain and reach his current position when descending.

(total amount/constant rate) = (time).

remember that it is a constant rate, so in total, when he gets back down the mountain, it will take the same amount of time as when he goes up. the total amount for the round trip is: 5 + 5 = 10.

~

(b) 240

step-by-step explanation:

first, we use the pythagorean formula using the given side lengths to find x. then we use x to find the side lengths. then we add the side lengths to find the perimeter.

pythagorean theorem formula:

a^2 + b^2 = c^2

(x - 20)^2 + (x - 40)^2 = x^2

x^2 - 40x + 400 + x^2 - 80x + 1600 = x^2

x^2 -120x + 2000 = 0

(x - 100)(x - 20) = 0

x - 100 = 0 or x - 20 = 0

x = 100 or x = 20

we see that the solution x = 20 must be discarded because it will give a negative side length and a side length of 0. the only valid solution is x = 100.

perimeter = sum of the lengths of the sides

perimeter = x + x - 20 + x - 40

perimeter = 3x - 60

replace x with 100.

perimeter = 3(100) - 60

perimeter = 300 - 60

perimeter = 240

what is

step-by-step explanation:

ratios represent a pattern or relationship that can be scaled.

equivalent ratios are pairs of ratios that represent the same relationship between quantities, and can be generated by multiplying or dividing both quantities in the ratio by the same factor.

in this lesson, students will use double number lines and ratio reasoning to solve ratio problems. by partitioning double number lines into proportionate intervals, students will build an understanding of equivalent ratio relationships as those that are generated by multiplying or dividing two related quantities by the same number. this lesson will build upon students' prior understanding of multiplicative comparisons, number line reasoning and equivalent fractions (while highlighting the difference between equivalent fractions and equivalent ratios); and will build toward future work in this unit on building ratio tables, and work in future grades in proportionality, scaling, and expressing and graphing linear functions.

step-by-step explanation:

ratios represent a pattern or relationship that can be scaled.

equivalent ratios are pairs of ratios that represent the same relationship between quantities, and can be generated by multiplying or dividing both quantities in the ratio by the same factor.

in this lesson, students will use double number lines and ratio reasoning to solve ratio problems. by partitioning double number lines into proportionate intervals, students will build an understanding of equivalent ratio relationships as those that are generated by multiplying or dividing two related quantities by the same number. this lesson will build upon students' prior understanding of multiplicative comparisons, number line reasoning and equivalent fractions (while highlighting the difference between equivalent fractions and equivalent ratios); and will build toward future work in this unit on building ratio tables, and work in future grades in proportionality, scaling, and expressing and graphing linear functions.