Jerome burns 4 cal/min walking and 10 cal/min running. he walks between 10 and 20 min each day and runs between 30 and 45 min each day. he never spends more than an hour running and walking together. how much time should he spend on each activity to maximize his calorie burning?

20 minutes should spend on walking and 30 minutes on running.

Step-by-step explanation:

Let x represents the minutes spent on walking and y represents the minutes spent on running,

∵ He burns 4 cal/min walking and 10 cal/min running,

So, the total calories burnt,

Z = 4x + 10y

Which has to be maximised.

Now, he walks between 10 and 20 min each day,

i.e. 10 < x < 20,

Also, he runs between 30 and 45 min each day,

i.e. 30 <  y < 45,

By graphing 10 < x < 20 and 30 <  y < 45,

We obtained the vertices of feasible region,

(10, 45), (20, 45), (10, 30) and (20, 30)

At (10, 45),

Z = 4(10) + 10(45) = 40 + 450 = 490,

At (20, 45),

Z = 4(20) + 10(45) = 80 + 450 = 530,

At (10, 30),

Z = 4(10) + 10(30) = 40 + 300 = 340,

At (20, 30)

Z = 4(20) + 10(30) = 80 + 300 = 380

Hence, maximum calories is 530 when 20 minutes spent on walking and 30 minutes spent on running.

the one on the bottom left

step-by-step explanation:

all of the other ones have a specific pattern

option a, c,d are correct.

step-by-step explanation:

from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have

tz=sz(given)

bz=zb(common)

therefore, by rhs rule,δtzb ≅δszb

by cpctc, sz≅tz

also, from δctz and δasz,

tz=sz(given)

∠tcz=∠saz(90°)

by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz

also,from δasz and δzsb,

zs=sz(common)

∠zbs=∠saz=90°

by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb

hence, option a, c,d are correct.