All categories

3 years ago

How do I solve this?

Mathematics

2 answers:

Aleksandr-060686 [28]3 years ago

8 0

First you have to set up your equation. 5x+80 and x+15.

Novosadov [1.4K]3 years ago

7 0

Answer:frog 1

Step-by-step explanation:

Frog 1 jumps 80cm in 5 sec

Frog 2 jumps 15cm in 1 sec

The rule says each frog must cross the finish line which is 5m away to win the race

If Frog 1 jumps 80cm in 5 sec, then frog 1 jumps 80/5 = 16cm in 1 sec

Converting from cm to m

1 m = 100cm

For frog 1, 16cm = 15/100 = 0.16m

For frog 2, 16cm = 15/100 = 0.15m

Frog 1 takes 5/0.16 = 31.25 sec to run the finish line

Frog 2 takes 5/0.15 = 33.33 sec to run the finish line

So frog 1 will win the race

You might be interested in

If you had 11 dollars and spent 7 what fraction of the money did you spend

tatyana61 [14]

Answer:

7/11

Step-by-step explanation:

3 0

2 years ago

Given the diagram below what is the length of segment EF

frosja888 [35]

Answer:

C is the closest.,

Step-by-step explanation:

I am assuming that the 2 trapezoids CBEF and EFDA are similar (they are similar if the 3 horizontal lines are parallel).

Corresponding sides are in the same ratio , so:

3.4 / EF = EF / 7.6

EF^2 = 3.4 * 7.6 = 25.84

EF = 5.08

6 0

3 years ago

Read 2 more answers

assume the graph of a function of the form y=asin(k(x+b)) is given below. which of the following are possible values for a, k, a

WITCHER [35]

Answer:

a=4, k=2, and b= -pi/4

Step-by-step explanation:

5 0

3 years ago

Q.5(b) The population {(P) in millions} of a country is estimated by the function, P=125e0.035t, t = time measured in years sinc

nekit [7.7K]

Answer:

Hello,

Step-by-step explanation:

Q.5(b) The population {(P) in millions} of a country is estimated by the function, P=125e0.035t, t = time measured in years since 1990. (a) what is the population expected to equal in year 2000 (b) determine the expression for the instantaneous rate of change in the population (c) what is the instantaneous rate of change in the population expected to equal in year 2000.

$P(t)=125*e^{0.035*t}\\a)\\P(2000)=125*e^{0.035*(2000-1990)}=177.38...\\\\b)\\P'(t)=125*0.035*e^{0.035*t}\\\\c)\\\\P'(2000)=125*0.035*e^{0.035*(2000-1990)}=6.2084...$

5 0

3 years ago

Need helpppppp!!!!!!

uysha [10]

These are 45-45-90 triangles, whose hypotenuses are $\sqrt{2}$ times their legs. Given a side of length 1, for example, the diagonal length is $\sqrt{2}\approx 1.4142.$ The first few ordered pairs $(x,y)$ in the table would be $(1,1.41),(2,2.83),(3,4.24),(4,5.66),...$

5 0

3 years ago