1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
elena-s [515]
3 years ago
7

Which runs faster: a cheetah that can run 60 feet in 6 seconds or a gazelle that can run 100 feet in 10 seconds?

Mathematics
2 answers:
bulgar [2K]3 years ago
5 0

Answer:

They run at the same speed

Step-by-step explanation:

First, you have to find a way seeing how fast both animals run in the same amount of time. You can find out how fast they both run in one second.

60 feet in 6 seconds is how many feet in one second? Well, you can do 60/6 (60 divided by 6) and you will see that in one second, the cheetah can run 10 feet

100 feet in 10 seconds is how many feet in one second? Same thing, you can do 100/10 (100 divided by 10) and you will see that the gazelle can run 10 feet in one second.

Finally, you compare the two answers from the two animals. For the cheetah, we learned that it can run 10 feet in one second, and for the gazelle, we learned that it can run 10 feet in one second. So, we know that they run at the same speed.

Thank you! Please let me know if this was helpful!

Vanyuwa [196]3 years ago
4 0

Answer:

Cheeta's speed = 10 feet per second

Gazelle's speed = 10 feet per second

Hence, They run at the same speed.

Step-by-step explanation:

Determining the Cheeta's speed:

Covered distance = 60 feet

Time taken = 6 seconds

Speed = Covered distance / Time taken

           = 60 / 6

           = 10 feet per second

Therefore, Cheeta's speed is 10 feet per second.

Determining the Gazelle's speed:

Covered distance = 100 feet

Time taken = 10 seconds

Speed = Covered distance / Time taken

           = 100 / 10

           = 10 feet per second

Therefore, Gazelle's speed is 10 feet per second.

Conclusion:

Cheeta's speed = 10 feet per second

Gazelle's speed = 10 feet per second

Hence, They run at the same speed.

You might be interested in
I need help with these 2 problems. Please and thank you :)
dsp73

Answer:

2,8

Step-by-step explanation:

the upper arrow goes down just like the other one so which every number you see its on go to that number and look up im at explaining but that is the answer for the 1st page

7 0
2 years ago
A football team loses 3 yards in 3 consecutive plays. What’s the total yardage gained?
Anni [7]
-3×3=-9 so gain is 0
4 0
3 years ago
Which of the following is a solution to 2cos^2x– 2 = 0 ?
Sphinxa [80]

Answer:

A and C

Step-by-step explanation:

Given

2cos²x - 2 = 0 ( add 2 to both sides )

2cos²x = 2 ( divide both sides by 2 )

cos²x = 1 ( take the square root of both sides )

cos x = ± \sqrt{1} = ± 1

cos x = 1 ⇒ x = cos^{-1}(1) = 0°

cos x  = - 1 ⇒ x = cos^{-1}(- 1) = 180°

8 0
3 years ago
Read 2 more answers
5x+6=61 Show Step By Step
Annette [7]
5x =61-6
5x= 55
X = 55/5
X = 11
4 0
3 years ago
Read 2 more answers
Simplify: 4x + 6(3x - 2) A) 10x - 12 B) 18x - 12 C) 22x - 12 D) 30x2 - 20x
makvit [3.9K]

the answer is defenitly c

that is the answer

4 0
3 years ago
Read 2 more answers
Other questions:
  • 1/2x+4=2/3x solve for x
    14·1 answer
  • How to solve this |9x-12|<-4
    6·1 answer
  • Alexandria has a collection of 34 dolls.A toy store has 15 times as many dolls as Alexandria.How many dolls are in the store
    11·2 answers
  • Find the mean 2.6, 0.0, 0.8 2.1 1.5
    10·1 answer
  • .. Tanner collected 360 cans and bottles while fundraising for his baseball team. This was
    10·1 answer
  • What percent is 23 of 92?
    6·2 answers
  • Whats 1+8............ ... ​
    11·1 answer
  • Help me plsss it’s due in five minuted
    13·1 answer
  • Simplily x+2x+7-5x+8<br><br>​
    10·2 answers
  • What is an equation of the line that passes through the point (−3,−1) and is perpendicular to the line x-2y=6?
    14·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!