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dezoksy [38]
2 years ago
14

Eighteen friends are going to the beach. If each car holds 5 people, how many cars need to be driven to the beach in order for a

ll of the friends to have a ride?
Mathematics
2 answers:
ANTONII [103]2 years ago
8 0

Answer:

4 cars

Step-by-step explanation:

18 people(friends)

5x4=20

Which means there is going to be 2 extra seats. Four cars with five seats 3x5=15. This means you would need 3 more seats so you need a whole other car. Which means you have 4 cars.

Hope this helps.

antiseptic1488 [7]2 years ago
6 0

Answer:

4

Step-by-step explanation:

4 because each holds 5 people. If there were 3 cars, only 15 people would be able to go. There will be two extra spaces in the car.

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Find the height of the prism if the length is 9 inches, the width is 5 inches, and the volume of the prism is 315 cubic inches.
trasher [3.6K]

Answer:

Height of prism = 7 inches

Step-by-step explanation:

Let h be the height of prism.

Hence,

h =  \frac{vol \: of \: prism}{l \times w}  \\  \\  =  \frac{315}{9 \times 5}  \\  \\  =  \frac{315}{45}  \\   \\ h= 7 \: inches

5 0
3 years ago
HELPPPPPP PLEASEEEEee 34 + 2 ⋅ 5 = ____. (Input whole numbers only.) (1 point)
yaroslaw [1]
The answer is 44
10+34
5 0
3 years ago
There are 14 teams in a basketball league.
Talja [164]

Answer:

Step-by-step explanation:

Given that :

Number of teams = 14

Each team plays every other team twice ;

Using the combination formula :

nCr = n! ÷ (n-r)! r!

14C2 = 14! ÷ (14 - 2)! 2!

14C2 = 14! ÷ (12)! 2!

14C2 = (14 * 13) ÷ 2 * 1

14C2 = 182 / 2

14C2 = 91

Hence, since they are going to be playing each other twice :

2(14C2)

2 * 91 = 182games

7 0
3 years ago
How many solutions does this linear system have?
vagabundo [1.1K]
It is no solution, bc the answer is (0,0)
7 0
3 years ago
Read 2 more answers
Please help logarithms!
nlexa [21]

Given:

\log_34\approx 1.262

\log_37\approx 1.771

To find:

The value of \log_3\left(\dfrac{4}{49}\right).

Solution:

We have,

\log_34\approx 1.262

\log_37\approx 1.771

Using properties of log, we get

\log_3\left(\dfrac{4}{49}\right)=\log_34-\log_349      \left[\because \log_a\dfrac{m}{n}=\log_am-\log_an\right]

\log_3\left(\dfrac{4}{49}\right)=\log_34-\log_37^2      

\log_3\left(\dfrac{4}{49}\right)=\log_34-2\log_37          [\log x^n=n\log x]

Substitute \log_34\approx 1.262 and \log_37\approx 1.771.

\log_3\left(\dfrac{4}{49}\right)=1.262-2(1.771)

\log_3\left(\dfrac{4}{49}\right)=1.262-3.542

\log_3\left(\dfrac{4}{49}\right)=-2.28

Therefore, the value of \log_3\left(\dfrac{4}{49}\right) is -2.28.

5 0
3 years ago
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