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3 years ago

How many minutes will it take the school bus to complete the 52-km trip back to the school at an average speed of 1.0 km/min?

Mathematics

1 answer:

Virty [35]3 years ago

6 0

Answer:ddgftuewdsvgasvdgvfchxcbncbhua

Step-by-step explanation: ME NEED POINTZ

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Halp plz i dont know the answwer

lisabon 2012 [21]

Answer:

-2.1 is the answer im pretty sure

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Solve for u.<br> u – 23 = 5<br> U =

VMariaS [17]

Answer:

U= 28

Step by step solution

add 23 to both sides

U-23+23=5+23

hence

U=28

3 0

3 years ago

73 m is equal to ____ dm. (Only input whole number answer.)<br><br> Numerical Answers please!

olasank [31]

Answer:

73 m is equal to 730 dm

Step-by-step explanation:

We Need to convert 73 m into dm

We know that 1 meter is equal to 10 decimeter

We are given 73 m. Multiply it with 10 and we will get value in decimeter

73*10

= 730 decimeter

So, 73 m is equal to 730 dm

7 0

3 years ago

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The marks scored by a group of students in English examination are as follows: 30,32, 33, 31, 32, 33, 34, 35, 33, 34 . Find the

Alecsey [184]

Answer:

Answer is 32

Step-by-step explanation:

Mode = no of stu +1/2

Mode =12+1/2

Mode =13/28=6.5

Mode =6th mark +7th mark /2

Mode=32+32/2

Mode=64/2

Mode=32

Hope it helps

Peace

Please mark me as brainliest

8 0

3 years ago

In the graph below, Point A represents Owen's house, Point B represents David's house and Point C represents the school. Who liv

ycow [4]

Answer:

• David

• 4 miles

Explanation:

In the graph:

The given locations are:

• Owen's House, A(11,3)

• David's House, B(15,13)

• School, C(3,18)

We determine both Owen's and David's distance from the school using the distance formula.

$Distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Owen's distance from school (AC)

$\begin{gathered} AC=\sqrt[]{(3-11)^2+(18-3)^2} \\ =\sqrt[]{(-8)^2+(15)^2} \\ =\sqrt[]{64+225} \\ =\sqrt[]{289} \\ AC=17\text{ miles} \end{gathered}$

David's distance from school (BC)

$\begin{gathered} BC=\sqrt[]{(3-15)^2+(18-13)^2} \\ =\sqrt[]{(-12)^2+(5)^2} \\ =\sqrt[]{144+25} \\ =\sqrt[]{169} \\ BC=13\text{ miles} \end{gathered}$

We see from the calculations that David lives closer to the school, and by 4 miles.

The graph below is attached for further understanding:

5 0

1 year ago