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3 years ago

8

The middle school is 2.72 Kilometers from Kim's house and 1.54 kilometers from Terri's house. How much farther in meter does Kim

live from the middle school than Terri

1 answer:

TiliK225 [7]3 years ago

4 0

Answer: 1,180 meters.

Step-by-step explanation:

Hi, to answer this question we have to subtract Terri's distance from the school (1.54) to Kim's distance from the school (2.72)

Mathematically speaking:

2.72 km - 1.54km = 1.18 km

Since the answer must be expressed in meters.

1000 meters = 1 kilometer

1.18 km x 1000 m/km = 1,180 meters.

Feel free to ask for more if needed or if you did not understand something.

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A regular hexagon has sides of 3 feet. What is the area of the hexagon?

frosja888 [35]

Answer:

$13.5\sqrt{3}\:\text{ft}^{2}$

Step-by-step explanation:

Given: The side of a regular hexagon is 3 feet.

To find: Area of the hexagon

Solution:

It is given that the side of a regular hexagon is 3 feet.

We know that the area of a regular hexagon whose side is a units is $\frac{3\sqrt{3} }{2} a^{2}$

Here, the side is 3 feet

So, area of the regular hexagon

$=\frac{3\sqrt{3} }{2} \times3^{2}$

$=\frac{3\sqrt{3} }{2} \times9$

$=\frac{27\sqrt{3} }{2}$

$=13.5\sqrt{3}\:\text{ft}^{2}$

Hence, area of the regular hexagon is $13.5\sqrt{3}\:\text{ft}^{2}$

6 0

3 years ago

(45pts) An elevator holds at most 1,400 pounds. A team of movers and their cart have a combined weight of 1,200 pounds. If each

emmainna [20.7K]

Let $x$ be the number of boxes the team brings with them. Their weight combined with the boxes can't exceed the capacity of 1400. Assuming the elevator runs fine at that exact weight, you want to find the number of boxes, each of which contributes 40 pounds. This is given by the equation

$1400=1200+40x$

Solving for $x$ , you have

$1400=1200-40x$
$200=40x$
$x=5$

So the team can bring *at most* 5 boxes at a time.

4 0

3 years ago

HELPP!!what is the answer to this picture above^

nasty-shy [4]

Answer:

$\fbox{ \fbox { \sf{Distance = \sf{15 \: units}}}}$

${ \fbox { \fbox { \sf{Midpoint = { \sf{ \: (12 \: , \: 10.5)}}}}}}$

Step-by-step explanation:

$\star{ \: \sf { \: Let \: the \: points \: be \: A \: and \: B}}$

$\star { \sf{Let \: A(6 \:, 6) \: be \: (x1 ,\: y1) \: and \: B(18 ,\: 15) \: be \: (x2 \:, y2)}}$

$\underline{ \underline{ \tt{Finding \: the \: distance}}}$

$\boxed{ \sf{Distance = \sqrt{ {(x2 - x1)}^{2} + {(y2 - y1)}^{2} } }}$

$\mapsto{ \sf{Distance = \sqrt{ {(18 - 6)}^{2} + {(15 - 6)}^{2} } }}$

$\mapsto{ \sf{Distance = \sqrt{ {(12)}^{2} + {(9)}^{2} } }}$

$\mapsto{ \sf{Distance = \sqrt{144 + 81}}}$

$\mapsto{ \sf{Distance = \sqrt{225} }}$

$\mapsto{ \sf{Distance = \sqrt{ {15}^{2} } }}$

$\mapsto{ \boxed{\sf{Distance = 15 \: units}}}$

$\underline{ \underline {\tt{Finding \: the \: Midpoint}}}$

$\boxed{ \sf{Midpoint = ( \frac{x1 + x2}{2} \: , \: \frac{y1 + y2}{2} )}}$

$\mapsto{ \sf{Midpoint = ( \frac{6 + 18}{2} \: , \: \frac{6 + 15}{2} )}}$

$\mapsto{ \sf{Midpoint = ( \frac{24}{2} \: , \: \frac{21}{2} )}}$

$\mapsto{ \boxed{ \sf{Midpoint = (12 \: , \: 10.5)}}}$

Hope I helped!

Best regards! :D

~ $\sf{TheAnimeGirl}$

3 0

3 years ago

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KATRIN_1 [288]

Answer:

Step-by-step explanation:

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Answer:

a

Step-by-step explanation:

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