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hichkok12 [17]
2 years ago
5

I need help pls! i’m kinda s t u p i d! lol

Mathematics
1 answer:
LiRa [457]2 years ago
3 0
The answer is b or no
i hope this helped
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Step-by-step explanation:

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3 years ago
Can someone help with this math question please
Alex73 [517]
Question 9

For this question, you work out the areas of the different shapes - the two triangles and the large rectangle.
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To get the area of the rectangle, you must use a ruler to measure out the:
 height of the shape and the base of the shape

(try to be as accurate as possible when measuring)

Then to get the area of the rectangle, multiply the Height by the Base. 
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For the triangles, you measure the out the base and the height .
Then you use the formula for the area of a triangle, which is:

Height x Base x \frac{1}{2}

So, you measure out the base and height of one triangle, then use the formula to work out the area,    then you do the same for the second triangle.
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Finally, you add up all of the areas:

Total area = area of rectangle + area of first triangle +area of second triangle

then you choose the option that is closest to your answer <span>(remember to be accurate with the ruler!)
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Question 10<span>

</span>Lets say we wanted to know how much 10 bottles of water cost

To do this, we would multiply the number of water bottles we want (which is 10)  by how much one water bottle costs.<span>

</span>Lets say one water bottle costs £1.  
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Now lets apply this knowledge to the question:

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Question 11

For this one, we are told that the company charges $40 for hooking a vehicle.
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$1.70m   means   $1.70   times the number of miles, and we are told that they charge 1.70 for every mile.  



8 0
3 years ago
What is the GCF of g^3 an g^15
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Answer:

The GCF of both is g^3

Step-by-step explanation:

Here, we are asked to give the greatest common factor of g^3 and g^15

In simpler terms we want to find that biggest term that could divide both values.

Mathematically, since g^3 is itself a factor of g^15, then we can conclude that the GCF of both is g^3

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2 years ago
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