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3 years ago

What is the correct answer for this problem?Explain

Mathematics

2 answers:

bearhunter [10]3 years ago

8 0

To solve this you would make a triangle, Assuming you had a piece of paper.

It would look like the picture included (where ever that is) sorry for the horrible drawing with mouse.

You can then you Pythagorean Theorem to solve this

$a^{2} + b^2 =c^2$

width will be (a)

put in your variables

$a^2 + 72^2 = 90^2$
$a^2 + 5184 = 8100$

Subtract 5184 over and get
$a^2 = 2916$

square root both sides and get
a=54

the field is 54 meters wide

melisa1 [442]3 years ago

5 0

It’s 54 meters wide if im not wrong.. im pretty sure it’s right tho.

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. 4.5w = 5.1W - 30 please help me with this question pls show the work

inn [45]

Answer: w = 50

Step-by-step explanation:

4.5w - 5.1w = -30

-0.6w = -30

0.6w = 30

W = 50

5 0

3 years ago

I need help............

Alexxandr [17]

Problem 35

The bar graph is shown below

You simply draw various rectangles such that the heights represent the frequency of each animal type.

Eg: there are 20 elephants, so the "elephant" bar is 20 units tall.

You can make the bar graph by hand, or use spreadsheet software. I recommend going with software (if you can).

=======================================================

Problem 36

1 book = 20 mm thickness

5 books = 5*20 = 100 mm thickness

1 paper = 0.016 mm thickness

5 papers = 5*0.016 = 0.08 mm thickness

total thickness = 100+0.08 = 100.08 mm

<h3>Answer: 100.08 mm</h3>

=======================================================

Problem 37

121/11 = 11

121 = 11*11

If we say 11+11+11+...+11, and have 11 copies of these values added, then we get to a sum of 121

This is probably the easiest way to get the answer assuming repeated values are allowed.

You can stop here if your teacher allows you to use repeated values. If not, then move onto the next section.

-----------

If your teacher requires you to add 11 <u>different</u> numbers, then you can follow this procedure

Write out eleven copies of "11" in a sequence
Subtract 2 from the first "11" (to get 9) and add it to the last copy of "11" (to get 13)
Subtract 4 from the second "11" (to get 7) and add it to the second to last copy of "11" (to get 15)
Subtract 6 from the third "11" (to get 5) and add it to the third to last copy of "11" (to get 17)
Subtract 8 from the fourth "11" (to get 3) and add it to the fourth to last copy of "11" (to get 19)
Finally, subtract 10 from the fifth "11" (to get 1) and add it to the fifth to last copy of "11" (to get 21)

After carefully following those steps, you'll get this sequence (in the exact order shown):

{9, 7, 5, 3, 1, 11, 21, 19, 17, 15, 13}

There are three properties to notice of this sequence

It's composed of two decreasing arithmetic sequences {9, 7, 5, 3, 1} and {21, 19, 17, 15, 13}
The 11 in the middle hasn't been changed from the original sequence of nothing but "11"s.
You should find that the terms of this new sequence add to 121.

That sequence we got sorts to {1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21}

and we can say 1+3+5+7+9+11+13+15+17+19+21 = 121

<h3>Answer: 1+3+5+7+9+11+13+15+17+19+21 = 121</h3>