All categories

3 years ago

A training field is formed by joining a rectangular and two semicircles. The rectangle is 95 m long and 74m wide. Find the area

Mathematics

1 answer:

Sergeeva-Olga [200]3 years ago

6 0

Answer: 7030m2
If you are finding the area you do L x W (Length x width) so 95m x 74m = 7020m2

You might be interested in

Drag the tiles to the boxes to form correct pairs. Not all tiles will be used. Match the resulting values to the corresponding l

belka [17]

The correct solution to the limits of x in the tiles can be seen below.

$\mathbf{ \lim_{x \to 9^+} (\dfrac{|x-9|}{-x^2-34+387}) }$ $\mathbf{ = -\dfrac{1}{52} }$

$\mathbf{ \lim_{x \to 8^-} (\dfrac{8-x}{|-x^2-63x+568|}) }$ $\mathbf{=\dfrac{1}{79} }$

$\mathbf{ \lim_{x \to 7^+} (\dfrac{|-x^2-17x+168| }{x-7}) }$ = -31

$\mathbf{ \lim_{x \to 6^-} (\dfrac{|x-6| }{-x^2-86x+552}) }$ $\mathbf{ =\dfrac{1}{98}}$

<h3>What are the corresponding limits of x?</h3>

The limits of x approaching a given number of a quadratic equation can be determined by knowing the value of x at that given number and substituting the value of x into the quadratic equation.

From the given diagram, we have:

$\mathbf{ \lim_{x \to 9^+} (\dfrac{|x-9|}{-x^2-34+387}) }$

So, x - 9 is positive when x → 9⁺. Therefore, |x -9) = x - 9

$\mathbf{ \lim_{x \to 9^+} (\dfrac{x-9}{-x^2-34+387}) }$

Simplifying the quadratic equation, we have:

$\mathbf{ \lim_{x \to 9^+} (-\dfrac{1}{x+43}) }$

Replacing the value of x = 9

$\mathbf{ = (-\dfrac{1}{9+43}) }$

$\mathbf{ = -\dfrac{1}{52} }$

$\mathbf{ \lim_{x \to 8^-} (\dfrac{8-x}{|-x^2-63x+568|}) }$

-x²-63x+568 is positive when x → 8⁻.

Thus |-x²-63x+568| = -x²-63x+568

$\mathbf{ \lim_{x \to 8^-} (\dfrac{1}{x+71}) }$

$\mathbf{=\dfrac{1}{8+71} }$

$\mathbf{=\dfrac{1}{79} }$

$\mathbf{ \lim_{x \to 7^+} (\dfrac{|-x^2-17x+168| }{x-7}) }$

x -7 is positive, therefore |x-7| = x - 7

$\mathbf{ \lim_{x \to 7^+} (\dfrac{-x^2-17x+168 }{x-7}) }$

$\mathbf{ \lim_{x \to 7^+} (-x-24)}$

$\mathbf{ \lim_{x \to 7^+} (-7-24)}$

= -31

$\mathbf{ \lim_{x \to 6^-} (\dfrac{|x-6| }{-x^2-86x+552}) }$

x-6 is negative when x → 6⁻. Therefore, |x-6| = -x + 6

$\mathbf{ \lim_{x \to 6^-} (\dfrac{-x+6 }{-x^2-86x+552}) }$

$\mathbf{ \lim_{x \to 6^-} (\dfrac{1}{x+92}) }$

$\mathbf{ \lim_{x \to 6^-} (\dfrac{1}{6+92}) }$

$\mathbf{ =\dfrac{1}{98}}$

Learn more about calculating the limits of x here:

brainly.com/question/1444047

#SPJ1

7 0

2 years ago

Hey guys, can yall help me out? I will give brain to whomever answers correctly

Svetlanka [38]

Answer:

A is the correct answer.

5 0

3 years ago

Read 2 more answers

In the math equation W=F*D what relationship is required for work to be done

DochEvi [55]

Kjakakakksskkajsjsjsksakak ididisisisisisisi dodososososissi

5 0

3 years ago

Please help!! i’ll mark as brainlest, 20 points

Lisa [10]

It’s either b or c. One of those two is the answer for sure

3 0

3 years ago

Whats three to the forth power

aalyn [17]

Answer:

1514

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers