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coldgirl [10]
3 years ago
8

A rectangular swimming pool is 19 m wide by 35 m long. what is the length of the diagonal of the pool round to the nearest 10th

Mathematics
2 answers:
pentagon [3]3 years ago
7 0

Answer:

the answer is 39.8m

Step-by-step explanation:

You divide the pool with the diagonal and you obtain a right angled triangle with its base as 35m and its height as 19m. then using pythagoras theorem, you find the length of the hypotenuse and this is the length of the diagonal of the pool.

Rina8888 [55]3 years ago
5 0

Answer:

39.8

Step-by-step explanation:

for this question, use the Pythagorean theorem which is:

a^2 + b^2 = c^3

a = 35

b = 19

c = the answer

lets solve it:

35^2 +19^2 = c^2

solve the exponents first

1225 + 361 = c^2

1586 = c^2

square root both to find c

(the answer is rounded)

39.824 = c

rounded to nearest 10th:

c = 39.8

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Simplify 2 a + 4 - 7 a + a
AlexFokin [52]
2a + 4 - 7a + a

2a -7a +a +4

-5a + a +4

-4a + 4

-4a + 4 is your answer

hope this helps
8 0
4 years ago
A ball is dropped straight down from a height of 200 feet. After 1 second? The hall is 184 feet above the ground. After 2 second
mote1985 [20]

The motion of the ball dropped from height is a free fall motion due to

gravitational acceleration.

<h3>Correct response;</h3>
  • The equation that models the height is; <u>y = -16·x²</u>

<h3>Method for arriving at the above equation;</h3><h3>Given values;</h3>

\begin{tabular}{|c|c|}Height (feet)&Time (s)\\200&0\\184&1\\136&2\end{array}\right]

<h3>Required:</h3>

To select the equation that models the height, <em>y</em>, of the ball <em>x</em> seconds after

its dropped;

<h3>Solution:</h3>

From the above table, we have that the first difference is not a constant

The second difference is = 48 - 16 = 32

Taking the second difference as a constant, we have the following

quadratic sequence;

y = a·x²  + b·x + c

Where;

x = The time in seconds

y = The height after <em>x</em> seconds

  • At x = 0, we have;

200 = a·0² + b × 0 + c

Therefore;

c = 0

  • At<em> x</em> = 1, we have;

184 = a × 1² + b × 1 + 200

184 = a + b + 200

a + b = -16

a = -16 - b

  • At <em>x</em> = 2, we have;

136 = a × 2² + b × 2 + 200

136 = 4·a + 2·b + 200

-64 = 4·a + 2×b

Therefore;

-64 = 4 × (-16 - b) + 2×b

-64 = -64 - 4·b + 2×b

b = 0

a + b = -16

Therefore;

a + 0 = -16

a = -16

The equation that models the height is; y = <u>-16·x²</u>

Learn more about quadratic function here;

brainly.com/question/2293136

4 0
2 years ago
Choose the equation that satisfies the data in the table.
White raven [17]

Answer:

C. y = 1½x + 9

Step-by-step explanation:

The quickest way is to know what intercepts are shown here and you are done, but if you want to, then all you have to do is plug in each number given to you into each solution to confirm the authentication.

I am joyous to assist you anytime.

3 0
3 years ago
What will the coordinate K(3, - 3) be after the translation 8 units to the left and 7 units up.
natita [175]

Answer:

(-5, 4)

Step-by-step explanation:

To find the new coordinate, subtract 8 from the x value and add 7 to the y value:

3 - 8 = -5

-5 will be the new x coordinate

-3 + 7 = 4

4 will be the new y coordinate

So, the coordinate K will be (-5, 4) after the translation

8 0
3 years ago
Read 2 more answers
Expand:<br><img src="https://tex.z-dn.net/?f=f%28z%29%20%3D%20%20%5Cfrac%7B1%7D%7Bz%28z%20-%202%29%7D%20" id="TexFormula1" title
Monica [59]

Expand f(z) into partial fractions:

\dfrac1{z(z-2)} = \dfrac12 \left(\dfrac1{z-2} - \dfrac1z\right)

Recall that for |z| < 1, we have the power series

\displaystyle \frac1{1-z} = \sum_{n=0}^\infty z^n

Then for |z| > 2, or |1/(z/2)| = |2/z| < 1, we have

\displaystyle \frac1{z-2} = \frac1z \frac1{1 - \frac2z} = \frac1z \sum_{n=0}^\infty \left(\frac 2z\right)^n = \sum_{n=0}^\infty \frac{2^n}{z^{n+1}}

So the series expansion of f(z) for |z| > 2 is

\displaystyle f(z) = \frac12 \left(\sum_{n=0}^\infty \frac{2^n}{z^{n+1}} - \frac1z\right)

\displaystyle f(z) = \frac12 \sum_{n=1}^\infty \frac{2^n}{z^{n+1}}

\displaystyle f(z) = \sum_{n=1}^\infty \frac{2^{n-1}}{z^{n+1}}

\displaystyle \boxed{f(z) = \frac14 \sum_{n=2}^\infty \frac{2^n}{z^n} = \frac1{z^2} + \frac2{z^3} + \frac4{z^4} + \cdots}

6 0
2 years ago
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