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2 years ago

Please help me in this Pythagoras theorem only part (d)

Mathematics

1 answer:

nikklg [1K]2 years ago

6 0

Answer:

d = 19.9m

Step-by-step explanation:

c² - a² = b²

24.7² - 14.5² = d²

610.09 - 210.25 = d

610.09 - 210.25 = 399.84

√399.84 = 19.9

d = 19.9m

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Mandarinka [93]

Answer:

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Step-by-step explanation:

8 -(x+5) = 5x - 3

-(x+5) ---> -1 (x +5) = -x - 5

8 - x - 5 = 5x -3 8 - 5 = 3

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2 years ago

2/3x+15=17 please and thank you

frozen [14]

Answer:

x=3

Step-by-step explanation:

2/3x+15=17

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2 years ago

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VikaD [51]

400,000,014,104

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2 years ago

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11. Which matrix represents the system of equations below?

dimulka [17.4K]

Answer:

$\left[\begin{array}{cccc}-12&-13&13&|15\\7&-10&-3&|11\\7&14&5&\:\:\:|-5\end{array}\right]$

Step-by-step explanation:

The system of equations is;

-12x-13y +13z =15

7x-10y-3z = 11

7x+14y +5z = -5

The coefficient matrix is $\left[\begin{array}{ccc}-12&-13&13\\7&-10&-3\\7&14&5\end{array}\right]$

The constant matrix is $\left[\begin{array}{c}15\\11\\-5\end{array}\right]$

The augmented matrix is $\left[\begin{array}{cccc}-12&-13&13&|15\\7&-10&-3&|11\\7&14&5&\:\:\:|-5\end{array}\right]$

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3 years ago

Solve using long division <br> Please

madreJ [45]

1. $Solution,\frac{2x^3+4x^2-5}{x+3}:\quad 2x^2-2x+6-\frac{23}{x+3}$

$Steps:$

$\mathrm{Divide}\:\frac{2x^3+4x^2-5}{x+3}:\quad \frac{2x^3+4x^2-5}{x+3}=2x^2+\frac{-2x^2-5}{x+3}$

$\mathrm{Divide}\:\frac{-2x^2-5}{x+3}:\quad \frac{-2x^2-5}{x+3}=-2x+\frac{6x-5}{x+3}$

$\mathrm{Divide}\:\frac{6x-5}{x+3}:\quad \frac{6x-5}{x+3}=6+\frac{-23}{x+3}$

$\mathrm{Simplify}, =2x^2-2x+6-\frac{23}{x+3}$

$\mathrm{The\:Correct\:Answer\:is\:2x^2-2x+6-\frac{23}{x+3}}$

2. $Solution, \frac{4x^3-2x^2-3}{2x^2-1}:\quad 2x-1+\frac{2x-4}{2x^2-1}$

$Steps:$

$\mathrm{Divide}\:\frac{4x^3-2x^2-3}{2x^2-1}:\quad \frac{4x^3-2x^2-3}{2x^2-1}=2x+\frac{-2x^2+2x-3}{2x^2-1}$

$\mathrm{Divide}\:\frac{-2x^2+2x-3}{2x^2-1}:\quad \frac{-2x^2+2x-3}{2x^2-1}=-1+\frac{2x-4}{2x^2-1}$

$\mathrm{The\:Correct\:Answer\:is\:2x-1+\frac{2x-4}{2x^2-1}}$

$\mathrm{Hope\:This\:Helps!!!}$

$\mathrm{-Austint1414}$

4 0

2 years ago

Please help me in this Pythagoras theorem only part (d)​

Please help me in this Pythagoras theorem only part (d)