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3 years ago

15 - (7 - 6)4 * 4 + 9

Mathematics

1 answer:

Naddik [55]3 years ago

3 0

Answer:

8 is your answer. good luck on your test

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Death Valley is the location of the lowest land elevation in the USA which is 282 feet below sea level.The location with the hig

il63 [147K]

Answer:

20,038

Step-by-step explanation:

20,320 - 282

7 0

3 years ago

Find the X COORDINATE OF THE intersection of 9x - 6y = 12 and 3x + 2y =8

yaroslaw [1]

Answer: $x=2$

Step-by-step explanation:

Given a System of linear equations, the solution of the system is the point $(x,y)$ in which the lines intersect, where "x" is the x-coordinate and "y" is the y-coordinate.

Having the following System of equations:

$\left \{ {{9x - 6y = 12} \atop {3x + 2y =8}} \right.$

You can use the Elimination method to find the value of "x":

-Multiply the second equation by 3.

- Add the equations.

- Solve for "x".

Then, you get:

$\left \{ {{9x - 6y = 12} \atop {9x +6y =24}} \right.\\.......................\\18x=36\\\\x=\frac{36}{18}\\\\x=2$

8 0

3 years ago

Solve for c: c - 12 = 3

vlabodo [156]

Steps to solve:

c - 12 = 3

~Add 12 to both sides

c = 15

Best of Luck!

8 0

4 years ago

Read 2 more answers

Complete the table for the given rule y=6x-4

larisa [96]

X. /. Y
—————

2. /. 8

4. /. 20

6. /. 32

8 0

3 years ago

Use the Newton-Raphson method to find the root of the equation f(x) = In(3x) + 5x2, using an initial guess of x = 0.5 and a stop

xxMikexx [17]

Answer with explanation:

The equation which we have to solve by Newton-Raphson Method is,

f(x)=log (3 x) +5 x²

$f'(x)=\frac{1}{3x}+10 x$

Initial Guess =0.5

Formula to find Iteration by Newton-Raphson method

$x_{n+1}=x_{n}-\frac{f(x_{n})}{f'(x_{n})}\\\\x_{1}=x_{0}-\frac{f(x_{0})}{f'(x_{0})}\\\\ x_{1}=0.5-\frac{\log(1.5)+1.25}{\frac{1}{1.5}+10 \times 0.5}\\\\x_{1}=0.5- \frac{0.1760+1.25}{0.67+5}\\\\x_{1}=0.5-\frac{1.426}{5.67}\\\\x_{1}=0.5-0.25149\\\\x_{1}=0.248$

$x_{2}=0.248-\frac{\log(0.744)+0.30752}{\frac{1}{0.744}+10 \times 0.248}\\\\x_{2}=0.248- \frac{-0.128+0.30752}{1.35+2.48}\\\\x_{2}=0.248-\frac{0.17952}{3.83}\\\\x_{2}=0.248-0.0468\\\\x_{2}=0.2012$

$x_{3}=0.2012-\frac{\log(0.6036)+0.2024072}{\frac{1}{0.6036}+10 \times 0.2012}\\\\x_{3}=0.2012- \frac{-0.2192+0.2025}{1.6567+2.012}\\\\x_{3}=0.2012-\frac{-0.0167}{3.6687}\\\\x_{3}=0.2012+0.0045\\\\x_{3}=0.2057$

$x_{4}=0.2057-\frac{\log(0.6171)+0.21156}{\frac{1}{0.6171}+10 \times 0.2057}\\\\x_{4}=0.2057- \frac{-0.2096+0.21156}{1.6204+2.057}\\\\x_{4}=0.2057-\frac{0.0019}{3.6774}\\\\x_{4}=0.2057-0.0005\\\\x_{4}=0.2052$

So, root of the equation =0.205 (Approx)

Approximate relative error

$=\frac{\text{Actual value}}{\text{Given Value}}\\\\=\frac{0.205}{0.5}\\\\=0.41$

Approximate relative error in terms of Percentage

=0.41 × 100

= 41 %

7 0

3 years ago