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2 years ago

What is the slope of the line that contains these points? (15,-10) (17,2) (19,14) (21,26)

Mathematics

1 answer:

umka2103 [35]2 years ago

6 0

The slopes of the line is 6

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PLEASE HELPPPPP M=2 N=16 P=1/3

andre [41]

Answer:

Step-by-step explanation:

Based on the information you have given me, p = 1/3.

All we need to do is substitute that value into the expression:

12p = 12(1/3) = 4

4 0

3 years ago

Read 2 more answers

18 more than the product of k and 5 Evaluate the expression for the value of the variable, k : 7 Please type your answer in the

Nastasia [14]

Answer:

The Answer Is 125.

Step-by-step explanation:

As, 18 More Than K Means

18 More Than 7

So, It's 25

Then Evaluate 25 With 5

It's 25(5)

Or 25 × 5

It's 125.

4 0

2 years ago

Find the value of X (in this picture)

alexgriva [62]

Answer:

x = 89

Step-by-step explanation:

The sum of the angles of a triangle add to 180 degrees

x+32+ 59 = 180

Combine like terms

x + 91 = 180

Subtract 91 from each side

x +91-91= 180-91

x = 89

6 0

3 years ago

Find the area of the rhombus. 7.5 yd 13 yd 13 yd 7.5 yd [ ? ] yd2

Doss [256]

Answer:

195

Step-by-step explanation:

4 0

2 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

2 years ago