All categories

3 years ago

A triangle has sides with lengths of 5 yards, 12 yards, and 13 yards. Is it a right triangle?

1 answer:

6 0

Yes because the hypotenuse would be the longest side (13) and when squared equals 169. now when you square 5 and 12 (25 and 144) and add them, you get the 169, which is 13. so yes it’s a right triangle.

You might be interested in

Which equation can be used to solve for in the following diagram?

Ivahew [28]

Answer:

the Answer is 3x+12)+X=180

5 0

3 years ago

Question Number 2!! Thank youu

rewona [7]

Answer:

56°

Step-by-step explanation:

Since triangle ABD and triangle CBD are congruent (SAS), so you can divide m∠ABC by 2 to get the m∠ABD and m∠CBD, or 56°.

3 0

2 years ago

Can anyone help me with this?

Alexxandr [17]

-1

The inequality is negative 1

6 0

3 years ago

Read 2 more answers

The product of a number, x, and 18 is 113.4. Which equation and value of x represent this relationship? x + 18 = I (113.4) = 95.

s344n2d4d5 [400]

equation

$x\times18=113.4$

value of x

$undefined$

3 0

1 year 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