All categories

3 years ago

PLEASE HELP ASAP,

Mathematics

1 answer:

rjkz [21]3 years ago

7 0

Answer:

29.11

Step-by-step explanation:

This is clearly cos. So make an equation of cos equals 9 / x. Then use cross product to get x times cos equals 9. Finally divide by cos and use the calcuator to get x equals 29.11.

You might be interested in

How. do i find the line of symmetry

kkurt [141]

The line of symmetry is at zero because the coordinates are X= -2 , X= 2

8 0

3 years ago

2³- (3x5²)<br> PLEASE HELP 20 POINTS , also show ur work, thank u sm!

Alexxx [7]

Answer:

-67

Step-by-step explanation:

2^3-(3*5^2)

simplify exponents

8-3*25

8-75

=-67

3 0

3 years ago

Read 2 more answers

15x+5y=30 with y by itself

SCORPION-xisa [38]

Answer:

15 x 1 + 5 x 3 = 30

Step-by-step explanation:

If you turn the x into a one then it multiplies 15 by one and then it remains at 15, then you multiply 5 and 3 so that's 15 and then 15 + 15 is 30.

3 0

3 years ago

Read 2 more answers

A is inversely proportional to b. If a=7.2 when b=-4, what will a equal when b=6?

Tpy6a [65]

Answer:

Step-by-step explanation:

To find the value of a when b=6 we must first find the relationship between them.

The statement

a is inversely proportional to b is written as

$a \: \: \alpha \: \: \frac{k}{b}$

where k is the constant of proportionality

When a = 7.2 , b = -4

So we have

$7.2= \frac{k}{ - 4}$

k = 7.2 × - 4

k = - 28.8

So the formula for the variation is

$a = - \frac{28.8}{b}$

When

b = 6

That's

$a = - \frac{28.8}{6}$

We have the final answer as

Hope this helps you

7 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