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4 years ago

What’s the surface area to 1,2,3 and 4

Mathematics

1 answer:

topjm [15]4 years ago

4 0

Wdym? Can you insert a photo or something.. I’m confuse uh..

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The graph shows two lines, A and B.

Harlamova29_29 [7]

Answer with explanation:

Part A:

One solution.

The number of points of intersection represents the number of solutions. Since the two lines only intersect at one point, there is only one solution.

Part B:

(3, 4)

The point(s) of intersection marks the solution(s) to the lines. Since lines A and B intersect at the point (3, 4), the solution to the equation of their lines is (3, 4), or $x=3, y=4$ , as coordinates are written as (x, y).

3 0

3 years ago

Margie is practicing for an upcoming tennis tournament. Her first serve is good 20 out of 30 times on average. Margie wants to k

mojhsa [17]

Answer:

0.6804

Step-by-step explanation:

Given that Margie is practicing for an upcoming tennis tournament. Her first serve is good 20 out of 30 times on average.

Since each trial is independent and there are two outcomes, X no of good serves is binomial with n=6 and p =2/3

Required probability

= Prob that atleast four of 6 times good serve

= $P(X\geq 4)$

= $P(x=4,5 or 6)\\=1-F(3)\\=0.6804$

Formula used:

P(X=r) = $6Cr (\frac{2}{3}^r )(\frac{1}{3}^{6-r} )$

7 0

3 years ago

Read 2 more answers

How to Convert 50°toradians

Gelneren [198K]

Answer:

50 degrees =0.873 radian

Formula 50 degrees x π/180= 0.8727 radians

6 0

3 years ago

3. 0.7kg : 800g I’m pretty bad with ratios lol

julsineya [31]

Answer:

7:8

Step-by-step explanation:

Convert o.7 Kg to Gram

It will be 700 grams

Divide it by 800

700

800

= 7

7 0

3 years ago

Read 2 more answers

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