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2 years ago

Lucy has a strawberry lace that is 36cm long. She needs pieces 18mm long to decorate a cake. How long is the 36cm one in mm

Mathematics

2 answers:

KonstantinChe [14]2 years ago

7 0

The strawberry lace is 36 cms long.
36 cms = 360 mm
(Because we multiply by 10, when we convert from cm to mm).

kirill [66]2 years ago

4 0

Given Lucy has a strawberry lace that is 36cm long.

The question is asking to convert 36 cm to mm.

1 centimeter is equal to 10 millimeters: 1 cm = 10 mm.

The distance 36 in millimeters (mm) is equal to the distance 36 in centimeters (cm) times 10

36 cm = 36x 10=360 mm

36 cm converted to mm is 360 mm.

The strawberry lace that is 36cm long in mm is equal to 360 mm.

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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

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Answer with explanation:

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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

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7 0

2 years ago

In mon, j, k, and l are midpoints. if jl = 11, lk = 13, and on = 20, and jl || mn, lk || mo, and jk || on, what is the length of

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2 years ago

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cestrela7 [59]

Answer:

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Step-by-step explanation:

Given

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Required

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This implies that, we name the polynomial based on the highest power of x

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When the highest power of x in a polynomial is 4, such polynomial is a quartic polynomial.

Hence, by degree; the polynomial is quartic

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