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2 years ago

36= W/4 what does w equal ???????????///

Mathematics

2 answers:

Eduardwww [97]2 years ago

8 0

Answer:

W=144

Step-by-step explanation:

36=w/4

36 times 4=w/4 times 4

144=W

Rus_ich [418]2 years ago

5 0

Answer:

144

Step-by-step explanation:

144/4=36 36x4=144

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Suppose that a cyclist began a 374 mi ride across a state at the western edge of the state, at the same time that a car traveli

kherson [118]

This is the concept of relative speed; We are required to calculate the speed of the car and the bicycle.
Distance between the car and Bicycle=374 miles
Time they met=5.5 hr
Speed traveled by bicycle=x
Speed traveled by car=x+33.4334
Relative speed=x+(x+33.4334)=(2x+33.4334) mph
Distance=speed*time
374=(2x+33.4334)*5.5
374=11x+183.8837
collecting like term we get:
374-183.8837=11x
11x=190.1163
thus;
x=(190.1163)/(11)
x=17.2833 mph
thus the speed of the bicycle was x=17.2833 mph
The speed of the car was (x+33.4334)=(17.2833+33.4334)=50.7167 mph

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3 years ago

Are all quadrilaterals either rectangles or squares?

Greeley [361]

No, there are lots of quadrilaterals. <span>Examples of other quadrilaterals are rhombus, kite, trapezoid, parallelogram etc</span>

5 0

3 years ago

with reference to a universal set, the inclusion of a subset in another is relation which is a) symmetric only b) transitive onl

Vlad [161]

Answer:

Step-by-step explanation:

I think it is D since a universal set would only be {1,2,3,4,5,6,...} and the subset would not satisfy any of the conditions of symmetric, transitive, and reflexive.

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2 years ago

4y-8-2y+5=0 help please

tia_tia [17]

4y - 8 - 2y + 5 = 0
2y - 3 = 0
2y = 3
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4 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

2 years ago