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katrin2010 [14]
3 years ago
8

A golden rectangle has side lengths in the ratio of about 1 : 1.618. If the long side of a golden rectangle is 35 cm, what is it

s area? Round your answer to the nearest tenth.
Mathematics
2 answers:
tangare [24]3 years ago
7 0

Answer:

area of the rectangle is ≈757.1 cm²

Step-by-step explanation:

To find the area we first need to find the other side length, since the first length is given as 35cm

But the question says they are in the ratio 1 : 1.618

Let x be the breadth of the golden box

1 : 1.618

x : 35cm

1/1.618 =x/35

cross multiply

1.618x = 35

Divide both-side of the equation by 1.618

1.618x/1.618 = 35/1.618

x = 21.632

Therefore the breadth is 21.632

Area of the rectangle = l × b

Area of the rectangle=35 × 21.632

Area of the rectangle = 757.12 cm²

Therefore; area of the rectangle is ≈757.1 cm² to the nearest tenth.

anyanavicka [17]3 years ago
4 0
1982.1cm^2. You multiply the side given by 1.618. 56.63. Then multiply 56.63 & 35
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Jim and bill each invest $15,000 into savings accounts that earn 3.5% interest. Jims account earns simple interests and bills ac
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Answer:

Bill will earn more interest

He will earn $ 20,448.67 from his investment

Step-by-step explanation:

Firstly let us calculate Jim's earnings based on simple interest

A = P(1 + rt)

Calculation:

First, converting R percent to r

a decimal

r = R/100 = 3.5%/100 = 0.035 per year.

Solving our equation:

A = 15000(1 + (0.035 × 25)) = 28125

A = $28,125.00

The total amount accrued, principal plus interest, from simple interest on a principal of $15,000.00 at a rate of 3.5% per year for 25 years is $28,125.00 for Jim

Now let's us calculate bill's investment based on compound interest

Equation

A = P(1 + r)^t

A=15000(1+0.035)^25

A=15000(1.035)^25

A=15000*2.36324498427

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We see that Bill will earn

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4 0
3 years ago
A 15 kilogram object is suspended from the end of a vertically hanging spring stretches the spring 1/3 meters. At time t = 0, th
Yuri [45]

Answer:

15\frac{d^{2}y(t)}{dt^{2} }  - 441.45y(t) = ± 170 cos(5t)

y(0)=0, y'(0)=0

Step-by-step explanation:

See the attached image

This problem involves Newton's 2nd Law which is: ∑F = ma, we have that the acting forces on the mass-spring system are: F_{r} (t) that correspond to the force of resistance on the mass by the action of the spring and F(t) that is an external force with unknown direction (that does not specify in the enounce).

For determinate F_{r} (t) we can use Hooke's Law given by the formula F_{r} (t) = k y(t) where k correspond to the elastic constant of the spring and y(t) correspond to  the relative displacement of the mass-spring system with respect of his rest state.

We know from the problem that an 15 Kg mass stretches the spring 1/3 m so we apply Hooke's law and obtain that...

k = \frac{F_{r}}{y} = \frac{mg}{y} = \frac{15 Kg (9.81 \frac{m}{s^{2} } )}{\frac{1}{3} m}  = 441.45 \frac{N}{m}

Now we apply Newton's 2nd Law and obtaint that...

F_{r} (t) ± F(t) = ma(t)

F_{r} (t) = ky(t) = 441.45y(t)

F(t) = 170 cos(5t)

m = 15 kg

a(t) = \frac{d^{2}y(t)}{dt^{2} }

Finally... 15\frac{d^{2}y(t)}{dt^{2} }  - 441.45y(t) = ± 170 cos(5t)

We know from the problem that there's not initial displacement and initial velocity, so... y(0)=0 and y'(0)=0

Finally the Initial Value Problem that models the situation describe by the problem is

\left \{ 15\frac{d^{2}y(t)}{dt^{2} }  - 441.45y(t) = \frac{+}{} 170 cos(5t) \atop {y(0)=0, y'(0)=0\right.

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