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3 years ago

A ramp AB, 4 metres long, is attached to the back of a lorry at a point, B 1.2 metres above the ground.

Mathematics

1 answer:

fomenos3 years ago

4 0

Answer:

AC=

$2 \sqrt{91 } \div 5$

The angle the ramp makes with the ground is approx. 17.4576

Step-by-step explanation:

see picture above

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Set up but do not solve for the appropriate particular solution yp for the differential equation y′′+4y=5xcos(2x) using the Meth

taurus [48]

Answer:

So, solution of the differential equation is

$y(t)=-\frac{5x^2}{4}\cot 2x\cdot \cos 2x+c_1e^{-2it}+c_2e^{2it}\\$

Step-by-step explanation:

We have the given differential equation: y′′+4y=5xcos(2x)

We use the Method of Undetermined Coefficients.

We first solve the homogeneous differential equation y′′+4y=0.

$y''+4y=0\\\\r^2+4=0\\\\r=\pm2i\\\\$

It is a homogeneous solution:

$y_h(t)=c_1e^{-2i t}+c_2e^{2i t}$

Now, we finding a particular solution.

$y_p(t)=A5x\cos 2x\\\\y'_p(t)=A5\cos 2x-A10x\sin 2x\\\\y''_p(t)=-A20\sin 2x-A20x\cos 2x\\\\\\\implies y''+4y=5x\cos 2x\\\\-A20\sin 2x-A20x\cos 2x+4\cdot A5x\cos 2x=5x\cos 2x\\\\-A20\sin 2x=5x\cos 2x\\\\A=-\frac{x}{4} \cot 2x\\$

we get

$y_p(t)=A5\cos 2x\\\\y_p(t)=-\frac{5x^2}{4}\cot 2x\cdot \cos 2x\\\\\\y(t)=y_p(t)+y_h(t)\\\\y(t)=-\frac{5x^2}{4}\cot 2x\cdot \cos 2x+c_1e^{-2it}+c_2e^{2it}\\$

So, solution of the differential equation is

$y(t)=-\frac{5x^2}{4}\cot 2x\cdot \cos 2x+c_1e^{-2it}+c_2e^{2it}\\$

7 0

3 years ago

Brooke made 10 liters of juice blend that is 60% juice for her brothers birthday party but her mother wants her to make it 80% j

ANEK [815]

The answer is 2 litters of juice

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3 years ago

What is the answer to f(x)=400(1/2)^4x

Vlad1618 [11]

Answer:

25x

Step-by-step explanation:

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3 years ago

Lisa’s car goes 300 miles on 15 gallons of gas. How fuel efficient is Lisa’s car?

sergejj [24]

300miles÷15=20
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4 0

3 years ago

Can y’all help me pls

krok68 [10]

Answer:

2. The change in expected height for every one additional centimeter of femur length.

Step-by-step explanation:

1. The expected height for someone with a femur length of 65 centimeters.

Doesn't make sense, that would be height value when centimeters = 65.

2. The change in expected height for every one additional centimeter of femur length.

Makes sense, for every increase in one additional centimeter, we can expect the height to be proportional to the slope.

3. The femur length for someone with an expected height of 2.5 centimeters.

Doesn't make sense, the linear relationship relies on the femur length to get the height.

4. The change in expected femur length for every one additional centimeter of height.

Doesn't make sense, again, the linear relationship relies on the femur length.

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2 years ago