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3 years ago

( − 7 , 21 ) ( - 7 , 21 ) and is parallel to the line y = 1 7 x − 56 ? y = 1 7 x - 56 ?

Mathematics

1 answer:

Nutka1998 [239]3 years ago

3 0

Answer:

9487593201-30498576

Step-by-step explanation:

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An object with a mass of 4 kg is hanging from an axle with a radius of 36 cm. if the wheel attached to the axle has a radius of

Airida [17]

The answer is of course 78cm

7 0

3 years ago

Rewrite the following equation as a function of x. 43 - 1/8y + 19 = 0

tamaranim1 [39]

Answer:

y = 496

Step-by-step explanation:

62 - 1/8y = 0

8 (62 - 1/8y ) = 8×0

496 - y = 0

496 = y

y = 496

5 0

2 years ago

Select all of the given trigonometric equations that are true statements. sin(180°) = cos(0°) cos(74°) = sin(16°) sin(20°) = cos

lara [203]

Answer:

Cos 74°= Sin 16°

Sin 20°= Cos 70°

Sin A°=Cos (90°- A°)

Step-by-step explanation:

This is because the three pairs are complementary i.e they add up to 90° and in trigonometric ratios, complementary ratios are equal.

3 0

3 years ago

Write down the explicit solution for each of the following: a) x’=t–sin(t); x(0)=1

Kay [80]

Answer:

a) x=(t^2)/2+cos(t), b) x=2+3e^(-2t), c) x=(1/2)sin(2t)

Step-by-step explanation:

Let's solve by separating variables:

$x'=\frac{dx}{dt}$

a) x’=t–sin(t), x(0)=1

$dx=(t-sint)dt$

Apply integral both sides:

$\int {} \, dx=\int {(t-sint)} \, dt\\\\x=\frac{t^2}{2}+cost +k$

where k is a constant due to integration. With x(0)=1, substitute:

$1=0+cos0+k\\\\1=1+k\\k=0$

Finally:

$x=\frac{t^2}{2} +cos(t)$

b) x’+2x=4; x(0)=5

$dx=(4-2x)dt\\\\\frac{dx}{4-2x}=dt \\\\\int {\frac{dx}{4-2x}}= \int {dt}\\$

Completing the integral:

$-\frac{1}{2} \int{\frac{(-2)dx}{4-2x}}= \int {dt}$

Solving the operator:

$-\frac{1}{2}ln(4-2x)=t+k$

Using algebra, it becomes explicit:

$x=2+ke^{-2t}$

With x(0)=5, substitute:

$5=2+ke^{-2(0)}=2+k(1)\\\\k=3$

Finally:

$x=2+3e^{-2t}$

c) x’’+4x=0; x(0)=0; x’(0)=1

Let $x=e^{mt}$ be the solution for the equation, then:

$x'=me^{mt}\\x''=m^{2}e^{mt}$

Substituting these equations in c)

$m^{2}e^{mt}+4(e^{mt})=0\\\\m^{2}+4=0\\\\m^{2}=-4\\\\m=2i$

This becomes the solution m=α±βi where α=0 and β=2

$x=e^{\alpha t}[Asin\beta t+Bcos\beta t]\\\\x=e^{0}[Asin((2)t)+Bcos((2)t)]\\\\x=Asin((2)t)+Bcos((2)t)$

Where A and B are constants. With x(0)=0; x’(0)=1:

$x=Asin(2t)+Bcos(2t)\\\\x'=2Acos(2t)-2Bsin(2t)\\\\0=Asin(2(0))+Bcos(2(0))\\\\0=0+B(1)\\\\B=0\\\\1=2Acos(2(0))\\\\1=2A\\\\A=\frac{1}{2}$

Finally:

$x=\frac{1}{2} sin(2t)$

7 0

3 years ago

Bonus: Mr. Rao is planning on making a shed in his backyard (see his plan in the image below). However, local laws require sheds

Leviafan [203]

Answer:

North 10 feet, and East 5 feet.

Step-by-step explanation:

So we know that North is ^ way. We also know that East is > way.

These are the two directions we must go, since the property line is both South and West of the shed.

The shed is 10ft East of the property line, and we must be 15ft in that direction. So we must go another 5ft East.

Now, the shed is 5ft North of the property line, and we must be 15ft in that direction. So we must go another 10 feet North.

Hope this helps!

6 0

2 years ago