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4 years ago

Describe the graph of the equation x = 3

Mathematics

1 answer:

DochEvi [55]4 years ago

4 0

Straight up and down at (3,0)
Slope is undefined.

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Ln bisects klm into two congruent angles measuring (3x-4) and (4x-27) Find klm

blsea [12.9K]

The answer would be 135 (B)

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3 years ago

What is the length of side a? Round to the nearest tenth of an inch. Enter your answer in the box.

Wittaler [7]

9514 1404 393

Answer:

14.4

Step-by-step explanation:

The Pythagorean theorem tells you ...

7² +a² = 16²

a² = 256 -49

a = √207

a ≈ 14.4

5 0

3 years ago

Help pleasee, its past due , im sufferingggg

jeyben [28]

Answer:

-11

Step-by-step explanation:

First we plug -2 in for c

4(-2)-3=-8-3=-11

5 0

3 years ago

A ladder 10 feet long rests against a vertical wall. Initially, the top of the ladder is 8 feet above the ground. If the top of

lora16 [44]

Step-by-step explanation:

Let vertical height of ladder from ground be y and

horizontal distance of the base of the ladder from the wall be x respectively.

Length of the ladder = l (constant) = 10 ft

Using Pythagoras theorem:

${l}^{2} = {y}^{2} + {x}^{2}$

Differentiate both sides w.r.t time

$0 = 2y \frac{dy}{dt} + 2x \frac{dx}{dt}$

$y \frac{dy}{dt} + x \frac{dx}{dt} = 0$

We know that (After 1 sec, y = 6 ft and x = 8 ft ; dy/dt = 2 ft/sec)

$6\times 2 + 8\frac{dx}{dt} = 0$

$\frac{dx}{dt} = - 1.5 ft \: per \: sec$

( Ignore - ive sign)

Therefore, bottom of the ladder is sliding away from the wall at a speed of 1.5 ft/sec one second after the ladder starts sliding.

8 0

3 years ago

Solve y" + y = tet, y(0) = 0, y'(0) = 0 using Laplace transforms.

irina1246 [14]

Answer:

The solution of the diferential equation is:

$y(t)=\frac{1}{2}cos(t)- \frac{1}{2}e^{t}+\frac{t}{2} e^{t}$

Step-by-step explanation:

Given $y" + y = te^{t}$ ; y(0) = 0 ; y'(0) = 0

We need to use the Laplace transform to solve it.

ℒ[y" + y]=ℒ[ $te^{t}$ ]

ℒ[y"]+ℒ[y]=ℒ[ $te^{t}$ ]

By using the Table of Laplace Transform we get:

ℒ[y"]=s²·ℒ[y]+s·y(0)-y'(0)=s²·Y(s)

ℒ[y]=Y(s)

ℒ[ $te^{t}$ ]= $\frac{1}{(s-1)^{2}}$

So, the transformation is equal to:

s²·Y(s)+Y(s)= $\frac{1}{(s-1)^{2}}$

(s²+1)·Y(s)= $\frac{1}{(s-1)^{2}}$

Y(s)= $\frac{1}{(s^{2}+1)(s-1)^{2}}$

To be able to separate in terms, we use the partial fraction method:

$\frac{1}{(s^{2}+1)(s-1)^{2}}=\frac{As+B}{s^{2}+1} +\frac{C}{s-1}+\frac{D}{(s-1)^2}$

1=(As+B)(s-1)² + C(s-1)(s²+1)+ D(s²+1)

The equation is reduced to:

1=s³(A+C)+s²(B-2A-C+D)+s(A-2B+C)+(B+D-C)

With the previous equation we can make an equation system of 4 variables.

The system is given by:

A+C=0

B-2A-C+D=0

A-2B+C=0

B+D-C=1

The solution of the system is:

A=1/2 ; B=0 ; C=-1/2 ; D=1/2

Therefore, Y(s) is equal to:

Y(s)= $\frac{s}{2(s^{2} +1)} -\frac{1}{2(s-1)} +\frac{1}{2(s-1)^{2}}$

By using the inverse of the Laplace transform:

ℒ⁻¹[Y(s)]=ℒ⁻¹[ $\frac{s}{2(s^{2} +1)}$ ]-ℒ⁻¹[ $\frac{1}{2(s-1)}$ ]+ℒ⁻¹[ $\frac{1}{2(s-1)^{2}}$ ]

$y(t)=\frac{1}{2}cos(t)- \frac{1}{2}e^{t}+\frac{t}{2} e^{t}$

8 0

3 years ago