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3 years ago

❗️5 points❗️ 5. state the slope of the line. A. -2 B. 0 C. 1 D. Undefined

Mathematics

2 answers:

wlad13 [49]3 years ago

5 0

Answer:

B. 0

Step-by-step explanation:

Use the following formula to solve for slope:

m (slope) = (y₂ - y₁)/(x₂ - x₁)

Let:

(x₁ , y₁) = (0 , -2)

(x₂ , y₂) = (2 , -2)

Plug in the corresponding numbers to the corresponding variables:

m = (-2 - (-2))/(2 - 0)

Simplify:

m = (-2 + 2)/(2 - 0)

m = 0/2 = 0

B. 0 is your slope.

Anna35 [415]3 years ago

4 0

Hey there! :)

Answer:

B. 0

Step-by-step explanation:

On the graph, the line is a horizontal line. This means that there is no change in the y-value across the domain.

Thus, the graph has a slope of 0.

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Priya has completed 9 exam questions. This is 60% of the questions on the exam.

Natali [406]

Answer:

Your answer is 15, 100% of the exam is 15 questions.

Step-by-step explanation:

60% x ? = 9

We need to do 9 divided by 60%

Which is actually 9 divided by 60/100.

Which is equal to; (100 x 9) divided by 6.

Which is then 900 divided by 60 which equals 15.

Answer: 15

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2 years ago

What is the solution of the system? Use elimination. 5x-6y=-32 3x+6y=48

vampirchik [111]

Answer:

x=2 y=7

Step-by-step explanation:

Add the equations in order to solve for the first variable. Plug this value into the other equations in order to solve for the remaining variables.

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2 years ago

Read 2 more answers

Trig questions:

navik [9.2K]

Question 1
Because the period is 2π, and the amplitude is 1obtain
f(x) = sin(x)
Because the horizontal shift is π, obtain
f(x) = sin(x - π)
Because the vertical shift is -4, obtain
f(x) = sin(x - π) - 4

Answer: 1. f(x) = sin(x - π) - 4

Question 2
The radius is 36/2 = 18 in.
1 revolution (360°) is the circumference, which is
2π(18) = 36π in
When the revolution is 62°, the distance traveled is
(62/360)*(36π) = (31/5)π in

Answer: 3. (31π)/5

Question 3.
Consider f(x) = 3cos(2x-π) - 1
f(0) = 3cos(-π) - 1 = -4
f(π/2) = 3cos(0) - 1 = 2
Rate of change = (2+4)/(π/2) = 12/π

From the graph, the rate of change of g(x) is
3/(π/2) = 6/π

Consider h(x) = sin(x) - 4
h(0) = 0 - 4 = -4
h(π/2) = 1 - 4 = -3
Rate of change = (-3+4)/(π/2) = 2/π
Therefore h(x) has the smallest rate of change

Answer: h(x)

8 0

3 years ago

What is the equation of the line that passes through the point (6,4) and has a slope of -2/3

Aleksandr [31]

Answer:

using the given points substitute it into the equation y=mx+c

therefore

y=4

m=-2/3

X=6

c=?

so we should c=intercept

Step-by-step explanation:

y=mx+c

4=-2/3(6)+c

4=-4+c

c=4+4=8

therefore equation of the graph is y=-2/3X+8

4 0

3 years ago

Using Laplace transforms, solve x" + 4x' + 6x = 1- e^t with the following initial conditions: x(0) = x'(0) = 1.

professor190 [17]

Answer:

The solution to the differential equation is

$X(s)=\cfrac 1{6} -\cfrac {1}{11}e^{t}+\cfrac {61}{66}e^{-2t}\cos(\sqrt 2t)+\cfrac {97}{66}\sqrt 2 e^{-2t}\sin(\sqrt 2t)$

Step-by-step explanation:

Applying Laplace Transform will help us solve differential equations in Algebraic ways to find particular solutions, thus after applying Laplace transform and evaluating at the initial conditions we need to solve and apply Inverse Laplace transform to find the final answer.

Applying Laplace Transform

We can start applying Laplace at the given ODE

$x''(t)+4x'(t)+6x(t)=1-e^t$

So we will get

$s^2 X(s)-sx(0)-x'(0)+4(sX(s)-x(0))+6X(s)=\cfrac 1s -\cfrac1{s-1}$

Applying initial conditions and solving for X(s).

If we apply the initial conditions we get

$s^2 X(s)-s-1+4(sX(s)-1)+6X(s)=\cfrac 1s -\cfrac1{s-1}$

Simplifying

$s^2 X(s)-s-1+4sX(s)-4+6X(s)=\cfrac 1s -\cfrac1{s-1}$

$s^2 X(s)-s-5+4sX(s)+6X(s)=\cfrac 1s -\cfrac1{s-1}$

Moving all terms that do not have X(s) to the other side

$s^2 X(s)+4sX(s)+6X(s)=\cfrac 1s -\cfrac1{s-1}+s+5$

Factoring X(s) and moving the rest to the other side.

$X(s)(s^2 +4s+6)=\cfrac 1s -\cfrac1{s-1}+s+5$

$X(s)=\cfrac 1{s(s^2 +4s+6)} -\cfrac1{(s-1)(s^2 +4s+6)}+\cfrac {s+5}{s^2 +4s+6}$

Partial fraction decomposition method.

In order to apply Inverse Laplace Transform, we need to separate the fractions into the simplest form, so we can apply partial fraction decomposition to the first 2 fractions. For the first one we have

$\cfrac 1{s(s^2 +4s+6)}=\cfrac As + \cfrac {Bs+C}{s^2+4s+6}$

So if we multiply both sides by the entire denominator we get

$1=A(s^2+4s+6) + (Bs+C)s$

At this point we can find the value of A fast if we plug s = 0, so we get

$1=A(6)+0$

So the value of A is

$A = \cfrac 16$

We can replace that on the previous equation and multiply all terms by 6

$1=\cfrac 16(s^2+4s+6) + (Bs+C)s$

$6=s^2+4s+6 + 6Bs^2+6Cs$

We can simplify a bit

$-s^2-4s= 6Bs^2+6Cs$

And by comparing coefficients we can tell the values of B and C

$-1= 6B\\B=-1/6\\-4=6C\\C=-4/6$

So the separated fraction will be

$\cfrac 1{s(s^2 +4s+6)}=\cfrac 1{6s} +\cfrac {-s/6-4/6}{s^2+4s+6}$

We can repeat the process for the second fraction.

$\cfrac1{(s-1)(s^2 +4s+6)}=\cfrac A{s-1} + \cfrac {Bs+C}{s^2+4s+6}$

Multiplying by the entire denominator give us

$1=A(s^2+4s+6) + (Bs+C)(s-1)$

We can plug the value of s = 1 to find A fast.

$1=A(11) + 0$

So we get

$A = \cfrac1{11}$

We can replace that on the previous equation and multiply all terms by 11

$1=\cfrac 1{11}(s^2+4s+6) + (Bs+C)(s-1)$

$11=s^2+4s+6 + 11Bs^2+11Cs-11Bs-11C$

Simplifying

$-s^2-4s+5= 11Bs^2+11Cs-11Bs-11C$

And by comparing coefficients we can tell the values of B and C.

$-s^2-4s+5= 11Bs^2+11Cs-11Bs-11C\\-1=11B\\B=-\cfrac{1}{11}\\5=-11C\\C=-\cfrac{5}{11}$

So the separated fraction will be

$\cfrac1{(s-1)(s^2 +4s+6)}=\cfrac {1/11}{s-1} + \cfrac {-s/11-5/11}{s^2+4s+6}$

So far replacing both expanded fractions on the solution

$X(s)=\cfrac 1{6s} +\cfrac {-s/6-4/6}{s^2+4s+6} -\cfrac {1/11}{s-1} -\cfrac {-s/11-5/11}{s^2+4s+6}+\cfrac {s+5}{s^2 +4s+6}$

We can combine the fractions with the same denominator

$X(s)=\cfrac 1{6s} -\cfrac {1/11}{s-1}+\cfrac {-s/6-4/6+s/11+5/11+s+5}{s^2 +4s+6}$

Simplifying give us

$X(s)=\cfrac 1{6s} -\cfrac {1/11}{s-1}+\cfrac {61s/66+158/33}{s^2 +4s+6}$

Completing the square

One last step before applying the Inverse Laplace transform is to factor the denominators using completing the square procedure for this case, so we will have

$s^2+4s+6 = s^2 +4s+4-4+6$

We are adding half of the middle term but squared, so the first 3 terms become the perfect square, that is

$=(s+2)^2+2$

So we get

$X(s)=\cfrac 1{6s} -\cfrac {1/11}{s-1}+\cfrac {61s/66+158/33}{(s+2)^2 +(\sqrt 2)^2}$

Notice that the denominator has (s+2) inside a square we need to match that on the numerator so we can add and subtract 2

$X(s)=\cfrac 1{6s} -\cfrac {1/11}{s-1}+\cfrac {61(s+2-2)/66+316 /66}{(s+2)^2 +(\sqrt 2)^2}\\X(s)=\cfrac 1{6s} -\cfrac {1/11}{s-1}+\cfrac {61(s+2)/66+194 /66}{(s+2)^2 +(\sqrt 2)^2}$

Lastly we can split the fraction one more

$X(s)=\cfrac 1{6s} -\cfrac {1/11}{s-1}+\cfrac {61(s+2)/66}{(s+2)^2 +(\sqrt 2)^2}+\cfrac {194 /66}{(s+2)^2 +(\sqrt 2)^2}$

Applying Inverse Laplace Transform.

Since all terms are ready we can apply Inverse Laplace transform directly to each term and we will get

$\boxed{X(s)=\cfrac 1{6} -\cfrac {1}{11}e^{t}+\cfrac {61}{66}e^{-2t}\cos(\sqrt 2t)+\cfrac {97}{66}\sqrt 2 e^{-2t}\sin(\sqrt 2t)}$

6 0

3 years ago