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lara [203]
3 years ago
14

Polygon ABCD goes through a sequence of rigid transformations to form polygon A′B′C′D′. The sequence of transformations involved

is a reflection across the
Mathematics
1 answer:
Valentin [98]3 years ago
4 0
X axis because of its sequence that it is involved with a reflection which is a transformation

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The table represents the linear function f(x), and the equation represents the linear function g(x).
Eva8 [605]

9514 1404 393

Answer:

  The slope of f(x) is greater than the slope of g(x). The y-intercept of f(x) is equal to the y-intercept of g(x).

Step-by-step explanation:

The y-intercept is the function value when x=0. The table shows f(0) = 1. The equation shows g(0) = 1, so the y-intercepts are equal.

__

The value of f(x) changes by (11 -1) = 10 when the value of x changes by (2 -0) = 2. That means the slope of f(x) is 10/2 = 5.

The slope of g(x) is the x-coefficient, 4. We note that 5 > 4, so the slope of f(x) is greater than for g(x).

The slope of f(x) is greater than the slope of g(x). The y-intercept of f(x) is equal to the y-intercept of g(x).

8 0
3 years ago
Read 2 more answers
Solve 5y'' + 3y' – 2y = 0, y(0) = 0, y'(0) = 2.8 y(t) = 0 Preview
mario62 [17]

Answer:  The required solution is

y(t)=-\dfrac{7}{3}e^{-t}+\dfrac{7}{3}e^{\frac{1}{5}t}.

Step-by-step explanation:   We are given to solve the following differential equation :

5y^{\prime\prime}+3y^\prime-2y=0,~~~~~~~y(0)=0,~~y^\prime(0)=2.8~~~~~~~~~~~~~~~~~~~~~~~~(i)

Let us consider that

y=e^{mt} be an auxiliary solution of equation (i).

Then, we have

y^prime=me^{mt},~~~~~y^{\prime\prime}=m^2e^{mt}.

Substituting these values in equation (i), we get

5m^2e^{mt}+3me^{mt}-2e^{mt}=0\\\\\Rightarrow (5m^2+3y-2)e^{mt}=0\\\\\Rightarrow 5m^2+3m-2=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mt}\neq0]\\\\\Rightarrow 5m^2+5m-2m-2=0\\\\\Rightarrow 5m(m+1)-2(m+1)=0\\\\\Rightarrow (m+1)(5m-1)=0\\\\\Rightarrow m+1=0,~~~~~5m-1=0\\\\\Rightarrow m=-1,~\dfrac{1}{5}.

So, the general solution of the given equation is

y(t)=Ae^{-t}+Be^{\frac{1}{5}t}.

Differentiating with respect to t, we get

y^\prime(t)=-Ae^{-t}+\dfrac{B}{5}e^{\frac{1}{5}t}.

According to the given conditions, we have

y(0)=0\\\\\Rightarrow A+B=0\\\\\Rightarrow B=-A~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(ii)

and

y^\prime(0)=2.8\\\\\Rightarrow -A+\dfrac{B}{5}=2.8\\\\\Rightarrow -5A+B=14\\\\\Rightarrow -5A-A=14~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{Uisng equation (ii)}]\\\\\Rightarrow -6A=14\\\\\Rightarrow A=-\dfrac{14}{6}\\\\\Rightarrow A=-\dfrac{7}{3}.

From equation (ii), we get

B=\dfrac{7}{3}.

Thus, the required solution is

y(t)=-\dfrac{7}{3}e^{-t}+\dfrac{7}{3}e^{\frac{1}{5}t}.

7 0
3 years ago
What is the solution to the linear equation d-10-2d+7=8+d-10-3d
bija089 [108]
D-10-2d+7 = 8 + d- 10 -3d
d-2d-10+7 = 8-10 +d -3d
-d-3=-2-2d
-d+2d=3-2
d=1

Final answer: C. d=1
8 0
3 years ago
Read 2 more answers
How can you represent a linear function in a way that reveals its slope and y-intercept
Alinara [238K]

Answer:

see explanation

Step-by-step explanation:

A linear function is a straight line so all you have to do is find a slope and a y-intercept

8 0
3 years ago
Please answer, thank you
cupoosta [38]

Answer:

Step-by-step explanation:

Since y = -3 we can plug straight into the other equation

--> -x + 2y = -6

-x + 2(-3) = -6

-x - 6 = -6

-x = 0

x = 0

To graph this draw a horizontal like across the page at y = -3 and a vertical line down the page to represent x = 0

The point where these lines intersect is (0,-3)

So that would be the ordered pair solution.

4 0
3 years ago
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