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3 years ago

WILL MARK BRAINLIEST ^^ picture included!! PLEASE and thx

Mathematics

1 answer:

Elenna [48]3 years ago

3 0

<h3>Answer: C) 3 units up and 4 units left</h3>

Explanation:

Like with the previous problems, the answer comes fairly straight forward from the values of c and d.

c = 4 tells us we have a shift of 4 units to the left

d = 3 indicates we're shifting 3 units up.

Note: we shift left instead of right because x+c = x+4 means the xy axis has been shifted 4 units to the right, giving the illusion the cosine curve is shifted 4 units to the left.

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The number of students in the period 7 study hall at jin's school is 4 times the number of students in jin's home room. how many

k0ka [10]

1/4 the students in study hall (per 7) = 16

16*4 = 64 students in study hall.

Hope I helped! (Pick my answer as brainliest!)

5 0

3 years ago

(10 points) Consider the initial value problem y′+3y=9t,y(0)=7. Take the Laplace transform of both sides of the given differenti

Rashid [163]

Answer:

The solution

$Y (s) = 9( -1 +3 t + e^{-3 t} ) + 7 e ^{-3 t}$

Step-by-step explanation:

Explanation:-

Consider the initial value problem y′+3 y=9 t,y(0)=7

Step(i):-

Given differential problem

y′+3 y=9 t

Take the Laplace transform of both sides of the differential equation

L( y′+3 y) = L(9 t)

Using Formula Transform of derivatives

L(y¹(t)) = s y⁻(s)-y(0)

By using Laplace transform formula

$L(t) = \frac{1}{S^{2} }$

Step(ii):-

Given

L( y′(t)) + 3 L (y(t)) = 9 L( t)

$s y^{-} (s) - y(0) + 3y^{-}(s) = \frac{9}{s^{2} }$

$s y^{-} (s) - 7 + 3y^{-}(s) = \frac{9}{s^{2} }$

Taking common y⁻(s) and simplification, we get

$( s + 3)y^{-}(s) = \frac{9}{s^{2} }+7$

$y^{-}(s) = \frac{9}{s^{2} (s+3}+\frac{7}{s+3}$

Step(iii):-

By using partial fractions , we get

$\frac{9}{s^{2} (s+3} = \frac{A}{s} + \frac{B}{s^{2} } + \frac{C}{s+3}$

$\frac{9}{s^{2} (s+3} = \frac{As(s+3)+B(s+3)+Cs^{2} }{s^{2} (s+3)}$

On simplification we get

9 = A s(s+3) +B(s+3) +C(s²) ...(i)

Put s =0 in equation(i)

9 = B(0+3)

B = 9/3 = 3

Put s = -3 in equation(i)

9 = C(-3)²

C = 1

Given Equation 9 = A s(s+3) +B(s+3) +C(s²) ...(i)

Comparing 'S²' coefficient on both sides, we get

9 = A s²+3 A s +B(s)+3 B +C(s²)

0 = A + C

put C=1 , becomes A = -1

$\frac{9}{s^{2} (s+3} = \frac{-1}{s} + \frac{3}{s^{2} } + \frac{1}{s+3}$

Step(iv):-

$y^{-}(s) = \frac{9}{s^{2} (s+3}+\frac{7}{s+3}$

$y^{-}(s) =9( \frac{-1}{s} + \frac{3}{s^{2} } + \frac{1}{s+3}) + \frac{7}{s+3}$

Applying inverse Laplace transform on both sides

$L^{-1} (y^{-}(s) ) =L^{-1} (9( \frac{-1}{s}) + L^{-1} (\frac{3}{s^{2} }) + L^{-1} (\frac{1}{s+3}) )+ L^{-1} (\frac{7}{s+3})$

By using inverse Laplace transform

$L^{-1} (\frac{1}{s} ) =1$

$L^{-1} (\frac{1}{s^{2} } ) = \frac{t}{1!}$

$L^{-1} (\frac{1}{s+a} ) =e^{-at}$

Final answer:-

Now the solution , we get

$Y (s) = 9( -1 +3 t + e^{-3 t} ) + 7 e ^{-3t}$

5 0

3 years ago

What are the coordinates of the point (-1, 6) after a reflection across the x-axis?

maks197457 [2]

Answer:

B(-1,-6)

Step-by-step explanation:

reflecting reflecting over the x axis changes the y outputs.

4 0

3 years ago

Read 2 more answers

63 multiplied by the difference of x and 5. (I would like an explanation too.)

vitfil [10]

63(x-5)
difference= subtraction
do you subtract x by 5, then multiply it by 63

8 0

3 years ago

Read 2 more answers

Need help with this page of my final . Have no idea how to do geometry . Anyone good at it ? I'd appreciate it !

Katen [24]

Answers:
21. Three noncollinear points determine 3 lines and 1 plane
22. If two quadrilaterals are similar, then they are squares
23. PR = 2.6
24. Midpoint is (3,-1)
25. Angle BXD = 108 degrees
26. The complement is 18 degrees
27. A) 30
28. Corresponding angles
29. False; Change "congruent" to "supplementary"
30. Neither
31. Equation is y = (-2/5)x + 9/5
32. Obtuse scalene triangle
----------------------------------------------------------------
Work Shown
Problem 21)
Three noncollinear points determine 3*2 = 6 pairings but half of those pairings are repeats, so we have 6/2 = 3 unique groups forming 3 lines (think of a triangle and its sides)
The three noncollinear points form a single plane. This is simply an axiom.
-----------------
Problem 22)
Original Conditional is in the form If P, then Q
The converse is the flip of that. So we go to If Q, then P.
So we have
Original Conditional: "If two quadrilaterals are squares, then they are similar"
Converse: "If two quadrilaterals are similar, then they are squares"
-----------------
Problem 23)
P is between Q and R. By the segment addition postulate, we know
QP+PR = QR
We're given PQ or QP to be 10.2 and we know that QR = 12.8, so this means,
QP+PR = QR
10.2+PR = 12.8
10.2+PR-10.2 = 12.8-10.2
PR = 2.6
-----------------
Problem 24)
Add up the x coordinates and divide by 2: (x1+x2)/2 = (8+(-2))/2 = 6/2 = 3
Add up the y coordinates and divide by 2: (y1+y2)/2 = (-6+4)/2 = -2/2 = -1
Therefore the midpoint is (3,-1)
-----------------
Problem 25)
Angle DXE = 36 (given)
Angle CXD = angle DXE (definition of bisection)
Angle CXD = 36
Angle CXE = (angle CXD)+(angle DXE)
Angle CXE = 36+36
Angle CXE = 72
Angle BXE = 2*(angle CXE) ... since XC bisects angle BXE
Angle BXE = 2*72
Angle BXE = 144
Angle BXD = (angle BXE) - (angle DXE)
Angle BXD = 144 - 36
Angle BXD = 108
-----------------
Problem 26)
From problem 25, we found that Angle CXE = 72. Since XC cuts angle BXE in half, and the other angle is BXC, this means
Angle BXE = angle CXE = 72 degrees
Now subtract that from 90
90 - (angle BXE) = 90 - 72 = 18
The complement is 18 degrees
-----------------
Problem 27)
A+B+C = 180
x+x+120 = 180
2x+120 = 180
2x+120-120 = 180-120
2x = 60
2x/2 = 60/2
x = 30
-----------------
Problem 28)
Angle 5 and angle 7 are corresponding angles. They are located on the same side of the transversal line. They both correspond to the same side of their respective parallel line counterparts. Both are on the right side of the parallel line they are attached to.
-----------------
Problem 29)
The statement in its current form is False. One way to fix it is to change the first underlined term from "congruent" to "supplementary". Angle 3 and angle 2 are same side interior angles which add up to 180 degrees.
-----------------
Problem 30)
Slope of AB = (y2-y1)/(x2-x1)
Slope of AB = (-2-6)/(-2-10)
Slope of AB = -8/(-12)
Slope of AB = 2/3
Slope of CD = (y2-y1)/(x2-x1)
Slope of CD = (2-6)/(6-(-6))
Slope of CD = -4/12
Slope of CD = -1/3
Multiply the slopes:
(Slope of AB)*(Slope of BC) = (2/3)*(-1/3) = -2/9
The result is NOT equal to -1, so the lines are NOT perpendicular
The two slopes are NOT equal, so the lines are NOT parallel
So the answer is "neither"
-----------------
Problem 31)
Anything parallel to 2x+5y = 12 is of the form 2x+5y = C where C is some fixed number
Plug in the given point (x,y) = (2,1) to find C
2x+5y = C
2*2+5*1 = C
4+5 = C
9 = C
C = 9
So we go from 2x+5y = C to 2x+5y = 9. Now solve for y
2x+5y = 9
2x+5y-2x = 9-2x
5y = -2x+9
5y/5 = (-2x+9)/5
y = (-2/5)x + 9/5
-----------------
Problem 32)
A+B+C = 180
16+B+64 = 180
B+80 = 180
B+80-80 = 180-80
B = 100
The angle B is 100 degrees, which is larger than 90 degrees. We have an obtuse triangle because of this fact.
All three angles (16, 64, 80) are different, so the side lengths are different. The three different side lengths means we have a scalene triangle.

7 0

3 years ago