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mixas84 [53]
3 years ago
9

Pls help. This is due today!

Mathematics
1 answer:
JulijaS [17]3 years ago
4 0

Answer:

goodluck

Step-by-step explanation:

I got it from another person I hope it helps

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(I need the answer right now please) ΔA'B'C' was constructed using ΔABC and line segment EH. 2 triangles are shown. Line E H is
irina1246 [14]

Answer:

The true statements are

1. BD = DB'

3. m∠EFA = 90°

4. The line of reflection, EH, is the perpendicular bisector of BB', AA', and

   CC'

Step-by-step explanation:

* Lets explain how to solve the problem

- Reflection is flipping an object over the line of reflection.

- The object and its image have the same shape and size, but the

  figures are in opposite directions from the line of reflection

- The objects appear as if they are mirror reflections, with right and left

  reversed

- The line of reflection is a perpendicular bisector for all lines joining

  points on the figure with their corresponding images

- Look to the attached figure for more understand

* Lets solve the problem

- ΔA'B'C' was constructed using ΔABC and line segment EH, where

 EH is the reflection line

- D is the mid-point of BB'

- F is the mid-point of AA'

- G is the mid-point of CC'

* Lets find from the answer the true statements

1. BD = DB'

∵ D is the mid point of BB'

- Point D divides BB' into two equal parts

∴ BD = DB' ⇒ <em>True</em>

2. DF = FG

- It depends on the size of the sides and angles of the triangle

∵ We can't prove that

∴ DF = FG ⇒ <em>Not true</em>

3. m∠EFA = 90°

∵ The line of reflection ⊥ the lines joining the points with their

   corresponding images

∴ EH ⊥ AA' and bisect it at F

∴ m∠EFA = 90° ⇒ <em>True</em>

4. The line of reflection, EH, is the perpendicular bisector of BB',

   AA', and CC'

- Yes the line of reflection is perpendicular bisectors of them

∴ The line of reflection, EH, is the perpendicular bisector of BB',

   AA', and CC' ⇒ <em>True</em>

5. ΔABC is not congruent to ΔA'B'C'

∵ In reflection the object and its image have the same shape and size

∴ Δ ABC is congruent to Δ A'B'C'

∴ ΔABC is not congruent to ΔA'B'C' ⇒ <em>Not true</em>

4 0
3 years ago
Read 2 more answers
I have enough money to but five regular priced CDs and have $6 left over. However, all CDs are on sale today for &amp;4 less tha
pshichka [43]
Think by this way. Equation

6 0
3 years ago
Please anyone answer me
ollegr [7]

Let's divide the shaded region into two areas:

area 1: x = 0 ---> x = 2

ares 2: x = 2 ---> x = 4

In area 1, we need to find the area under g(x) = x and in area 2, we need to find the area between g(x) = x and f(x) = (x - 2)^2. Now let's set up the integrals needed to find the areas.

Area 1:

A\frac{}{1}  = ∫g(x)dx =  ∫xdx =  \frac{1}{2}  {x}^{2}  | \frac{2}{0}  = 2

Area 2:

A\frac{}{2}  = ∫(g(x) - f(x))dx

= ∫(x -  {(x - 2)}^{2} )dx

=  ∫( - {x}^{2}  + 5x - 4)dx

= ( - \frac{1}{3}{x}^{3} +   \frac{5}{2} {x}^{2}  - 4x)   | \frac{4}{2}

= 2.67 - ( - 0.67) = 3.34

Therefore, the area of the shaded portion of the graph is

A = A1 + A2 = 5.34

3 0
3 years ago
Graph the system of equations to find the solutions of x3 + 6x2 - 40x = 192. y = x3 + 6x2 - 40x y = 192 Select all of the soluti
STatiana [176]
It would be x=6,−4,−<span>8, good luck</span>

5 0
3 years ago
Read 2 more answers
5. If position of object x = 3 sinΘ – 7 cosΘ then motion of object is bounded between position.​
lesya692 [45]

9514 1404 393

Answer:

  ±√58 ≈ ±7.616

Step-by-step explanation:

The linear combination of sine and cosine functions will have an amplitude that is the root of the sum of the squares of the individual amplitudes.

  |x| = √(3² +7²) = √58

The motion is bounded between positions ±√58.

_____

Here's a way to get to the relation used above.

The sine of the sum of angles is given by ...

  sin(θ+c) = sin(θ)cos(c) +cos(θ)sin(c)

If this is multiplied by some amplitude A, then we have ...

  A·sin(θ+c) = A·sin(θ)cos(c) +A·cos(θ)sin(c)

Comparing this to the given expression, we find ...

  A·cos(c) = 3   and   A·sin(c) = -7

We know that sin²+cos² = 1, so the sum of the squares of these values is ...

  (A·cos(c))² +(A·sin(c))² = A²(cos(c)² +sin(c)²) = A²(1) = A²

That is, A² = (3)² +(-7)² = 9+49 = 58. This tells us the position function can be written as ...

  x = A·sin(θ +c) . . . . for some angle c

  x = (√58)sin(θ +c)

This has the bounds ±√58.

3 0
3 years ago
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