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3 years ago

In each diagram abc has been tranfromed to yield abc

Mathematics

1 answer:

Lady_Fox [76]3 years ago

5 0

What is your question? and Where is your graph?

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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' ⇒ True

2. DF = FG

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

∵ We can't prove that

∴ DF = FG ⇒ Not true

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° ⇒ True

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' ⇒ True

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' ⇒ Not true

4 0

4 years ago

Read 2 more answers

The length of a rectangle is the width minus 8 units. The area of the rectangle is 9 units. What is the length, in units, of the

frez [133]

Length of rectangle is 1 unit

Solution: 

Given that

Length of a rectangle is width minus 8 units.

Area of rectangle = 9 square units

Need to calculate the length of rectangle.

Let us assume width of rectangle = x

As Length is width minus 8,

Length of the rectangle = x – 8

$\text{ Area of rectangle }=\text{ width of rectangle }\times \text{ Length of the rectangle }$

$\text{ Area of rectangle }= x\times (x-8)$

$\text { Area of rectangle }=\left(x^{2}-8 x\right)$

As given that area of rectangle = 9 square units

$\begin{array}{l}{\Rightarrow x^{2}-8 x=9} \\\\ {\Rightarrow x^{2}-8 x-9=0}\end{array}$

On solving above quadratic equation for x, we get

$\begin{array}{l}{\Rightarrow x^{2}-9 x+x-9=0} \\\\ {\Rightarrow x(x-9)+1(x-9)=0} \\\\ {\Rightarrow (x-9)(x+1)=0}\end{array}$

When $x-9 =9, x = 9$

When $x + 1 = 0, x = -1$

As width cannot be negative, so considering x = 9

$\text{ Length of rectangle }= x-8 = 9-8 = 1 \text{ unit }$

Hence, Length of rectangle is 1 unit.

5 0

3 years ago

Problem 4: Let F = (2z + 2)k be the flow field. Answer the following to verify the divergence theorem: a) Use definition to find

Viktor [21]

Given that you mention the divergence theorem, and that part (b) is asking you to find the downward flux through the disk $x^2+y^2\le3$ , I think it's same to assume that the hemisphere referred to in part (a) is the upper half of the sphere $x^2+y^2+z^2=3$ .

a. Let $C$ denote the hemispherical cap $z=\sqrt{3-x^2-y^2}$ , parameterized by

$\vec r(u,v)=\sqrt3\cos u\sin v\,\vec\imath+\sqrt3\sin u\sin v\,\vec\jmath+\sqrt3\cos v\,\vec k$

with $0\le u\le2\pi$ and $0\le v\le\frac\pi2$ . Take the normal vector to $C$ to be

$\vec r_v\times\vec r_u=3\cos u\sin^2v\,\vec\imath+3\sin u\sin^2v\,\vec\jmath+3\sin v\cos v\,\vec k$

Then the upward flux of $\vec F=(2z+2)\,\vec k$ through $C$ is

$\displaystyle\iint_C\vec F\cdot\mathrm d\vec S=\int_0^{2\pi}\int_0^{\pi/2}((2\sqrt3\cos v+2)\,\vec k)\cdot(\vec r_v\times\vec r_u)\,\mathrm dv\,\mathrm du$

$\displaystyle=3\int_0^{2\pi}\int_0^{\pi/2}\sin2v(\sqrt3\cos v+1)\,\mathrm dv\,\mathrm du$

$=\boxed{2(3+2\sqrt3)\pi}$

b. Let $D$ be the disk that closes off the hemisphere $C$ , parameterized by

$\vec s(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec\jmath$

with $0\le u\le\sqrt3$ and $0\le v\le2\pi$ . Take the normal to $D$ to be

$\vec s_v\times\vec s_u=-u\,\vec k$

Then the downward flux of $\vec F$ through $D$ is

$\displaystyle\int_0^{2\pi}\int_0^{\sqrt3}(2\,\vec k)\cdot(\vec s_v\times\vec s_u)\,\mathrm du\,\mathrm dv=-2\int_0^{2\pi}\int_0^{\sqrt3}u\,\mathrm du\,\mathrm dv$

$=\boxed{-6\pi}$

c. The net flux is then $\boxed{4\sqrt3\pi}$ .

d. By the divergence theorem, the flux of $\vec F$ across the closed hemisphere $H$ with boundary $C\cup D$ is equal to the integral of $\mathrm{div}\vec F$ over its interior:

$\displaystyle\iint_{C\cup D}\vec F\cdot\mathrm d\vec S=\iiint_H\mathrm{div}\vec F\,\mathrm dV$

We have

$\mathrm{div}\vec F=\dfrac{\partial(2z+2)}{\partial z}=2$

so the volume integral is

$2\displaystyle\iiint_H\mathrm dV$

which is 2 times the volume of the hemisphere $H$ , so that the net flux is $\boxed{4\sqrt3\pi}$ . Just to confirm, we could compute the integral in spherical coordinates:

$\displaystyle2\int_0^{\pi/2}\int_0^{2\pi}\int_0^{\sqrt3}\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi=4\sqrt3\pi$

4 0

4 years ago

Can u please help me with this math assignment

Troyanec [42]

Hard to read sideways but what I got from that is that you have a right triangle and you are trying to explain the other angles?
Both of the other two angles would be 45 because an angle is 180 total. If it is a right angle, that means that there is a 90 degree angle. 180-90=90. In a right angle, the other two angles are the same in degree. Therefor 90/2 is 45. 45 applies to both of the acute angles. I hope this helps! Good luck :)

6 0

3 years ago

In investing $6,200 of a couple's money, a financial planner put some of it into a savings account paying 4% annual simple inter

ddd [48]

Answer:

$ 2,600 was invested at 4% and $ 3,600 was invested at 9%.

Step-by-step explanation:

Given that in investing $ 6,200 of a couple's money, a financial planner put some of it into a savings account paying 4% annual simple interest, and the rest was invested in a riskier mini-mall development plan paying 9% annual simple interest, and the combined interest earned for the first year was $ 428, to determine how much money was invested at each rate, the following calculation must be performed:

3000 x 0.04 + 3200 x 0.09 = 408

2500 x 0.04 + 3700 x 0.09 = 433

2600 x 0.04 + 3600 x 0.09 = 428

Therefore, $ 2,600 was invested at 4% and $ 3,600 was invested at 9%.

7 0

3 years ago