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3 years ago

Which transformation shows a reflection of ΔDEF?

Mathematics

2 answers:

Helen [10]3 years ago

8 0

Answer:

The image B with purple figures is correct.

Step-by-step explanation:

We are given the figure DEF and we are required to find the figure which is obtained after reflecting it.

Now, reflection of a figure means to flip the figure over a line.

In the green figure, the points D and E are interchanged which is not possible in reflection.

In the blue figure, there is no change.

In the purple figure, the triangle is flipped over the line DE and so it is the correct answer.

Hence, the purple figure is obtained after reflecting the figure DEF.

m_a_m_a [10]3 years ago

7 0

The answer is B or the purple one

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Explain and correct the error made by a student who simplified this expression.

blagie [28]

Answer:

7x+54

Step-by-step explanation:

because you distribute the 7 to the x and the 8 first then you bring that down which becomes 7x+56 then you to the addition so you do 56+-2 which would give you 54.

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3 years ago

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3 years ago

Read 2 more answers

If ABCD is an A4 sheet and BCPO is the square, prove that △OCD is an isosceles triangle. And find the angles marked as 1 to 8 wi

Dmitry [639]

Answer:

The diagram for the question is missing, but I found an appropriate diagram fo the question:

Proof:

since OC = CD = 297mm Therefore, Δ OCD is an isoscless triangle

∠BCO = 45°

∠BOC = 45°

∠PCO = 45°

∠POC = 45°

∠DOP = 22.5°

∠PDO = 67.5°

∠ADO = 22.5°

∠AOD = 67.5°

Step-by-step explanation:

Given:

AB = CD = 297 mm

AD = BC = 210 mm

BCPO is a square

∴ BC = OP = CP = OB = 210mm

Solving for OC

OCB is a right anlgled triangle

using Pythagoras theorem

(Hypotenuse)² = Sum of square of the other two sides

(OC)² = (OB)² + (BC)²

(OC)² = 210² + 210²

(OC)² = 44100 + 44100

OC = √(88200

OC = 296.98 = 297

OC = 297mm

An isosceless tringle is a triangle that has two equal sides

Therefore for △OCD

CD = OC = 297mm; Hence, △OCD is an isosceless triangle.

The marked angles are not given in the diagram, but I am assuming it is all the angles other than the 90° angles

Since BC = OB = 210mm

∠BCO = ∠BOC

since sum of angles in a triangle = 180°

∠BCO + ∠BOC + 90 = 180

(∠BCO + ∠BOC) = 180 - 90

(∠BCO + ∠BOC) = 90°

since ∠BCO = ∠BOC

∴ ∠BCO = ∠BOC = 90/2 = 45

∴ ∠BCO = 45°

∠BOC = 45°

∠PCO = 45°

∠POC = 45°

For ΔOPD

$Tan\ \theta = \frac{opposite}{adjacent}\\ Tan\ (\angle DOP) = \frac{87}{210} \\(\angle DOP) = Tan^-1(0.414)\\(\angle DOP) = 22.5 ^{\circ}$

Note that DP = 297 - 210 = 87mm

∠PDO + ∠DOP + 90 = 180

∠PDO + 22.5 + 90 = 180

∠PDO = 180 - 90 - 22.5

∠PDO = 67.5°

∠ADO = 22.5° (alternate to ∠DOP)

∠AOD = 67.5° (Alternate to ∠PDO)

3 0

3 years ago

Maxine spent 15 hours doing her homework last week. This week she spent 18 hours doing homework. She says that she spent 120% mo

Vadim26 [7]

Answer:

she is correct beacuse if she worked more harder this week than last week it shows beacuse she did 15 hours

and she did 18 hours as well

5 0

2 years ago

Jack can build 1/5 of a shed in the same time Kyle build 5/8 of a shed. How much of a shed will jack have built when Kyle has fi

nadya68 [22]

Answer:

Step-by-step explanation:

If Kyle can get 5/8 of a shed built in x amount of time, he can get 62.5% done; if Jack can get 1/5 of the shed built in the same time, he can 20% of it done. As a ratio:

$\frac{Kyle}{Jack}:\frac{62.5}{20}$ Kyle gets the whole thing done, 100% of the way, leaving Jack lagging behind at a somewhat lower percentage of the work being done.

$\frac{Kyle}{Jack}:\frac{62.5}{20}=\frac{100}{x}$ and cross multiply to solve for the percentage of the shed built by Jack by the time Kyle gets it completed:

62.5x = 2000 so

x = 32%

4 0

3 years ago