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sergij07 [2.7K]
3 years ago
12

Quadrilaterals ABCD and EFGH are similar. If the perimeter of the quadrilateral ABCD is equal to 4y² what is the perimeter of th

e quadrilateral EFGH?
Mathematics
2 answers:
Elodia [21]3 years ago
8 0

Answer:

Below.

Step-by-step explanation:

Similar figures have the same size but not necessarily the same size.

The respective sides will be in the same ratio in a similar figure.

So from the information given you can't say what the measure of the perimeter of quadrilateral EFGH is. All you can say is the ratio of the perimeters is the same as the ratio of the respective sides. If the ratio is 1:2 for example then the perimeter of EFGH is 2*4y^2 = 8y^2.

UNO [17]3 years ago
4 0

Answer:

Can't be determined.

Step-by-step explanation:

From the fact that the quadrilaterals are similar:

Corresponding sides are in proportion and

Corresponding angles are congruent.

We can't create a proportion, so there's nothing else that we can determine.

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Is DEF a right triangle? True or false
umka2103 [35]
Short answer = false
Remark

If it is a right triangle, it will obey the Pythagorean Equation. a^2 + b^2 = c^2. Let's see if it does.

Equation
a^2 + b^2= c^2
c must be the longest line. (Hypotenuse)

Givens
a = 24
b = 17
c = 40

Sub and Solve
a^2 + b^2 = c^2
17^2 + 24^2 = c^2 [We'll see if this comes out to 40]

289 + 576 = c^2
865 = c^2 which is no where near what 40^2 equals. 40^2 = 1600

Short answer False

Square root of 865 = (865)^1/2 = 29.4
5 0
4 years ago
Rule start at 5 and 1/2 add 1 1/5
Margaret [11]

Answer:

The solution is 6\frac{7}{10}.

Step-by-step explanation:

It is given that rule start at 5\frac{1}{2} add 1\frac{1}{5}.

From the given information, we get the expression

5\frac{1}{2}+1\frac{1}{5}

Simplify the complex fraction.

\frac{10+1}{2}+\frac{5+1}{5}

\frac{11}{2}+\frac{6}{5}

Taking LCM, we get

\frac{11\times 5+6\times 2}{2\times 5}

\frac{55+12}{10}

\frac{67}{10}

6\frac{7}{10}

Therefore the he solution is 6\frac{7}{10}.

4 0
3 years ago
If two lines are cut by a transversal such that corresponding angles are NOT congruent, what must be true? Justify your response
Ivahew [28]

Answer and Step-by-step explanation:

what must be true is that this particular two lines are not a set of PARALLEL lines. When the lines are parallel, the corresponding angles are congruent  and the then the pairs of consecutive angles formed are supplementary .  

If we are takling about corresponding angles, they will not be congruent if the lines are not parallel

6 0
4 years ago
PLEASE HELP! URGENT WILL GIVE BRAINLEST! PLEASE ANSWER ALL QUESTIONS!!! IF YOU DON'T I WILL REPORT.
Nesterboy [21]

Answer:

1.  b)  135°, 225°

2.  c)  0°, 180°

3.  a)  π/2

4.  b)  30°, 150°, 210°, 330°

5.  a)  1.25 radians and 4.39 radians

Step-by-step explanation:

<u>Question 1</u>

\begin{aligned}\cos \theta & = -\dfrac{\sqrt{2}}{2}\\\implies \theta & = \cos^{-1}\left(-\dfrac{\sqrt{2}}{2}\right)\\\\& = 135^{\circ} \pm 360^{\circ}n, 225^{\circ} \pm 360^{\circ}\\\\& = 135^{\circ}, 225^{\circ}\quad \textsf{for}\:0^{\circ}\leq \theta < 360^{\circ}\end{aligned}

<u>Question 2</u>

\begin{aligned}\cos^2 \theta-1 & = 0\\\implies \cos^2 \theta & = 1\\\cos \theta & = \pm1\\\theta & = \cos^{-1}(\pm1)\\& = 0^{\circ} \pm 360^{\circ}n, 180^{\circ} \pm360^{\circ}n\\& = 0^{\circ}, 180^{\circ}\quad \textsf{for}\:0^{\circ}\leq \theta < 360^{\circ}\end{aligned}

<u>Question 3</u>

\begin{aligned}\sin \theta & = 1\\\implies \theta & = \sin^{-1}(1)\\\theta & = \dfrac{\pi}{2}\pm2\pi n\\\theta & = \dfrac{\pi}{2}\quad \textsf{for}\:0 \leq \theta < 2 \pi \end{aligned}

<u>Question 4</u>

\begin{aligned}3 \tan^2 \theta-1 & = 0\\\implies \tan^2 \theta & = \dfrac{1}{3}\\\tan \theta & = \pm\dfrac{1}{\sqrt{3}}\\\theta & = \tan^{-1}\left(\pm\dfrac{1}{\sqrt{3}}\right)\\& = 30^{\circ}\pm180^{\circ}n, 150^{\circ}\pm180^{\circ}n\\& = 30^{\circ}, 150^{\circ}, 210^{\circ}, 330^{\circ}\quad \textsf{for}\:0^{\circ}\leq \theta < 360^{\circ}\end{aligned}

<u>Question 5</u>

\begin{aligned}2 \tan \theta -6 & = 0\\\implies \tan \theta & = 3\\\theta & = \tan^{-1}(3)\\\theta & = 1.25\pm \pi n\\\theta & = 1.25, 4.39 \quad \textsf{for}\:0^{\circ}\leq \theta < 2 \pi \end{aligned}

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