3x + 2(x - k) + 3 = 5x

If two triangles have equal perimeters, then they could also have which of the following? Select all that apply.

Radda [10]

Answer: The answers is (B) equal areas.

Step-by-step explanation: Given that two triangles have equal perimeters.

As shown in the attached figure, let us consider two right-angles triangles, ΔABC and ΔDEF, with sides AB = 3 cm, BC = 4 cm, AC = 5 cm, DE = 4 cm, EF = 3 cm and DF = 5 cm.

So the perimeters of both the triangles = 3 + 4 + 5 = 4 + 3 + 5 = 12 cm.

Since volume term is not valid in case of triangles, so they cannot have equal volumes. Therefore, option (A) is incorrect.

Area of ΔABC is

$A_{ABC}=\dfrac{1}{2}\times AB \times BC = \dfrac{1}{2}\times 3\times 4=6~\textup{cm}^2,$

and area of ΔDEF is

$A_{DEF}=\dfrac{1}{2}\times DE\times EF=\dfrac{1}{2}\times 4\times 3=6~\textup{cm}^2.$

Therefore, they may have equal areas and so option (B) is correct.

If the triangles have equal bases, then the heights will also be equal and both the triangles will be same. Similar is the case with equal heights. So, options (C) and (D) are incorrect.

Thus, the correct option is (B). equal areas.

4 0

3 years ago

Carolina leaves her house and walks 6 blocks she turns and heads east 8 blocks until she reaches matties house what is the short

s2008m [1.1K]

Answer:

10 blocks

Step-by-step explanation:

Given

$First = 6\ blocks$

$Second = 8\ blocks$ east

Required

Determine the shortest possible distance

To better explain my solution, I've added an attachment that illustrates her movement.

Using the attachment as a point of reference, the shortest distance is calculated by calculating the length of the hypotenuse using Pythagoras theorem.

So, we have:

$x^2 = 6^2 + 8^2$

$x^2 = 36 + 64$

$x^2 =100$

Take positive square root of both sides

$x = \sqrt{ 100$

$x = 10$

5 0

3 years ago

Hi please help me thanksss<br>pls provide workings too thanks :)

anygoal [31]

Step-by-step explanation:

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Class 11

>>Maths

>>Trigonometric Functions

>>Trigonometric Equations

>>Solve for x: sin x + sin 2x...

Question

Bookmark

Solve for x:sinx+sin2x+sin3x=cosx+cos2x+cos3x in the interval 0≤x≤2π.

Medium

Solution

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Verified by Toppr

We write the given equation as

(sinx+sin3x)+sin2x=(cosx+cos3x)+cos2x

or 2sin2xcosx+sin2x=2cos2xcosx+cos2x

or sin2x(2cosx+1)=cos2x(2cosx+1)

or (sin2x−cos2x)(2cosx+1)=0

∴sin2x−cos2x=0 or 2cosx+1=0

If sin2x−cos2x=0, then tan2x=1,

Hence 2x=nπ+π/4

or x=(4n+1)

8

π

.(1)

If 2cosx+1=0, then cosx=−1/2 (2)

∴x=2nπ±

3

2π

or x=

3

6n±2

π

We seek values of x in the interval 0≤x≤2π

In this interval (1) gives

x=π/8,5π/8,9π/8,13π/8. (n=0,1,2,3)

and (2) gives x=2π/3,4π/3. (for n=0,1)

Thus we get the answer

x=π/8,5π/8,2π/3,9π/8,4π/3,13π/8.

4 0

2 years ago

is 0.75 less or greater than 7 over 8

Oksanka [162]

0.75 is 3/4 or 6/8, which is less than 7/8

4 0

3 years ago

Read 2 more answers

x2 + y2 − 4x + 12y − 20 = 0 (x − 6)2 + (y − 4)2 = 56 x2 + y2 + 6x − 8y − 10 = 0 (x − 2)2 + (y + 6)2 = 60 3x2 + 3y2 + 12x + 18y −

snow_tiger [21]

For this case, what we must do is fill squares in all the expressions until we find the correct result.
We have then:

x2 + y2 − 4x + 12y − 20 = 0 x2 + y2 − 4x + 12y = 20
x2 − 4x + y2 + 12y = 20
x2 − 4x + (12/2)^2 + y2 + 12y + (-4/2)^2 = 20 + (12/2)^2 + (-4/2)^2
x2 − 4x + (6)^2 + y2 + 12y + (-2)^2 = 20 + (6)^2 + (-2)^2
x2 − 4x + 36 + y2 + 12y + 4 = 20 + 36 + 4
(x − 2)2 + (y + 6)2 = 60

3x2 + 3y2 + 12x + 18y − 15 = 0
x2 + y2 + 4x + 6y − 5 = 0
x2 + y2 + 4x + 6y = 5
x2 + 4x + (4/2)^2 + y2 + 6y + (6/2)^2 = 5 + (4/2)^2 + (6/2)^2
x2 + 4x + (2)^2 + y2 + 6y + (3)^2 = 5 + (2)^2 + (3)^2
x2 + 4x + 4 + y2 + 6y + 9 = 5 + 4 + 9
(x + 2)2 + (y + 3)2 = 18

2x2 + 2y2 − 24x − 16y − 8 = 0
x2 + y2 − 12x − 8y − 4 = 0
x2 + y2 − 12x − 8y = 4
x2 − 12x + (-12/2)^2 + y2 − 8y + (-8/2)^2 = 4 + (-12/2)^2 + (-8/2)^2
x2 − 12x + (-6)^2 + y2 − 8y + (-4)^2 = 4 + (-6)^2 + (-4)^2
x2 − 12x + 36 + y2 − 8y + 16 = 4 + 36 + 16
(x − 6)2 + (y − 4)2 = 56

x2 + y2 + 2x − 12y − 9 = 0
x2 + y2 + 2x - 12y = 9
x2 + 2x + y2 - 12y = 9
x2 + 2x + (2/2)^2 + y2 - 12y + (-12/2)^2 = 9 + (2/2)^2 + (-12/2)^2
x2 + 2x + (1)^2 + y2 - 12y + (-6)^2 = 9 + (1)^2 + (-6)^2
x2 + 2x + 1 + y2 - 12y + 36 = 9 + 1 + 36
(x + 1)2 + (y − 6)2 = 46

7 0

3 years ago

3x + 2(x - k) + 3 = 5x - 5