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3 years ago

Someone plz help me

Mathematics

2 answers:

Alchen [17]3 years ago

6 0

If each triangle in the figure is equilateral, the perimeter of the figure would be 18 inches.

Setler79 [48]3 years ago

5 0

Answer:

18 inches

Step-by-step explanation:

1. Multiply 3 by 6

3 x 6 = 18

2. Get answer of 18

18 inches

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Which of the following sets are subspaces of R3 ?

Ratling [72]

Answer:

The following are the solution to the given points:

Step-by-step explanation:

for point A:

$\to A={(x,y,z)|3x+8y-5z=2} \\\\\to for(x_1, y_1, z_1),(x_2, y_2, z_2) \varepsilon A\\\\ a(x_1, y_1, z_1)+b(x_2, y_2, z_2) = (ax_1+bx_2,ay_1+by_2,az_1+bz_2)$

$=3(aX_l +bX_2) + 8(ay_1 + by_2) — 5(az_1+bz_2)\\\\=a(3X_l+8y_1- 5z_1)+b (3X_2+8y_2—5z_2)\\\\=2(a+b)$

The set A is not part of the subspace $R^3$

for point B:

$\to B={(x,y,z)|-4x-9y+7z=0}\\\\\to for(x_1,y_1,z_1),(x_2, y_2, z_2) \varepsilon B \\\\\to a(x_1, y_1, z_1)+b(x_2, y_2, z_2) = (ax_1+bx_2,ay_1+by_2,az_1+bz_2)$

$=-4(aX_l +bX_2) -9(ay_1 + by_2) +7(az_1+bz_2)\\\\=a(-4X_l-9y_1+7z_1)+b (-4X_2-9y_2+7z_2)\\\\=0$

$\to a(x_1,y_1,z_1)+b(x_2, y_2, z_2) \varepsilon B$

The set B is part of the subspace $R^3$

for point C: $\to C={(x,y,z)|x$

In this, the scalar multiplication can't behold

$\to for (-2,-1,2) \varepsilon C$

$\to -1(-2,-1,2)= (2,1,-1)$ ∉ C

this inequality is not hold

The set C is not a part of the subspace $R^3$

for point D:

$\to D={(-4,y,z)|\ y,\ z \ arbitrary \ numbers)$

The scalar multiplication s is not to hold

$\to for (-4, 1,2)\varepsilon D\\\\\to -1(-4,1,2) = (4,-1,-2)$ ∉ D

this is an inequality, which is not hold

The set D is not part of the subspace $R^3$

For point E:

$\to E= {(x,0,0)}|x \ is \ arbitrary) \\\\\to for (x_1,0 ,0) ,(x_{2},0 ,0) \varepsilon E \\\\\to a(x_1,0,0) +b(x_{2},0,0)= (ax_1+bx_2,0,0)\\$

The $x_1, x_2$ is the arbitrary, in which $ax_1+bx_2$ is arbitrary

$\to a(x_1,0,0)+b(x_2,0,0) \varepsilon E$

The set E is the part of the subspace $R^3$

For point F:

$\to F= {(-2x,-3x,-8x)}|x \ is \ arbitrary) \\\\\to for (-2x_1,-3x_1,-8x_1),(-2x_2,-3x_2,-8x_2)\varepsilon F \\\\\to a(-2x_1,-3x_1,-8x_1) +b(-2x_1,-3x_1,-8x_1)= (-2(ax_1+bx_2),-3(ax_1+bx_2),-8(ax_1+bx_2))$

The $x_1, x_2$ arbitrary so, they have $ax_1+bx_2$ as the arbitrary $\to a(-2x_1,-3x_1,-8x_1)+b(-2x_2,-3x_2,-8x_2) \varepsilon F$

The set F is the subspace of $R^3$

5 0

3 years ago

Please answer this question ..

Yanka [14]

Answer:

plz mark brainliest if it is correct ✌️✌️

6 0

3 years ago

Read 2 more answers

Identified the pattern for the following sequence find the next three terms in the sequence -1 3 -9 27

Nastasia [14]

Each term after the first is obtained by multiplying the previous one by -3 so the answer is -81 , 243

7 0

3 years ago

Fatima wants to find the value of sine theta, given Cotangent theta = four-sevenths. Which identity would be best for Fatima to

marysya [2.9K]

Answer: D, or 1+cotangent squared theta= cosecant squared theta

Step-by-step explanation: Took the test it is right

3 0

3 years ago

Read 2 more answers

x f(x) 3.0 4.0 3.5 -0.2 4.0 -0.8 4.5 0.1 5.0 0.6 5.5 0.7 For the given table of values for a polynomial function, where must the

torisob [31]

Answer:

A. between 3.0 and 3.5 and between 4.0 and 4.5

Step-by-step explanation:

The idea here is that if a function is continuous and you have points on opposite sides of the x-axis, then there must be a zero-crossing between those points. As the table and graph show, ...

... (3.0, 4.0) and (3.5, -0.2)

are points on opposite sides of the x-axis. So, there must be a zero crossing between x=3.0 and x=3.5.

Likewise, ...

... (4.0, -0.8) and (4.5, 0.1)

are points on opposite sides of the x-axis. So, there must be a zero-crossing between x=4.0 and x=4.5

The appropriate answer choice lists both of these possible zero crossings.

5 0

3 years ago