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3 years ago

Can someone explain how to get the right answers thx

Mathematics

1 answer:

Free_Kalibri [48]3 years ago

4 0

Idk exactly what those angles are called. I think supplementary interior. Which means they add up to 180°
So, (4x-2)+(6x-8)=180
Simplify. 10x-10=180
Now we need to get x by itself so add 10 to both sides to get 10x=190
Then divide by 10 to get x by itself x=190/10
So x=19

Hope this helped!!!

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Like charges repel and I like charges attract Coulomb’s law states that the force F of attraction or repulsion between two charg

olga_2 [115]

Step-by-step explanation:

For the charges that have same sign of charges will repel each other while for the charges that have different charges will attract each other. So, we can say that like charges repel and unlike charges attract each other.

The Coulomb's law of attraction of repulsion states that force between charges is directly proportion to the product of charges and inversely proportional to the square of distance between them. Mathematically, it is given by :

$F=\dfrac{kq_1q_2}{r^2}$

Hence, all the given statements are true.

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3 years ago

A) 5<br> 11 + 4<br> 3<br> add the following rational number

mel-nik [20]

Step-by-step explanation:

the question is not correct

edit it and I will answer u in comments

5 0

2 years ago

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

Rewrite in radical form (-243)2/5

nadezda [96]

Answer:

243 and 2 over 5

Multiply

(−243) and 2 over 5 . −486 over 5.

Move the negative in front of the fraction.

−486 over 5

Exact Form:

-486 over 5

Decimal Form:

-97.2

Mixed Number Form:

-97 and 1 over 5

7 0

3 years ago

Does anyone have a website to cheat on

lorasvet [3.4K]

Answer:

No i do for other stuff but not this and the answer is 18

Step-by-step explanation:

5 0

3 years ago