Use a property of equality to solve this equation 4.5x=18

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madam [21]

Answer:

B

Step-by-step explanation:

8 0

3 years ago

URGENT

Svetlanka [38]

let's solve for y :

$2y - 2x = 4$

$2(y - x) = 4$

$y - x = \dfrac{4}{2}$

$y - x = 2$

$y = x + 2$

5 0

3 years ago

Read 2 more answers

I need help ASAP <br> Integrated math II

Molodets [167]

Answer:

$\boxed{ \bold{ \sf{x = 16}}}$

$\boxed{ \bold{ \sf{m < AFD = 82}}}$

Step-by-step explanation:

$\sf{5x + 18 = 3x + 50}$

( Being vertically opposite angles)

Vertically opposite angles are always equal.

Move variable to L.H.S and change it's sign

Similarly, Move constant to R.H.S and change it's sign

⇒ $\sf{5x - 3x = 50 - 18}$

Collect like terms

⇒ $\sf{2x = 50 - 18}$

Subtract 18 from 50

⇒ $\sf{2x = 32}$

Divide both sides of the equation by 2

⇒ $\sf{ \frac{2x}{2} = \frac{32}{2} }$

Calculate

⇒ $\sf{x = 16}$

The value of x is 16

Now, let's find value of m<AFD :

$\sf{m < afd + 5x + 18 = 180}$ ( sum of angle in straight line )

plug the value of x

⇒ $\sf{m < AFD + 5 \times 16 + 18 = 180}$

Multiply the numbers

⇒ $\sf{m < AFD + 80 + 18 = 180}$

Add the numbers

⇒ $\sf{m < AFD + 98 = 180}$

Move constant to R.H.S and change it's sign

⇒ $\sf{m < AFD = 180 - 98}$

Subtract 98 from 180

⇒ $\sf{m < AFD = 82}$

Value of m<AFD = 82

Hope I helped!

Best regards!!

3 0

3 years ago

The figure below shows a straight line AB intersected by another straight line t:

BaLLatris [955]

When two straight lines intersect, the pairs of nonadjacent angles in opposite posi-tions are known as vertical angles.

If a segment AB is intersected by a transversal labeled t, then ∠1 and ∠3 and ∠2 and ∠4 are vertically angles formed by the transversal t on the segment AB.

Angles ∠1 and ∠2 can be described as adjacent and supplementary angles, so

$m\angle 1+m\angle 2=180^{\circ}$ .

Angles ∠3 and ∠2 can be also described as adjacent and supplementary angles, so

$m\angle 3+m\angle 2=180^{\circ}$ .

Subtract from the first equation the second equation:

$m\angle 1+m\angle 2-(m\angle 3+m\angle 2)=180^{\circ}-180^{\circ},\\ m\angle 1-m\angle 3=0,\\ m\angle 1=m\angle 3$ .

Similarly you can prove that $m\angle 2=m\angle 4$ .

4 0

3 years ago

Find equations of the spheres with center(3, −4, 5) that touch the following planes.a. xy-plane b. yz- plane c. xz-plane

postnew [5]

Answer:

(a) (x - 3)² + (y + 4)² + (z - 5)² = 25

(b) (x - 3)² + (y + 4)² + (z - 5)² = 9

(c) (x - 3)² + (y + 4)² + (z - 5)² = 16

Step-by-step explanation:

The equation of a sphere is given by:

(x - x₀)² + (y - y₀)² + (z - z₀)² = r² ---------------(i)

Where;

(x₀, y₀, z₀) is the center of the sphere

r is the radius of the sphere

Given:

Sphere centered at (3, -4, 5)

=> (x₀, y₀, z₀) = (3, -4, 5)

(a) To get the equation of the sphere when it touches the xy-plane, we do the following:

i. Since the sphere touches the xy-plane, it means the z-component of its centre is 0.

Therefore, we have the sphere now centered at (3, -4, 0).

Using the distance formula, we can get the distance d, between the initial points (3, -4, 5) and the new points (3, -4, 0) as follows;

$d = \sqrt{(3-3)^2+ (-4 - (-4))^2 + (0-5)^2}$

$d = \sqrt{(3-3)^2+ (-4 + 4)^2 + (0-5)^2}$

$d = \sqrt{(0)^2+ (0)^2 + (-5)^2}$

$d = \sqrt{(25)}$

d = 5

This distance is the radius of the sphere at that point. i.e r = 5

Now substitute this value r = 5 into the general equation of a sphere given in equation (i) above as follows;

(x - 3)² + (y - (-4))² + (z - 5)² = 5²

(x - 3)² + (y + 4)² + (z - 5)² = 25

Therefore, the equation of the sphere when it touches the xy plane is:

(x - 3)² + (y + 4)² + (z - 5)² = 25

(b) To get the equation of the sphere when it touches the yz-plane, we do the following:

i. Since the sphere touches the yz-plane, it means the x-component of its centre is 0.

Therefore, we have the sphere now centered at (0, -4, 5).

Using the distance formula, we can get the distance d, between the initial points (3, -4, 5) and the new points (0, -4, 5) as follows;

$d = \sqrt{(0-3)^2+ (-4 - (-4))^2 + (5-5)^2}$

$d = \sqrt{(-3)^2+ (-4 + 4)^2 + (5-5)^2}$

$d = \sqrt{(-3)^2 + (0)^2+ (0)^2}$

$d = \sqrt{(9)}$

d = 3

This distance is the radius of the sphere at that point. i.e r = 3

Now substitute this value r = 3 into the general equation of a sphere given in equation (i) above as follows;

(x - 3)² + (y - (-4))² + (z - 5)² = 3²

(x - 3)² + (y + 4)² + (z - 5)² = 9

Therefore, the equation of the sphere when it touches the yz plane is:

(x - 3)² + (y + 4)² + (z - 5)² = 9

(b) To get the equation of the sphere when it touches the xz-plane, we do the following:

i. Since the sphere touches the xz-plane, it means the y-component of its centre is 0.

Therefore, we have the sphere now centered at (3, 0, 5).

Using the distance formula, we can get the distance d, between the initial points (3, -4, 5) and the new points (3, 0, 5) as follows;

$d = \sqrt{(3-3)^2+ (0 - (-4))^2 + (5-5)^2}$

$d = \sqrt{(3-3)^2+ (0+4)^2 + (5-5)^2}$

$d = \sqrt{(0)^2 + (4)^2+ (0)^2}$

$d = \sqrt{(16)}$

d = 4

This distance is the radius of the sphere at that point. i.e r = 4

Now substitute this value r = 4 into the general equation of a sphere given in equation (i) above as follows;

(x - 3)² + (y - (-4))² + (z - 5)² = 4²

(x - 3)² + (y + 4)² + (z - 5)² = 16

Therefore, the equation of the sphere when it touches the xz plane is:

(x - 3)² + (y + 4)² + (z - 5)² = 16

3 0

3 years ago