"A gumball machine is in the shape of a sphere with a radius of 6 inches. A store manager wants to fill up the machine with mini

189 tickets were purchased on Saturday. The ratio of children's tickets to adult tickets is 8:1, indicating that 8 times as many children's tickets were sold compared to adult tickets. Let c represent the number of children's tickets and a the number of adult tickets. Therefore, 8a = a + 147. By subtracting a from both sides, we find 7a = 147. Upon dividing both sides by 7, we find a = 21 adult tickets. By multiplying the number of adult tickets by 8, we discover that 21 * 8 = 168 children's tickets. Adding these together gives a total of 168 + 21 = 189 tickets sold on Saturday.

Answer:

mCEA = 90ᴼ, as CEA forms a right angle, and by definition, right angles measure 90ᴼ.

The angle CEF is classified as a straight angle as it combines two right angles (CEA and AEF), equating to 180ᴼ altogether. Straight lines are defined to measure 180ᴼ.

AEF is determined to be a right angle as CEA is already a right angle, and since CEF is a straight line, AEF must also be a right angle.

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Response:

The measure of mHLK is "(204)°".

Step-by-step breakdown:

Given values include:

mJI = (3x+2)°

mHLK = (15x-36)°

and,

m∠HML = (8x-1)°

then,

What is mHLK?

Now,

Utilizing the chord-chord angle formula, we find

Inserting the known values into the equation gives us

⇒  

By carrying out cross-multiplication, we arrive at

⇒  

⇒  

By subtracting "18x" from both sides, we obtain

⇒  

⇒  

Upon adding "2" to both sides, we end up with

⇒  

⇒  

⇒  

⇒  

By substituting the value of "x" into mHLK = (15x-36)°, we calculate

⇒ (15x-36)° = (15×16-36)°

⇒                = (240-36)°

⇒                = (204)°

Thus, mHLK = (204)°

Answer:

Step-by-step explanation:

Given:

KL ║ NM,

LM = 45

m∠M = 50°

KN ⊥ NM  

NL ⊥ LM

To determine: KN and KL

1. Analyzing triangle NLM, we see it is a right triangle due to NL ⊥ LM. In this context,

LM = 45

m∠M = 50°

Consequently,

It is also true that

(angles LNM and M are complementary).

2. Now considering triangle NKL, it also forms a right triangle as KN ⊥ NM. Within this triangle,

(alternate interior angles)

(angles KNL and KLN are complementary).

Thus,

and