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VARVARA [1.3K]
3 years ago
10

Plz help me with 3,4,and 6.Also what are the numbers on the sides called?

Mathematics
1 answer:
Elis [28]3 years ago
4 0
3. 1*2* 1/3= 2/3 yd

4. 1/4 * 1/8 *2= 1/16in

6. 12 *4= 48ft

the numbers on the sides are the length, width and height. you multiply them all together to find the volume
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HOW do you do 126% of 350?
podryga [215]
100% of 350 is 350.
What you need to do is find 26% of 350, and add it to 350 (100%). 

To accomplish that, change 26% to a decimal by moving the decimal point two places to the left. 26% = 0.26

Next, multiply 350 by 0.26 

350 • 0.26 = 91

Finally, just add 350 and 91 to get 126%

350 + 91 = 441

126% of 350 = 441
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3 years ago
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The local diner offers a meal combination consisting of an appetizer, a soup, a main course, and a dessert. There are three appe
ololo11 [35]

Answer:

number of ways to chose a three course meal is 90

Step-by-step explanation:

Given the data in the question;

Number of appetizer = 3

Number of soups = 5

Number of main courses = 3

Number of desserts = 3

Now,

Number of ways that appetizer can be chosen = ³C₁ = 3!/(1!(3-1)!) = 3

Number of ways that soups can be chosen = ⁵C₁ = 5!/(1!(5-1)!) = 5

Number of ways that main courses can be chosen = ³C₁ = 3!/(1!(3-1)!) = 3

Number of ways that desserts can be chosen = ³C₁ = 3!/(1!(3-1)!) = 3

So,

First we chose an appetizer, soup and main course.

Number of ways will be;

⇒ 3 × 5 × 3 = 45

Next,  we chose a dessert, a soup & a main course.

Number of ways will be;

⇒ 3 × 5 × 3 = 45

Total number of ways to chose a three course meal

⇒ 45 + 45 = 90 ways

Therefore, number of ways to chose a three course meal is 90

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In circle O, diameter AB has been drawn. Locate point C anywhere between A and B (on either side of the diameter). Draw in AC an
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Answer:

Right Triangle.

Step-by-step explanation:

Given AB is a diameter of a circle O.  

Point C is located on the circle's circumference as shown in the attached diagram.

\angle AOB=180^\circ

Theorem: The angle at the centre is twice the angle at the circumference.

The angle formed at centre O by arc AB is \angle AOB

The angle formed at the circumference at point C by arc AB is \angle ACB

By the theorem above:

\angle AOB=2*\angle ACB\\180^\circ=2*\angle ACB\\\angle ACB=90^\circ

Therefore, Triangle ACB is a Right Triangle with the right angle at C.

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