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jarptica [38.1K]
2 years ago
14

Divide : (8a^2b^4 - 20a^4b^5 + 48a^6b^6) by 4a^2b^4​

Mathematics
1 answer:
Leno4ka [110]2 years ago
4 0

Hey there ! please refer to the attachment .

I hope its help you !

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I GIVE BRAINLIEST FOR EXPLANATION AND CORRECT ANSWER EXTRA POINTS
madreJ [45]

Answer:x=4.2643

Step-by-step explanation:

tan35degrees=x/9

9(tan35degrees)=x

4.2643=x

6 0
3 years ago
Read 2 more answers
107,600 115,220 104,910 median, mode, mean
san4es73 [151]

Answer:

<u>1. Mean =  342.7 (Rounding to the nearest tenth)</u>

<u>2. Median =  167.5 </u>

<u>3. Mode = There isn't a mode for this set of numbers because there isn't a data value that occur more than once. </u>

Step-by-step explanation:

Given this set of numbers: 107, 600,  115, 220,  104, 910, find out these measures of central tendency:

1. Mean = 107 + 600  + 115 + 220 + 104 + 910/6 = <u>342.7</u> (Rounding to the nearest tenth)

2. Median. In this case, we calculate it as the average between the third and the fourth element, this way:

115 + 220 =335

335/2 = <u>167.5 </u>

3. Mode = <u>There isn't a mode for this set of numbers because there isn't a data value that occur more than once. All the data values occur only once.</u>

5 0
3 years ago
Two families are planning a trip to Disney. The Smith family bought tickets for 2 adults and 3 children for $557. The Jones fami
emmainna [20.7K]

<em>An adult ticket costs $205 and a child ticket costs $49.</em>

<h2>Explanation:</h2>

Hello! Recall you have to write complete questions in order to find exact answers. Here I'll assume the complete question as:

<em>Two families are planning a trip to Disney. The Smith family bought tickets for 2 adults and 3 children for $557. The Jones family bought tickets for 2 adults and 1 child </em><em>for $459</em><em>. How much does and adult and child ticket cost?</em>

To solve this problem, we need to write a system of linear equations in two variables. So, we know some facts:

  • Two families are planning a trip to Disney.
  • The Smith family bought tickets for 2 adults and 3 children for $557.
  • The Jones family bought tickets for 2 adults and 1 child for $459.

Let:

x:Cost \ of \ ticket \ per \ adult \\ \\ y: Cost \ of \ ticket \ per \ child

For the Smith family:

Cost for the 2 adults:

2x

Cost for the 3 children:

3y

Total cost:

2x+3y=557

For the Jones family:

Cost for the 2 adults:

2x

Cost for the 1 child:

y

Total cost:

2x+y=459

So we have the following system of linear equations:

\begin{array}{c}(1)\\(2)\end{array}\left\{ \begin{array}{c}2x+3y=557\\2x+y=459\end{array}\right.

Subtracting (2) from (1):

\begin{array}{c}(1)\\(2)\end{array}\left\{ \begin{array}{c}2x+3y=557\\-(2x+y=459)\end{array}\right. \\ \\ \\ \begin{array}{c}(1)\\(2)\end{array}\left\{ \begin{array}{c}2x+3y=557\\-2x-y=-459\end{array}\right. \\ \\ \\ (2x-2x)+(3y-y)=557-459 \\ \\ 2y=98 \\ \\ y=49 \\ \\ \\ Finding \ x \ from \ (1): \\ \\ 2x+3(49)=557 \\ \\ 2x=557-147 \\ \\ 2x=410 \\ \\ x=205

Finally, <em>an adult ticket costs $205 and a child ticket costs $49.</em>

<em></em>

<h2>Learn more:</h2>

System of linear equations: brainly.com/question/13799715

#LearnWithBrainly

3 0
4 years ago
One third of the difference of 20 and 2 <br> 1/3 x (20-2)
Alchen [17]
ANSWER:
6

Explanation:
20-2 = 18
1/3 = 0.33333 (3 repeating)
1/3 x 18 = 6
4 0
3 years ago
How many elements are in the union of four sets if
12345 [234]

Answer:

20

Step-by-step explanation:

Let A, B, C and D sets.

Let |A| the number of elements of A.

Remember, the quantity of elements that have the set A\cup B\cup C\cup D is

|A\cup B\cup C\cup D|=|A|+ |B|+ |C|+|D|-|A\cap B|-|A\cap C|-|A\cap D|-|B\cap C|-|B\cap D|-|C\cap D|-|A\cap B\cap C|-|A\cap B\cap D|-|B\cap C\cap D|-|A\cap B\cap C \cap D|

1. Since each set has 100 elements,

2.  each pair of the sets shares 50 elements

3. each three of the sets share 25 elements,

4. the four sets share 5 elements

Then

|A\cup B\cup C\cup D|=100+100+100+100-50-50-50-50-50-50-25-25-25-5=20

So, there are 20 elements in the union of the sets A, B, C and D

6 0
3 years ago
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