MATHEMATICS HIGH SCHOOL

Add the polynomials (7x3−2x2−12)+(−3x3−8x2+10x)

Answers

Answer 1

Answer:

Answer:

Addition:::::4x^3-10x^2+10x-12

Answer 2

Answer:

See below

Step-by-step explanation:

● (7x^3 - 2x^2 -12) + (-3x^3 - 8x^2 +10x)

● 7x^3 - 2x^2 - 12- 3x^3 - 8x^2 + 10x

Combine like terms

● 7x^3 - 3x^3 -2x^2 - 8x^2 -12 + 10x

● 4x^3 - 10x^3 +10x -12

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Add 1/6+3/4+2/3. Simplify the answer and write it as a mixed number, if possible. A)
1 14/24

B)
1/2

C)
19/12

D)
1 7/12

Answers

To simplify the answer and write it as a mixed number is D)1 7/12

How to Add 1/6+3/4+2/3.

To add the fractions 1/6, 3/4, and 2/3, we need to find a common denominator.

The common denominator for 6, 4, and 3 is 12.

Converting the fractions to have a denominator of 12:

1/6 = 2/12

3/4 = 9/12

2/3 = 8/12

Now we can add the fractions:

2/12 + 9/12 + 8/12 = 19/12

The sum of the fractions is 19/12.

To simplify the answer and write it as a mixed number, we divide the numerator (19) by the denominator (12):

19 ÷ 12 = 1 remainder 7

Therefore, the answer is 1 7/12.

Learn more about common denominator. at brainly.com/question/1098206

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Find the lowest common denominator for the 3 fractions. It is 12.
Rewrite each fraction with the denominator 12.
1/6 = 2/12
3/4 = 9/12
2/3 = 8/12
2/12 + 9/12 + 8/12 = 19/12
This can be written as the mixed number 1 7/12.
The answer is D) 1 7/12

A rectangle has an area of 40 square units. The length is 6 units greater than the width.

Answers

Let

x------> the length of the rectangle

y------> the width of the rectangle

A-----> area of the rectangle

we know that

The area of rectangle is equal to

$A=x*y\nA=40\ units^(2)$

$40=x*y$ --------> equation $1$

$x=6+y$ --------> equation $2$

Substitute equation $2$ in equation $1$

$40=[6+y]*y\n 40=6y+y^(2) \ny^(2) +6y-40=0$

using a graph tool-----> to resolve the second order equation

see the attached figure

the solution is

$y=4$

Find the value of x

$x=6+y\n \n x=6+4=10\ units$

therefore

the answer is

The length of the rectangle is equal to $10\ units$

The width of the rectangle is equal to $4 \ units$

possible dimensions:
40 x 1
20 x 2
4 x 10
5 x 8

since 10 is 6 more than 4, the length is 10 and the width is 4

What is the area of length being 32 and width 24