Mathematics: One and seven eighths times three and one third how would you do this problem?

One and seven eighths times three and one third

Ответ:

fabio37

10.01.2022

Mathematics

15/24

Step-by-step explanation:

1) Change the mixed numbers into improper fractions: 15/8 and leave 1/3 alone.

2) Then multiply the numerators and denominators: 15/24

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Ответ:

ayoismeisalex

26.02.2022

Mathematics

Yes , the points (-4,-1), (2,1) and (11,4) are collinear.

Step-by-step explanation:

We know that if three points (x_1,y_1),(x_2,y_2) and (x_3,y_3) are collinear, then their area must be zero.

The area of triangle passes through points (x_1,y_1),(x_2,y_2) and (x_3,y_3) is given by :-

$\text{Area}=\dfrac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|$

Given points : (-4,-1), (2,1) and (11,4)

Then, the area of ΔABC will be :-

$\text{Area}=\dfrac{1}{2}|-4(1-4)+(2)(4-(-1))+(11)(-1-1)|\\\\\Rightarrow\text{Area}=\dfrac{1}{2}|-4(-3)+(2)(5)+(11)(-2)||\\\\\Rightarrow\text{Area}=\dfrac{1}{2}|12+10-22|\\\\\Rightarrow\text{Area}=\dfrac{1}{2}|0|=0$

Hence, the points (-4,-1), (2,1) and (11,4) are collinear.

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One and seven eighths times three and one third how would you do this problem?

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