How many equal sections divide a directed line segment if it is to be partitioned with a 5:7 ratio

Mathematics

2 answers:

Evgen [1.6K]3 years ago

7 0

Answer: The line segment can be divided into 12 equal sections.

Step-by-step explanation: We are given to find the number of equal sections that divide a directed line segment if it is to be partitioned with a 5 : 7 ratio.

We know that

if a line directed line segment is partitioned in the ratio a : b, then we can divide the line segment in (a + b) equal sections.

So, the number of equal sections in which the given line segment can be divided is given by

$N_s=5+7=12.$

Thus, the line segment can be divided into 12 equal sections.

natima [27]3 years ago

5 0

Well you see partition means to separate or divide. A line segment with end points a and b can be partioneded by another point by a given ratio, and a ratio is a comparison of two numbers. For other ratios besides 1:1 , it is necessary to determine the total number of parts that the line segment must be divided into. So in your case the point is 5:7 so you to determine the point that divides your line segment you just simply add the figure 5 a and the figure 7 together and you should end up with the sum of 12.

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The whole sales price for a chair is $165. A certain furniture store marks up the wholesalers price by 15%. Find the price of th

vichka [17]

Answer:

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Step-by-step explanation:

Whenever you want to find the percent of something, you need to turn the percent into a decimal. In this case, we need to find 115% of the wholesale price because there is a markup, so the cost will be higher.

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3 years ago

Write the equation of a polynomial of degree 3, with zeros 1, 2 and -1 where f(0)=2

drek231 [11]

<u>Answer:</u>

The equation of a polynomial of degree 3, with zeros 1, 2 and -1 is $x^{3}-2 x^{2}-x+2=0$

<u>Solution:</u>

Given, the polynomial has degree 3 and roots as 1, 2, and -1. And f(0) = 2.

We have to find the equation of the above polynomial.

We know that, general equation of 3rd degree polynomial is

$F(x)=x^{3}-(a+b+c) x^{2}+(a b+b c+a c) x-a b c=0$

where a, b, c are roots of the polynomial.

Here in our problem, a = 1, b = 2, c = -1.

Substitute the above values in f(x)

$F(x)=x^{3}-(1+2+(-1)) x^{2}+(1 \times 2+2(-1)+1(-1)) x-1 \times 2 \times(-1)=0$

$\begin{array}{l}{\rightarrow x^{3}-(3-1) x^{2}+(2-2-1) x-(-2)=0} \\ {\rightarrow x^{3}-(2) x^{2}+(-1) x-(-2)=0} \\ {\rightarrow x^{3}-2 x^{2}-x+2=0}\end{array}$