Which of the following is true given that 2.1 < 2.8?

Mathematics

2 answers:

Firlakuza [10]3 years ago

8 0

Answer:1 is to the left of 2.8 on a horizontal number line

Step-by-step explanation:

sattari [20]3 years ago

5 0

Answer:

$\large \boxed{\mathrm{2.1 \ is \ to \ the \ left \ of \ 2.8 \ on \ a \ horizontal \ number \ line}}$

Step-by-step explanation:

$2.1$

$\sf 2.1 \ is \ lesser \ than \ 2.8.$

$\sf A \ number \ that \ is \ lesser \ than \ another \ number \ is \ to \ the \ left \\ \ of \ that \ number \ on \ a \ horizontal \ number \ line.$

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Suppose a and b are both non zero real numbers. Find real numbers c and d such that 1/a+ib= c+id

Thepotemich [5.8K]

$\begin{gathered} c=\frac{a}{a^2+b^2} \\ d=\frac{-b}{a^2+b^2} \end{gathered}$

Explanation

$\frac{1}{a+bi}=c+di$

Step 1

multiplicate by the conjugate

$\begin{gathered} \frac{1}{a+bi}\cdot\frac{a-bi}{a+bi}=\frac{a-bi}{(a+bi)(a-bi)}=\frac{a-bi}{a^2-(bi)^2} \\ \frac{1}{a+bi}\cdot\frac{a-bi}{a+bi}=\frac{a-bi}{(a+bi)(a-bi)}=\frac{a-bi}{a^2-(-b^2)}=\frac{a-bi}{a^2+b^2} \end{gathered}$

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$\begin{gathered} \frac{1}{a+bi}=\frac{a-bi}{a^2+b^2}=\frac{a}{a^2+b^2}-\frac{b}{a^2+b^2}i \\ \frac{a}{a^2+b^2}-\frac{b}{a^2+b^2}i=c+di \\ so \\ \end{gathered}$ $\begin{gathered} c=\frac{a}{a^2+b^2} \\ d=\frac{-b}{a^2+b^2} \end{gathered}$

I hope this helsp you

6 0

1 year ago

Which of the following statements are true ? Check all that apply

Elden [556K]

Answer:

Options A) and E)

Step-by-step explanation:

If you use Ruffini's method for polynomials, you can find the roots.

The given picture shows us the first root of the polynomial $5x^{2}-16x+12=0$ wich is 2.