Write a decimal that is less than 3.45 but greater than 3.0. helppp!

Mathematics

2 answers:

Lapatulllka [165]3 years ago

5 0

Answer:

any decimal between 3.0 and 3.45. You can,use 3.25 if u don't wanna think about it

gtnhenbr [62]3 years ago

3 0

Answer:There are mutipal answers but here’s one 3.1

Step-by-step explanation:

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PLEEEEEEEAAAAAAASSSSSSSSSSEEEEEEEEEEE help me!!!!!!!!!!

defon

Answer:

$\boxed{\sf \ \ \ \dfrac{18}{(x-7)(x+12)(x+30)} \ \ \ }$

Step-by-step explanation:

hello,

first of all we will study the quadratic expressions

we can write that, (the different answers provide good clues :-) )

$x^2+5x-84=(x-7)(x+12)$

and

$x^2+23x-210=(x-7)(x+30)$

so first of all as we cannot divide by 0 we need to take x different from 7, -12 and -30 and then we can write

$\dfrac{1}{x^2+5x-48}-\dfrac{1}{x^2+23x-210}=\dfrac{1}{(x-7)(x+12)}-\dfrac{1}{(x-7)(x+30)}\\\\=\dfrac{(x+30) -(x+12)}{(x-7)(x+12)(x+30)}=\dfrac{x+30-x-12}{(x-7)(x+12)(x+30)}\\\\=\dfrac{18}{(x-7)(x+12)(x+30)}$

hope this helps

7 0

3 years ago

For the matrices below. Find all (real) eigenvalues. A= [7 9] [0 9]

Jet001 [13]

Answer:

The answer is "9 and 7".

Step-by-step explanation:

Given:

$A=\left[\begin{array}{cc}7&9\\0&9\end{array}\right]$

Using formula:

$|A-\lambda \cdot I|= 0\\\\$

$\to |\left[\begin{array}{cc}7&9\\0&9\end{array}\right]-\lambda \left[\begin{array}{cc}1&0\\0&1\end{array}\right] |=0\\\\\\ \to |\left[\begin{array}{cc}7&9\\0&9\end{array}\right]- \left[\begin{array}{cc}\lambda&0\\0&\lambda\end{array}\right] |=0\\\\\\\to|\left[\begin{array}{cc}7-\lambda &9\\0&9-\lambda\end{array}\right]|=0\\\\\\\to|(7-\lambda)(9-\lambda)|=0\\\\\to (7-\lambda)(9-\lambda)=0\\\\\to 7-\lambda=0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 9-\lambda=0\\\\$