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stiks02 [169]
3 years ago
11

1.Prove that the roots of x^2 + (1 - k).x+k-3 = 0 are real for all real values of k​

Mathematics
1 answer:
hammer [34]3 years ago
4 0

Step-by-step explanation:

{x}^{2}  + (1 - k)x + k - 3 = 0

{x}^{2}    + x - 3 = 0

Use the discramnt formula.

{ {b}^{2} - 4ac }

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Out of numbers: 5, 7, 21, 25, 28, 35, 42, 56, 75, and 80, choose those which are not factors of 42
Blizzard [7]

Answer:

5, 25, 28, 35, 56, 75 80 are not the factors of 42

6 0
3 years ago
Read 2 more answers
Evaluate 6t−20−32u when t=6 and u=1/4
UNO [17]
6(6)-20-32(1/4)
36-20-8
16-8
8
4 0
3 years ago
If f(x) = 1 – x2 + x3, then f(-1) =
Monica [59]

Answer:

Step-by-step explanation:

f(-1)= 1 - (-1)^2 + (-1)^3 = 1 - 1 - 1 = 0 - 1 = -1

4 0
3 years ago
What is the peremeter of this polygon? (With picture)
lyudmila [28]
Check the picture below.

so.. simply, use the distance formula, to get their length an add them up, and that's the perimeter of the polygon.


\bf \textit{distance between 2 points}\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
%  (a,b)
&({{ -1}}\quad ,&{{ 2}})\quad 
%  (c,d)
&({{ 2}}\quad ,&{{ 4}})\\
&({{ 2}}\quad ,&{{ 4}})\quad 
%  (c,d)
&({{ 3}}\quad ,&{{ -2}})\\
&({{ 3}}\quad ,&{{ -2}})\quad 
%  (c,d)
&({{ -2}}\quad ,&{{ -3}})\\
&({{ -2}}\quad ,&{{ -3}})\quad 
%  (c,d)
&({{ -1}}\quad ,&{{ 2}})
\end{array}\qquad 
%  distance value
d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}

\bf -------------------------------\\\\
d=\sqrt{[2-(-1)]^2+(4-2)^2}\implies d=\sqrt{(2+1)^2+(2)^2}
\\\\\\
d=\sqrt{3^2+2^2}\implies \boxed{d=\sqrt{13}}\\\\
-------------------------------\\\\
d=\sqrt{(3-2)^2+(-2-4)^2}\implies d=\sqrt{1^2+(-6)^2}\implies \boxed{d=\sqrt{37}}\\\\
-------------------------------\\\\
d=\sqrt{(-2-3)^2+[-3-(-2)]^2}\implies d=\sqrt{(-5)^2+(-3+2)^2}
\\\\\\
d=\sqrt{(-5)^2+(-1)^2}\implies \boxed{d=\sqrt{26}}

\\\\
-------------------------------\\\\
d=\sqrt{[-1-(-2)]^2+[2-(-3)]^2}\implies d=\sqrt{(-1+2)^2+(2+3)^2}
\\\\\\
d=\sqrt{(1)^2+(5)^2}\implies \boxed{d=\sqrt{26}}

so, those are their lengths, sum them all up, that's the polygon's perimeter.

4 0
3 years ago
Three sweets have a mass of 9g. How many sweets have a mass of 24g?
konstantin123 [22]
<h2> <u>Answer </u><u>:</u><u>-</u></h2>

\tt \pink{8 \: sweet}

<h2><u>Given </u><u>:</u><u>-</u></h2>

Mass of 3 sweets = 9 g

<h2><u>To </u><u>Find</u><u> </u><u>:</u><u>-</u></h2>

Total sweets of 24 g

<h2><u>Solution</u><u> </u><u>:</u><u>-</u></h2>

Firstly we will find weight of 1 sweet

Weight of one sweet = 9/3 = 3 g

Total sweet of 24 g = 24/3 = 8 sweet

3 0
3 years ago
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