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3 years ago

Prove that the roots of the equation X^2 - 2px + p^2 - q^2 =0 where p and q are constants, are real.

Mathematics

1 answer:

lora16 [44]3 years ago

3 0

Answer:

Step-by-step explanation:

Sorry I am not sure and don’t want to give you the wrong answer sorry again

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The question was not clear for me! Help :(

Vsevolod [243]

Answer:

0.1 , 0.2 , 0.05

Step-by-step explanation:

*each column adds to 1*

20 x 1/2 = 10

1/2 x 20 x 0.1 = 1

top left box = 0.1

; 5 x 2 x 0.1 = 1

1/2 x 10 = 5

; 0.2 x 1/2 x 10 = 1

bottom middle box = 0.2

; 5 x 1 x 0.2 = 1

; 2 x 10 x 0.05 = 1

middle right box = 0.05

; 20 x 1 x 0.05

5 0

3 years ago

Read 2 more answers

Need help !!!!

iren2701 [21]

Answer:

4(2x+3)=(x) +2(3x+7)

4(2x+3)-2(3x+7)=

4-x+10=6

3 0

3 years ago

PLEASE HELP!!!!<br> Which of these pair of functions are inverse functions?

Mamont248 [21]

Answer:

Option B and C are correct.

Step-by-step explanation:

Inverse function: If both the domain and the range are R for a function f(x), and if f(x) has an inverse g(x) then:

$f(g(x)) = g(f(x)) = x$ for every x∈R.

Let $f(x) = \frac{1}{2}(\ln(\frac{x}{2}) -1)$ and $g(x) = 2e^{2x+1}$

Use logarithmic rules:

$ln e^a = a$

$e^{lnx} = x$

$\ln a^b = b\ln a$

then, by definition;

$f(g(x)) = f(2e^{2x+1}) =\frac{1}{2}(\ln(\frac{2e^{2x+1}}{2})-1)$ = $\frac{1}{2}(\ln(e^{2x+1}}){-1) = \frac{1}{2} (2x+1-1) =\frac{1}{2}(2x) = x$

$g(f(x)) = g(\frac{1}{2}(\ln(\frac{x}{2}) -1)) = 2e^{2({\frac{1}{2}(\ln(\frac{x}{2}) -1})+1$ $2e^{(\ln(\frac{x}{2}) -1+1}=2e^{\ln(\frac{x}{2})} =2\cdot \frac{x}{2} = x$

Similarly;

for $f(x) = \frac{4 \ln(x^2)}{e^2}$ and $g(x) = e^{\frac{e^2 \cdot x}{8} }$

then, by definition;

$f(g(x)) = f(e^{\frac{e^2 \cdot x}{8}}) =\frac{4 \ln {(\frac{e^2 \cdot x}{8})^2}}{e^2}$ = $\frac{8 \ln {(\frac{e^2 \cdot x}{8})}}{e^2} =\frac{8\frac{e^2\cdot x}{8} }{e^2}=\frac{8e^2 \cdot x}{8e^2}=x$