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4 years ago

X-y = 10 x^2-y^2 = 60 x^2+2xy+y^2=?

1 answer:

8 0

$\displaystyle\\x-y=\boxed{10}\\\\----------\\\\x^2-y^2 = 60\\\\(x-y)(x+y)=60\\\\10(x+y)=60\\\\x+y=\frac{60}{10}=\boxed{6}\\\\----------\\\\x^2+2xy+y^2=(x+y)^2=6^2=\boxed{\bf36}$

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What is the value of m in the figure below? In this diagram, ΔABD ~ ΔBCD.

Kryger [21]

Answer:

Option D is the right option.

solution,

$\frac{ac}{bc} = \frac{bc}{dc} \\ or \: \frac{18}{ m} = \frac{m}{7} \\ or \: m \times \: m = 18 \times 7(cross \: multiplication) \\ or \: {m}^{2} = 126 \\ m = \sqrt{126}$

Hope this helps..

Good luck on your assignment.

Hope

6 0

3 years ago

Nine scores have an average of 8. Scores x and x+1 are added, and these increase the average to 9. Find x

Degger [83]

Answer:

x must be 13

Step-by-step explanation:

Represent the nine scores by a, b, c, d, e, f, g, h, i.

Write the following equation to indicate that the average of these nine scores is 8:

a+b+c+d+e+f+g+h+i

----------------------------- = 8, or <em>a+b+c+d+e+f+g+h+i = 72</em> if we multiply both

9 sides of this equation by 9.

Next, we add scores x and x+ 1 to

a+b+c+d+e+f+g+h+i + x + x + 1

and divide the result by 11, as there are now 11 scores instead of just 9:

a+b+c+d+e+f+g+h+i + x + x + 1

----------------------------------------------

Earlier, we saw that a+b+c+d+e+f+g+h+i = 72. We can now replace

a+b+c+d+e+f+g+h+i in the above fraction by 72:

72 + 2x + 1

------------------- = 9 (as the new average is 9)

Multiplying both sides of this equation by 11 results in:

73 + 2x = 99, or 2x = 26.

Then x must be 13, and x + 1 must be 14.

3 0

3 years ago

The solution to the equation x + 11 = 18 is x = 7.<br><br><br> True<br><br> False

Usimov [2.4K]

Answer:

True

Step-by-step explanation:

X+11=18

subtract 11 from both sides

18-11=7

x=7

5 0

3 years ago

Read 2 more answers

Which one of these is the correct answer?

sukhopar [10]

Answer:

the top answer.............

8 0

3 years ago

How are percent transmittance and absorbance related algebraically?

statuscvo [17]

Answer:

So, if all the light passes through a solution without any absorption, then absorbance is zero, and percent transmittance is 100%. If all the light is absorbed, then percent transmittance is zero, and absorption is infinite.

Absorbance is the inverse of transmittance so,

A = 1/T

Beer's law (sometimes called the Beer-Lambert law) states that the absorbance is proportional to the path length, b, through the sample and the concentration of the absorbing species, c:

A ∝ b · c

As Transmittance, $T =\dfrac{P}{P_0}$

% Transmittance, $\%T=100\times T$

Absorbance,

$A=\log_{10} \dfrac{P_0}{P}\\\\A =\log_{10}\times \dfrac{1}{T}\\\\A=\dfrac{\log_{10} 100}{\%T}\\\\A=(2 - log_{10})\times \%T$

Hence, $A=\log_{10}\times \dfrac{1}{T}$ is the algebraic relation between absorbance and transmittance.

8 0

3 years ago