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3 years ago

I'm stuck :/ help hurry

Mathematics

1 answer:

Free_Kalibri [48]3 years ago

4 0

Answer: i think A is negative -2 and B is -19 hope this help im not a hundred percent sure if i got it wrong please let know.

Step-by-step explanation:

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The plc as value of 6 in12.6 is the tenths place

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3 years ago

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If 1/√a-√b=1/3 and 1/√a+√b=1/2, then find the difference of a and b.

kow [346]

<u>ANSWER:</u>

If $\frac{1}{\sqrt{a}-\sqrt{b}}=\frac{1}{3}$ and $\frac{1}{\sqrt{a}+\sqrt{b}}=\frac{1}{2}$ then the difference of a and b is 6

<u>SOLUTION:</u>

Given, $\frac{1}{\sqrt{a}-\sqrt{b}}=\frac{1}{3}$ → $\sqrt{a}-\sqrt{b}=3$ ----- (1)

And $\frac{1}{\sqrt{a}+\sqrt{b}}=\frac{1}{2}$ → $\sqrt{a}+\sqrt{b}=2$ --- (2)

We have to find difference of a and b.

Now, add (1) and (2)

$\sqrt{a}-\sqrt{b}=3$

$\sqrt{a}+\sqrt{b}=2$

Adding above two equations, we get,

$2 \sqrt{a}+0=2+3$

$\begin{array}{l}{2 \sqrt{a}=5} \\\\ {\sqrt{a}=\frac{5}{2}} \\\\ {a=\frac{25}{4}}\end{array}$

substitute $\sqrt{a}$ value in (2)

$\begin{array}{l}{\frac{5}{2}+\sqrt{b}=2} \\\\ {\sqrt{b}=\frac{2}{\sin \frac{5}{2}}} \\\\ {\sqrt{b}=\frac{4-5}{2}} \\\\ {\sqrt{b}=\frac{-1}{2}} \\\\ {b=\frac{1}{4}}\end{array}$

Now, difference of a and b is a – b = $\frac{25}{4}-\frac{1}{4}=\frac{24}{4}=6$

Hence, the difference of a and b is 6.

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3 years ago

Determine whether the Mean Value Theorem applies to the function f(x)=2x^1/5 on the interval [-32, 32]

Anettt [7]

Answer

given,

$f(x) = 2 x^{1/5}$

interval = [-32, 32]

differentiating the given equation

$f'(x) = \dfrac{d}{dx}(2 x^{1/5})$

$f'(x) = \dfrac{2}{5x^{4/5}}$

hence, the above solution is not defined at x = 0

and x = 0 lie in the given interval i.e. [-32, 32]

so, at x = 0 the function is not differentiable.

Hence, mean value theorem does not apply to the given function.

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3 years ago

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2 years ago

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nydimaria [60]

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I'm stuck :/ help hurry​

I'm stuck :/ help hurry