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3 years ago

Simply this equation

Mathematics

2 answers:

Liono4ka [1.6K]3 years ago

3 0

Answer:

b= x

Step-by-step explanation:

$\boxed{ {x}^{m} \times {x}^{n} = {x}^{m + n} }$

Applying the law of indices above:

$b = {x}^{ \frac{1}{4} } \times {x}^{ \frac{3}{4} }$

$b = {x}^{ \frac{1}{4} + \frac{3}{4} }$

$b = {x}^{ \frac{4}{4} }$

$b = {x}^{1}$

∴b= x

Montano1993 [528]3 years ago

3 0

Answer:

It is b = x

Step-by-step explanation:

From law of indices:

${a}^{b} \times {a}^{c} = {a}^{(b + c)}$

for, multiplication: same coefficients, we sum up the powers.

$b = {x}^{ \frac{1}{4} } \times {x}^{ \frac{3}{4} } \\ b = {x}^{( \frac{1}{4} + \frac{3}{4} ) } \\ b = {x}^{ \frac{4}{4} } \\ b = {x}^{1} \\ b = x$

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Find the area of the part of the paraboloid z = 9 - x^2 - y^2 that lies above the xy-plane.

svetoff [14.1K]

Parameterize this surface (call it $S$ ) by

$\vec r(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec\jmath+(9-u^2)\,\vec k$

with $0\le u\le3$ and $0\le v\le2\pi$ . Take the normal vector to $S$ to be

$\vec r_u\times\vec r_v=2u^2\cos v\,\vec\imath+2u^2\sin v\,\vec\jmath+u\,\vec k$

Then the area of $S$ is

$\displaystyle\iint_S\mathrm dA=\iint_S\|\vec r_u\times\vec r_v\|\,\mathrm du\,\mathrm dv$

$=\displaystyle\int_0^{2\pi}\int_0^3u\sqrt{1+4u^2}\,\mathrm du\,\mathrm dv$

$=\displaystyle2\pi\int_0^3u\sqrt{1+4u^2}\,\mathrm du=\boxed{\frac{37\sqrt{37}-1}6\pi}$

3 0

4 years ago

29.

Elis [28]

Answer:

Sum of their present ages = 20 years

Step-by-step explanation:

Present age :

Neha's age = x years

Reemas' age = x - 4

After 5 years:

Neha's age = (x +5) years

Reemas' age = (x - 4 + 5 ) = x + 1

Sum of the ages of Reema and Neha will be 30

Therefore,

x + 5 + x + 1 = 30

x + x + 5 + 1 = 30

2x + 6 = 30

2x = 30 - 6

2x = 24

x = 24/2

x = 12

Neha's age = 12 years

Reema's age = 12 - 4 = 8 years

Sum of their present ages = 12 + 8 = 20 years

4 0

3 years ago

7. Construct a pie chart to depict the data given below:

marta [7]

aku gak tau apa Jawaban nya Jadi aku Minta maaf

8 0

3 years ago

Did I do this right? (Accounting)

levacccp [35]

Yes you did, congratulations ;-)

4 0

3 years ago

I honestly need help with these

Brilliant_brown [7]

9. The curve passes through the point (-1, -3), which means

$-3 = a(-1) + \dfrac b{-1} \implies a + b = 3$

Compute the derivative.

$y = ax + \dfrac bx \implies \dfrac{dy}{dx} = a - \dfrac b{x^2}$

At the given point, the gradient is -7 so that

$-7 = a - \dfrac b{(-1)^2} \implies a-b = -7$

Eliminating $b$ , we find

$(a+b) + (a-b) = 3+(-7) \implies 2a = -4 \implies \boxed{a=-2}$

Solve for $b$ .

$a+b=3 \implies b=3-a \implies \boxed{b = 5}$

10. Compute the derivative.

$y = \dfrac{x^3}3 - \dfrac{5x^2}2 + 6x - 1 \implies \dfrac{dy}{dx} = x^2 - 5x + 6$

Solve for $x$ when the gradient is 2.

$x^2 - 5x + 6 = 2$

$x^2 - 5x + 4 = 0$

$(x - 1) (x - 4) = 0$

$\implies x=1 \text{ or } x=4$

Evaluate $y$ at each of these.

$\boxed{x=1} \implies y = \dfrac{1^3}3 - \dfrac{5\cdot1^2}2 + 6\cdot1 - 1 = \boxed{y = \dfrac{17}6}$

$\boxed{x = 4} \implies y = \dfrac{4^3}3 - \dfrac{5\cdot4^2}2 + 6\cdot4 - 1 \implies \boxed{y = \dfrac{13}3}$

11. a. Solve for $x$ where both curves meet.

$\dfrac{x^3}3 - 2x^2 - 8x + 5 = x + 5$

$\dfrac{x^3}3 - 2x^2 - 9x = 0$

$\dfrac x3 (x^2 - 6x - 27) = 0$

$\dfrac x3 (x - 9) (x + 3) = 0$

$\implies x = 0 \text{ or }x = 9 \text{ or } x = -3$

Evaluate $y$ at each of these.

$A:~~~~ \boxed{x=0} \implies y=0+5 \implies \boxed{y=5}$

$B:~~~~ \boxed{x=9} \implies y=9+5 \implies \boxed{y=14}$

$C:~~~~ \boxed{x=-3} \implies y=-3+5 \implies \boxed{y=2}$

11. b. Compute the derivative for the curve.

$y = \dfrac{x^3}3 - 2x^2 - 8x + 5 \implies \dfrac{dy}{dx} = x^2 - 4x - 8$

Evaluate the derivative at the $x$ -coordinates of A, B, and C.

$A: ~~~~ x=0 \implies \dfrac{dy}{dx} = 0^2-4\cdot0-8 \implies \boxed{\dfrac{dy}{dx} = -8}$

$B:~~~~ x=9 \implies \dfrac{dy}{dx} = 9^2-4\cdot9-8 \implies \boxed{\dfrac{dy}{dx} = 37}$

$C:~~~~ x=-3 \implies \dfrac{dy}{dx} = (-3)^2-4\cdot(-3)-8 \implies \boxed{\dfrac{dy}{dx} = 13}$

12. a. Compute the derivative.

$y = 4x^3 + 3x^2 - 6x - 1 \implies \boxed{\dfrac{dy}{dx} = 12x^2 + 6x - 6}$

12. b. By completing the square, we have

$12x^2 + 6x - 6 = 12 \left(x^2 + \dfrac x2\right) - 6 \\\\ ~~~~~~~~ = 12 \left(x^2 + \dfrac x2 + \dfrac1{4^2}\right) - 6 - \dfrac{12}{4^2} \\\\ ~~~~~~~~ = 12 \left(x + \dfrac14\right)^2 - \dfrac{27}4$

so that

$\dfrac{dy}{dx} = 12 \left(x + \dfrac14\right)^2 - \dfrac{27}4 \ge 0 \\\\ ~~~~ \implies 12 \left(x + \dfrac14\right)^2 \ge \dfrac{27}4 \\\\ ~~~~ \implies \left(x + \dfrac14\right)^2 \ge \dfrac{27}{48} = \dfrac9{16} \\\\ ~~~~ \implies \left|x + \dfrac14\right| \ge \sqrt{\dfrac9{16}} = \dfrac34 \\\\ ~~~~ \implies x+\dfrac14 \ge \dfrac34 \text{ or } -\left(x+\dfrac14\right) \ge \dfrac34 \\\\ ~~~~ \implies \boxed{x \ge \dfrac12 \text{ or } x \le -1}$

13. a. Compute the derivative.

$y = x^3 + x^2 - 16x - 16 \implies \boxed{\dfrac{dy}{dx} = 3x^2 - 2x - 16}$

13. b. Complete the square.

$3x^2 - 2x - 16 = 3 \left(x^2 - \dfrac{2x}3\right) - 16 \\\\ ~~~~~~~~ = 3 \left(x^2 - \dfrac{2x}3 + \dfrac1{3^2}\right) - 16 - \dfrac13 \\\\ ~~~~~~~~ = 3 \left(x - \dfrac13\right)^2 - \dfrac{49}3$

Then

$\dfrac{dy}{dx} = 3 \left(x - \dfrac13\right)^2 - \dfrac{49}3 \le 0 \\\\ ~~~~ \implies 3 \left(x - \dfrac13\right)^2 \le \dfrac{49}3 \\\\ ~~~~ \implies \left(x - \dfrac13\right)^2 \le \dfrac{49}9 \\\\ ~~~~ \implies \left|x - \dfrac13\right| \le \sqrt{\dfrac{49}9} = \dfrac73 \\\\ ~~~~ \implies x - \dfrac13 \le \dfrac73 \text{ or } -\left(x-\dfrac13\right) \le \dfrac73 \\\\ ~~~~ \implies \boxed{x \le 2 \text{ or } x \ge \dfrac83}$

5 0

2 years ago

Simply this equation ​

Simply this equation