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3 years ago

What is the value of 20+30(7+4)+5+2(7+99?

Mathematics

2 answers:

KATRIN_1 [288]3 years ago

5 0

Answer:

the answer is 567

Step-by-step explanation:

20+30(11)+5+2(106)

20+330+5+212

= 567

Luda [366]3 years ago

4 0

The answer is 567 woot

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Need help with my homework

Volgvan

Answer:

$\displaystyle y=\frac{16-9x^3}{2x^3 - 3}$

$\displaystyle y=-\frac{56}{13}$

Step-by-step explanation:

<u>Equation Solving</u>

We are given the equation:

$\displaystyle x=\sqrt[3]{\frac{3y+16}{2y+9}}$

To make y as a subject, we need to isolate y, that is, leaving it alone in the left side of the equation, and an expression with no y's to the right side.

We have to make it in steps like follows.

Cube both sides:

$\displaystyle x^3=\left(\sqrt[3]{\frac{3y+16}{2y+9}}\right)^3$

Simplify the radical with the cube:

$\displaystyle x^3=\frac{3y+16}{2y+9}$

Multiply by 2y+9

$\displaystyle x^3(2y+9)=\frac{3y+16}{2y+9}(2y+9)$

Simplify:

$\displaystyle x^3(2y+9)=3y+16$

Operate the parentheses:

$\displaystyle x^3(2y)+x^3(9)=3y+16$

$\displaystyle 2x^3y+9x^3=3y+16$

Subtract 3y and $9x^3$ :

$\displaystyle 2x^3y - 3y=16-9x^3$

Factor y out of the left side:

$\displaystyle y(2x^3 - 3)=16-9x^3$

Divide by $2x^3 - 3$ :

$\mathbf{\displaystyle y=\frac{16-9x^3}{2x^3 - 3}}$

ii) To find y when x=2, substitute:

$\displaystyle y=\frac{16-9\cdot 2^3}{2\cdot 2^3 - 3}$

$\displaystyle y=\frac{16-9\cdot 8}{2\cdot 8 - 3}$

$\displaystyle y=\frac{16-72}{16- 3}$

$\displaystyle y=\frac{-56}{13}$

$\mathbf{\displaystyle y=-\frac{56}{13}}$

8 0

3 years ago

Help math question derivative!

atroni [7]

Let $f(x)=\sec^{-1}x$ . Then $\sec f(x)=x$ , and differentiating both sides with respect to $x$ gives

$(\sec f(x))'=\sec f(x)\tan f(x)\,f'(x)=1$
$f'(x)=\dfrac1{\sec f(x)\tan f(x)}$

Now, when $x=\sqrt2$ , you get

$(\sec^{-1})'(\sqrt2)=f'(\sqrt2)=\dfrac1{\sec\left(\sec^{-1}\sqrt2\right)\tan\left(\sec^{-1}\sqrt2\right)}$

You have $\sec^{-1}\sqrt2=\dfrac\pi4$ , so $\sec\left(\sec^{-1}\sqrt2\right)=\sqrt2$ and $\tan\left(\sec^{-1}\sqrt2\right)=1$ . So $(\sec^{-1})'(\sqrt2)=\dfrac1{\sqrt2\times1}=\dfrac1{\sqrt2}$