according to the ratio, if every other chocolates Jodi ate is 36.5 calories, then if she ate 5 chocolates then she would have eaten 157.5 calories. What varies means is some chocolates may not exactly be 36.5 calories.

Step-by-step explanation:

8 0

3 years ago

Find the length of the curve. R(t) = cos(8t) i + sin(8t) j + 8 ln cos t k, 0 ≤ t ≤ π/4

arsen [322]

we are given

$R(t)=cos(8t)i+sin(8t)j+8ln(cos(t))k$

now, we can find x , y and z components

$x=cos(8t),y=sin(8t),z=8ln(cos(t))$

Arc length calculation:

we can use formula

$L=\int\limits^a_b {\sqrt{(x')^2+(y')^2+(z')^2} } \, dt$

$x'=-8sin(8t),y=8cos(8t),z=-8tan(t)$

now, we can plug these values

$L=\int _0^{\frac{\pi }{4}}\sqrt{(-8sin(8t))^2+(8cos(8t))^2+(-8tan(t))^2} dt$

now, we can simplify it

$L=\int _0^{\frac{\pi }{4}}\sqrt{64+64tan^2(t)} dt$

$L=\int _0^{\frac{\pi }{4}}8\sqrt{1+tan^2(t)} dt$

$L=\int _0^{\frac{\pi }{4}}8\sqrt{sec^2(t)} dt$

$L=\int _0^{\frac{\pi }{4}}8sec(t) dt$

now, we can solve integral

$\int \:8\sec \left(t\right)dt$

$=8\ln \left|\tan \left(t\right)+\sec \left(t\right)\right|$

now, we can plug bounds

and we get

$=8\ln \left(\sqrt{2}+1\right)-0$

so,

$L=8\ln \left(1+\sqrt{2}\right)$ ..............Answer

5 0

3 years ago

(iii) (3x + 4)(3x - 5)

kiruha [24]

Answer:

$x_{1}=-\frac{4}{3}\\x_{2}=\frac{5}{3}\\$

Step-by-step explanation:

$(3x+4)(3x-5)=9x^{2} -15x+12x-20=9x^{2} -3x-20\\\\9x^{2} -3x-20=0\\\\D=b^{2} -4ac=(-3)^{2}-4*9*(-20)=9+720=27^{2} \\\\x_1_,_2=\frac{-b +_- \sqrt{D} }{2a}\\\\x_1=\frac{3-27}{18}=\frac{-24}{18} =-\frac{4}{3}\\\\x_2=\frac{3+27}{18}=\frac{30}{18} =\frac{5}{3} \\\\$

6 0

3 years ago