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3 years ago

9

Help please. i am desperate

1 answer:

Nana76 [90]3 years ago

4 0

Answer:

lol

Step-by-step explanation:

whyyy u say that hahahahhaa

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Please help me with this problem!): (#5)

astra-53 [7]

5a.

18/9 = 12/x

18x = 12(9)...cross multiply

18x = 108

x=108/18

x=6

6 0

3 years ago

Solve using substitution.<br> y = -10x-10<br> y = -3x + 4

Nikolay [14]

-10x-10 = -3x+4
+3x +3x
————————
-7x-10 = 4
+10 +10
————————
-7x = 14
then divide -7 on both side of the equal sign
you’ll get x=-2

then plug in -2 as x into one of the equations ( it doesn’t matter which one)

y= -10(-2)-10 = 20-10= 10

so y=10

4 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

(Question 2 : Select the best answer for the question.)

iris [78.8K]

Answer:

B is the answer i think

Step-by-step explanation:

because its lenear and positive...

5 0

3 years ago

Mrs. Perez has 28 students and 5

solniwko [45]

Answer:39 1/5

Step-by-step explanation:

3 0

3 years ago

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