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3 years ago

Help please :) I'd greatly appreciate it

Mathematics

1 answer:

azamat3 years ago

6 0

= 4i (2i) √6
= 8i^2 √6 but i^2 = -1
= - 8√6

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Solve for x help me please will give brainliest answer!!!

schepotkina [342]

Answer:

$x = 7 \frac{1}{3}$

6 0

3 years ago

( - 8)( - 4)(12) I need help with this :)

Alisiya [41]

Answer:

= 384

Step-by-step explanation:

(-8)(-4)(12)

= (32)(12)

=384

8 0

2 years ago

Read 2 more answers

The can of tomatoes is three times as tall but one-third as wide as the can of tuna. What is true about the areas of their label

yulyashka [42]

Problem One
Call the radius of the second can = r
Call the height of the second can = h

Then the radius of the first can = 1/3 r
The height of the first can = 3*h

A1 / A2 = (2*pi*(1/3r)*(3h)] / [2*pi * r * h]

Here's what will cancel. The twos on the right will cancel. The 3 and 1/3 will multiply to one. The 2 r's will cancel. The h's will cancel. Finally, the pis will cancel

Result A1 / A2 = 1/1
The labels will be shaped differently, but they will occupy the same area.

Problem Two
It seems like the writer of the problem put some lids on the new solid that were not implied by the question.

If I understand the problem correctly, looking at it from the top you are sweeping out a circle for the lid on top and bottom, plus the center core of the cylinder.

One lid would be pi r^2 = pi w^2 and so 2 of them would be 2 pi w^2
The region between the lids would be 2 pi r h for the surface area which is 2pi w h

Put the 2 regions together and you get
Area = 2 pi w^2 + 2 pi w h

Answer: Upper left corner <<<<< Answer

8 0

3 years ago

Evaluate the interval (Calculus 2)

Darya [45]

Answer:

$2 \tan (6x)+2 \sec (6x)+\text{C}$

Step-by-step explanation:

Fundamental Theorem of Calculus

$\displaystyle \int \text{f}(x)\:\text{d}x=\text{F}(x)+\text{C} \iff \text{f}(x)=\dfrac{\text{d}}{\text{d}x}(\text{F}(x))$

If differentiating takes you from one function to another, then integrating the second function will take you back to the first with a constant of integration.

Given indefinite integral:

$\displaystyle \int \dfrac{12}{1-\sin (6x)}\:\:\text{d}x$

$\boxed{\begin{minipage}{5 cm}\underline{Terms multiplied by constants}\\\\$\displaystyle \int a\:\text{f}(x)\:\text{d}x=a \int \text{f}(x) \:\text{d}x$\end{minipage}}$

If the terms are multiplied by constants, take them outside the integral:

$\implies 12\displaystyle \int \dfrac{1}{1-\sin (6x)}\:\:\text{d}x$

Multiply by the conjugate of 1 - sin(6x) :

$\implies 12\displaystyle \int \dfrac{1}{1-\sin (6x)} \cdot \dfrac{1+\sin(6x)}{1+\sin(6x)}\:\:\text{d}x$

$\implies 12\displaystyle \int \dfrac{1+\sin(6x)}{1-\sin^2(6x)} \:\:\text{d}x$

$\textsf{Use the identity} \quad \sin^2 x+ \cos^2 x=1:$

$\implies \sin^2 (6x) + \cos^2 (6x)=1$

$\implies \cos^2 (6x)=1- \sin^2 (6x)$

$\implies 12\displaystyle \int \dfrac{1+\sin(6x)}{\cos^2(6x)} \:\:\text{d}x$

Expand:

$\implies 12\displaystyle \int \dfrac{1}{\cos^2(6x)}+\dfrac{\sin(6x)}{\cos^2(6x)} \:\:\text{d}x$

$\textsf{Use the identities }\:\: \sec \theta=\dfrac{1}{\cos \theta} \textsf{ and } \tan\theta=\dfrac{\sin \theta}{\cos \theta}:$

$\implies 12\displaystyle \int \sec^2(6x)+\dfrac{\tan(6x)}{\cos(6x)} \:\:\text{d}x$

$\implies 12\displaystyle \int \sec^2(6x)+\tan(6x)\sec(6x) \:\:\text{d}x$

$\boxed{\begin{minipage}{5 cm}\underline{Integrating $\sec^2 kx$}\\\\$\displaystyle \int \sec^2 kx\:\text{d}x=\dfrac{1}{k} \tan kx\:\:(+\text{C})$\end{minipage}}$

$\boxed{\begin{minipage}{6 cm}\underline{Integrating $ \sec kx \tan kx$}\\\\$\displaystyle \int \sec kx \tan kx\:\text{d}x= \dfrac{1}{k}\sec kx\:\:(+\text{C})$\end{minipage}}$