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3 years ago

Every 3 seconds , 4 cubic feet of water pass under the bridge. what is the unit rate?

Mathematics

1 answer:

svetoff [14.1K]3 years ago

7 0

12 would be the unit rate you mulitply 3 × 4

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544ml^2 is how many liters show your work

ivolga24 [154]

Answer:

295.936

Step-by-step explanation:

First find out what 544^2 is. That is 295,936. There are 100 ml in 1 liter. So divide 295,936 by 100 to see how many liters are in 295,936 ml.

295,936÷100=295.936

7 0

2 years ago

I don't even know what I'm soposto do

seropon [69]

Its A or C because Angle AEC has to be bigger then 90 degrees

3 0

3 years ago

Read 2 more answers

Look at the diagram. If RS= 8y + 4, ST = 4y + 8, and RT = 36, find the value of y.

vladimir2022 [97]

RS= 8y + 4, ST = 4y + 8, and RT = 36

RS + ST = RT
8y + 4 + 4y + 8 = 36
12y + 12 = 36
12y = 24
y = 2

answer
y = 2 (first choice)

6 0

3 years ago

Read 2 more answers

Simplify (23)^–2 genuinely confused

statuscvo [17]

Answer:

1/529

Step-by-step explanation:

$23^{-2}\\\\\mathrm{Apply\:exponent\:rule}:\quad \:a^{-b}=\frac{1}{a^b}\\23^{-2}=\frac{1}{23^2}\\\\23^2=529\\\\=\frac{1}{529}$

7 0

2 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})$\end{minipage}}$

$\implies 12 \left[\dfrac{1}{6} \tan (6x)+\dfrac{1}{6} \sec (6x) \right]+\text{C}$

Simplify:

$\implies \dfrac{12}{6} \tan (6x)+\dfrac{12}{6} \sec (6x)+\text{C}$

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

Learn more about indefinite integration here:

brainly.com/question/27805589

brainly.com/question/28155016

3 0

1 year ago