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Trava [24]
3 years ago
10

4 tomatoes cost $3.40 what is the unit rate

Mathematics
1 answer:
IRINA_888 [86]3 years ago
7 0
The unit rat is the cost per 1 unit
cost per 1 tomatoe

3.4/4=1.7/2=0.85/1

unit rate is $0.85 per tomatoe
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Answer:

5

Step-by-step explanation:

10 = 2 * 5

125 = 5 * 5 * 5

Taking out the duplicates from the  2 lists :-  GCF = 5

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A function is a relation in which every input has exactly one output. Which choice represents a function?
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9/4 divided by 3/8 what the answer?
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Evaluate the interval (Calculus 2)
Darya [45]

Answer:

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

Step-by-step explanation:

<u>Fundamental Theorem of Calculus</u>

\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

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QUESTION IN PICTURE
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Answer:

3 6/5

Step-by-step explanation:

the fraction is its kinda blurry

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