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Georgia [21]
3 years ago
13

What is the greatest number that divides 14, 2 and -8

Mathematics
2 answers:
KATRIN_1 [288]3 years ago
8 0
| 2 | I think that’s right
Scrat [10]3 years ago
7 0

Answer:

14, 2 and -8 all can be divided by the greatest number, 2.

Step-by-step explanation:

If you do not want a decimal number for each, then first you would choose an even number. Odd numbers will likely make 14, 2 and -8 a decimal.

Dividing by 1 makes the numbers 14, 2 and -8. 1 is an odd number, but it does not affect any of the numbers. However, there may be a better number.

Dividing by 2 makes the numbers 7, 1 and -4. So far this number works to divide them all.

Dividing by 4 makes the numbers 3.5, 0.5 and -2. Only one number is not a decimal, so this does not work.    

Dividing by 6 makes the numbers  2.333..., 0.333... and -1.333... All of these numbers are decimals. This will not work either.

We can conclude that any number greater than 2 will not make all the numbers whole. They will be made irrational and decimal.

So 2 is the greatest number that can divide into all of the numbers.

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then if you have the midpoint and one endpoint, find the distance between them to find the other endpoint.

so, for the third example, start with the x. how fair is -8 from -1? The answer is 7. So, we must ask, what number is 7 more than -1? Answer is 6. Then repeat for the y. 6 is 5 units away from 1, so then we do 1-5=-4. So it should be (6,-4) as point B

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AVprozaik [17]

Answer:

\sqrt{37} units

Step-by-step explanation:

If we draw a perpendicular from point (6,2) on the line 6x - y = - 3, then we have to find the length of the perpendicular.  

We know the formula of length of perpendicular from a point (x_{1}, y_{1} ) to the straight line ax + by + c = 0 is given by  

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\frac{|6(6)-2+3|}{\sqrt{6^{2} +(-1)^{2} } } = \frac{37}{\sqrt{37} } = \sqrt{37}  units. (Answer)

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3 years ago
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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