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2 years ago

Use the least common denominator to enter an equivalent fraction for each fraction

Mathematics

1 answer:

adelina 88 [10]2 years ago

5 0

Answer:

The least common denominator of the fraction is 14

1/2= 7/14

1/7= 2/14

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Choose the equation that is equivalent to: 4(8x+6)=20

KengaRu [80]

Answer:

Step-by-step explanation:

32x + 24 = 20

24 - 24= 0

20 - 24= -4

32x = -4

32x/32

-4/32= -8

X= -1/8

7 0

1 year ago

Read 2 more answers

Please help fast with correct answer no guessing answer. Only answer if actually know the answer

lawyer [7]

Answer:

it is d inverse

Step-by-step explanation:

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3 years ago

Mr. Smith is unhappy with the 2 foot concrete walkway around the edge of the pool. He decides to

nirvana33 [79]

Given that the total area of the walkway and the pool is w²+17w+66, and the width w, will be found as follows:
A=length×width
A=w²+17w+66
to get the new width we need to factorize the above quadratic form
A=w²+17w+66
=w²+11w+6w+66
=w(w+11)+6(w+11)
=(w+11)(w+6)
From the answer, the width=6 units, length=11 units

5 0

2 years ago

Derivative, by first principle<br><img src="https://tex.z-dn.net/?f=%20%5Ctan%28%20%5Csqrt%7Bx%20%7D%20%29%20" id="TexFormula1"

vampirchik [111]

$\displaystyle\lim_{h\to0}\frac{\tan\sqrt{x+h}-\tan x}h$

Employ a standard trick used in proving the chain rule:

$\dfrac{\tan\sqrt{x+h}-\tan x}{\sqrt{x+h}-\sqrt x}\cdot\dfrac{\sqrt{x+h}-\sqrt x}h$

The limit of a product is the product of limits, i.e. we can write

$\displaystyle\left(\lim_{h\to0}\frac{\tan\sqrt{x+h}-\tan x}{\sqrt{x+h}-\sqrt x}\right)\cdot\left(\lim_{h\to0}\frac{\sqrt{x+h}-\sqrt x}h\right)$

The rightmost limit is an exercise in differentiating $\sqrt x$ using the definition, which you probably already know is $\dfrac1{2\sqrt x}$ .

For the leftmost limit, we make a substitution $y=\sqrt x$ . Now, if we make a slight change to $x$ by adding a small number $h$ , this propagates a similar small change in $y$ that we'll call $h'$ , so that we can set $y+h'=\sqrt{x+h}$ . Then as $h\to0$ , we see that it's also the case that $h'\to0$ (since we fix $y=\sqrt x$ ). So we can write the remaining limit as

$\displaystyle\lim_{h\to0}\frac{\tan\sqrt{x+h}-\tan\sqrt x}{\sqrt{x+h}-\sqrt x}=\lim_{h'\to0}\frac{\tan(y+h')-\tan y}{y+h'-y}=\lim_{h'\to0}\frac{\tan(y+h')-\tan y}{h'}$

which in turn is the derivative of $\tan y$ , another limit you probably already know how to compute. We'd end up with $\sec^2y$ , or $\sec^2\sqrt x$ .

So we find that

$\dfrac{\mathrm d\tan\sqrt x}{\mathrm dx}=\dfrac{\sec^2\sqrt x}{2\sqrt x}$

7 0

3 years ago

Find the measure of this angle m 1 = 123 m 8

Anit [1.1K]

If it is parallel lines cut by a transversal then they are probably equal. Alternate exterior.

6 0

2 years ago