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3 years ago

Can anyone do my imagine math and get me 6 certificates (pass the lessons) pleaseeeeeee I need it today before 5

1 answer:

5 0

IF ANYONE knows Imagine Math or has the site, Please help, she needs it and were trying to figure the questions....AGAIN please answer ASAP! :)

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Write the standard form for 25

pishuonlain [190]

Answer:

twenty fifth

Step-by-step explanation:

3 0

3 years ago

Simplify : 5y-2y+4=10

Mnenie [13.5K]

Answer:

See below.

Step-by-step explanation:

5y - 2y + 4 = 10

5y - 2y = 6

3y = 6

y = 2

Check:

5y - 2y + 4 = 10

5(2) - 2(2) + 4 = 10

10 - 4 + 4 = 10

6 + 4 = 10

10 = 10

The value of y = 2

7 0

3 years ago

Read 2 more answers

The side of a square is x, and the area is A. Express A in terms of x. Draw the graph which shows how A depends on x.

Alexandra [31]

a=x squared or x times itself

4 0

3 years ago

mya average 86 points on five tests. what would she have to score on the sixth test to bring her average up by one point

Ray Of Light [21]

Answer:

The answer is 88.

She would have to score 88

Step-by-step explanation:

86 + 88 = 174

174 ÷ 2 = 87

4 0

3 years ago

Read 2 more answers

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