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3 years ago

If the slope of line m is 2/5, what is the slope of a line that is parallel

Mathematics

1 answer:

kramer3 years ago

5 0

2/5. Two line have the same slope if they are parallel to each other.

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Lucy collects data from a random sample of seventh-graders. Out of 40 respondents, 7 attend after-school programs. Of the 200 se

Shalnov [3]

Answer:

Step-by-step explanation:

For every 40 students, you add 7

Ratio:

Students : Students in afterschool programs

40:7

80:14

120:21

160:28

200:35

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3 years ago

Can someone be my friend

marta [7]

Answer: yes

Step-by-step explanation:

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2 years ago

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If sinX= 0.9, what is the value of cos X?<br><br> Round the value to the nearest thousandth.

SSSSS [86.1K]

Answer: sqrt(19)/10

Step-by-step explanation:

sinx=0.9

x=arcsin(0.9)

Insert x for cos(x):

cos(arcsin(0.9)=

Use identify $cos(arcsin(x))=\sqrt{1-x^{2} }$ , so

$\sqrt{1-(\frac{9}{10})^{2}} =\sqrt{\frac{19}{100} } =\frac{\sqrt{19}}{10}$

Sorry if this is confusing.

6 0

3 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

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

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3 years ago

Whats 1x600 plssssssssssssssssssssssssssssssssssss

Andre45 [30]

Answer:

600

Step-by-step explanation:

anything times 1 is itself

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3 years ago

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