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2 years ago

Help please I need help with 15,16,17,18

Mathematics

1 answer:

N76 [4]2 years ago

4 0

For 15 its just the shape, if you draw a shape based on the dots, compare them. Like left one is a bit taller and wider.

16: for left one median is 8 i think. Right one is 9.5 i think
17: for left one, 13 is an outlier, for right one, 11.5 is outliner. You just write the beginning and ends including and not including those outliers.
18: people in group A were kids and people in group B were adults because their shoe sizes are much higher.

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2 years ago

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2 years ago

A BUSHEL OF APPLES WEIGHS ABOUT 42 POUNDS. THERE ARE 4 PECKS OF BUSHELS. IT TAKES 2 POUNDS OF APPLES TO MAKE ONE PIE. HOW MANY P

madam [21]

Answer:

84 pies

Step-by-step explanation:

if you multiply 42 pounds by4 pecks. You get 168 pounds. and then you divide that by 2 and you can make 84 pies

7 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

Workings please......

attashe74 [19]

Answer:

21.142

Step-by-step explanation:

not sure it's correct

3 0

2 years ago