All categories

3 years ago

I WILL GIVE CROWN PLS HELP NEED RIGHT ANSWER

Mathematics

2 answers:

FromTheMoon [43]3 years ago

4 0

Answer:

Step-by-step explanation:

Surely you can figure out how many stairs she has to climb. She has to go down 4, back up to 0, and then up another 4, then back to 0

aev [14]3 years ago

3 0

Answer: 21.4 ft

Step-by-step explanation: considering all levels are same size, ÷ depth of 7th, the deepest level with number of levels. 149.8 ÷ 7 = 21.4

You might be interested in

Help Me With This Question Please!

Georgia [21]

Answer:

60°

Step-by-step explanation:

x + 160° + 140° = 360°

x + 300° = 360°

x = 360° - 300°

x = 60°

8 0

3 years ago

Quest-Room ist die beste Unternehmung für Gruppen in Köln und liegt nur wenige Minuten vom Hauptbahnhof und dem Kölner Dom entfe

kramer

Idk...........................................................................................

8 0

3 years ago

Determine whether the three points are collinear.<br> (0, - 10). (-3,- 13), (2,-8)

LUCKY_DIMON [66]

Answer:

<h2>YES. These points are collinear.</h2>

Step-by-step explanation:

If three points are collinear, then the slopes are the same.

The formula of a slope:

$m=\dfrac{y_2-y_1}{x_2-x_1}$

For (0, -10) and (-3, -13):

$m=\dfrac{-13-(-10)}{-3-0}=\dfrac{-13+10}{-3}=\dfrac{-3}{-3}=1$

For (-3, -13) and (2, -8):

$m=\dfrac{-8-(-13)}{2-(-3)}=\dfrac{-8+13}{2+3}=\dfrac{5}{5}=1$

We can check the last pair (0, -10) and (2, -8):

$m=\dfrac{-8-(-10)}{2-0}=\dfrac{-8+10}{2}=\dfrac{2}{2}=1$

5 0

3 years ago

Justin weighted 8lb 12oz when he was born at his two week check-up he had gained 8oz what was his new weight in pounds and ounce

avanturin [10]

The answer is 9 lbs and 4 oz

5 0

3 years ago

Read 2 more answers

Help math question derivative!

atroni [7]

Let $f(x)=\sec^{-1}x$ . Then $\sec f(x)=x$ , and differentiating both sides with respect to $x$ gives

$(\sec f(x))'=\sec f(x)\tan f(x)\,f'(x)=1$
$f'(x)=\dfrac1{\sec f(x)\tan f(x)}$

Now, when $x=\sqrt2$ , you get

$(\sec^{-1})'(\sqrt2)=f'(\sqrt2)=\dfrac1{\sec\left(\sec^{-1}\sqrt2\right)\tan\left(\sec^{-1}\sqrt2\right)}$

You have $\sec^{-1}\sqrt2=\dfrac\pi4$ , so $\sec\left(\sec^{-1}\sqrt2\right)=\sqrt2$ and $\tan\left(\sec^{-1}\sqrt2\right)=1$ . So $(\sec^{-1})'(\sqrt2)=\dfrac1{\sqrt2\times1}=\dfrac1{\sqrt2}$

5 0

3 years ago