All categories

2 years ago

Help please, need to show work.

Mathematics

1 answer:

stiks02 [169]2 years ago

8 0

9514 1404 393

Answer:

3. m∠A < 80°

Step-by-step explanation:

The exterior angle NGS, marked as 80°, has a measure that is the sum of the measures of the remote interior angles: GAN, GNA.

If we assume that angle measures cannot be zero, then we must have ...

m∠A +m∠N = 80°

m∠A = 80° -m∠N

Since m∠N > 0, the measure of angle A must be less than 80°.

This conclusion matches choice 3.

You might be interested in

Which one of these numbers could form a triangle?

padilas [110]

Answer: non of those numbers

Step-by-step explanation:

all the numbers couldn't form a triangle

7 0

2 years ago

While planning a hiking trip, you examine a map of the trail you are going on hike. The scale on the map shows that 2 inches rep

Marrrta [24]

Answer:

8.0

Step-by-step explanation:

eight inch big ooo thats the pick

7 0

3 years ago

Which equation describes a line perpendicular to the line described by y= 1/2x -5?

Aneli [31]

Answer:

y= 2/1x -5

Step-by-step explanation:

flip it!

7 0

3 years ago

Derivative of tan(2x+3) using first principle

kodGreya [7K]

$f(x)=\tan(2x+3)$

The derivative is given by the limit

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

You have

$\displaystyle\lim_{h\to0}\frac{\tan(2(x+h)+3)-\tan(2x+3)}h$
$\displaystyle\lim_{h\to0}\frac{\tan((2x+3)+2h)-\tan(2x+3)}h$

Use the angle sum identity for tangent. I don't remember it off the top of my head, but I do remember the ones for (co)sine.

$\tan(a+b)=\dfrac{\sin(a+b)}{\cos(a+b)}=\dfrac{\sin a\cos b+\cos a\sin b}{\cos a\cos b-\sin a\sin b}=\dfrac{\tan a+\tan b}{1-\tan a\tan b}$

By this identity, you have

$\tan((2x+3)+2h)=\dfrac{\tan(2x+3)+\tan2h}{1-\tan(2x+3)\tan2h}$

So in the limit you get

$\displaystyle\lim_{h\to0}\frac{\dfrac{\tan(2x+3)+\tan2h}{1-\tan(2x+3)\tan2h}-\tan(2x+3)}h$
$\displaystyle\lim_{h\to0}\frac{\tan(2x+3)+\tan2h-\tan(2x+3)(1-\tan(2x+3)\tan2h)}{h(1-\tan(2x+3)\tan2h)}$
$\displaystyle\lim_{h\to0}\frac{\tan2h+\tan^2(2x+3)\tan2h}{h(1-\tan(2x+3)\tan2h)}$
$\displaystyle\lim_{h\to0}\frac{\tan2h}h\times\lim_{h\to0}\frac{1+\tan^2(2x+3)}{1-\tan(2x+3)\tan2h}$
$\displaystyle\frac12\lim_{h\to0}\frac1{\cos2h}\times\lim_{h\to0}\frac{\sin2h}{2h}\times\lim_{h\to0}\frac{\sec^2(2x+3)}{1-\tan(2x+3)\tan2h}$

The first two limits are both 1, and the single term in the last limit approaches 0 as $h\to0$ , so you're left with

$f'(x)=\dfrac12\sec^2(2x+3)$

which agrees with the result you get from applying the chain rule.

7 0

3 years ago

What is the common ratio of the following sequence? 80, 20, 5, ... <br> A) -4 <br> B) 1/4 <br> C) 4

Stolb23 [73]

$\bf 80~~,~~20~~,~~5~\hfill \cfrac{20}{80}\implies \boxed{\cfrac{1}{4}}~~,~~ \cfrac{5}{20}\implies \boxed{\cfrac{1}{4}}~\hfill \stackrel{\textit{common ratio}}{\boxed{\cfrac{1}{4}}}$

6 0

3 years ago