All categories

3 years ago

Emily is building a square bookshelf. She wants to add a diagonal support beam to the back to strengthen it. The

Mathematics

1 answer:

Rina8888 [55]3 years ago

7 0

Answer: 4√2

The sides of a 45-45-90 triangle are in the ratio 1 : 1 : √2. In this case, the ratio is 4 : 4 : 4√2.

i hope this helped! :D

You might be interested in

alice rode her bike the same number every day for 17 days she rode a tottle of 68miles during thes 17 days at this rate how mane

s344n2d4d5 [400]

4 miles!

68 divided by 17 = 4

7 0

3 years ago

Read 2 more answers

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

V2 - 15 = -2v Solve the Quadratic. What are the 2 solutions?

ki77a [65]

Answer:

v=-5 and v=3

Step-by-step explanation:

We are given that

$v^2-15=-2v$

We have to find two solutions of quadratic equation.

$v^2+2v-15=0$

Using addition property of equality

$v^2+5v-3v-15=0$ (By using factorization method)

$v(v+5)-3(v+5)=0$

$(v+5)(v-3)=0$

Substitute each factor equal to 0

$v+5=0$ and $v-3=0$

$v=-5$ and $v=3$

Hence, two solutions of quadratic equation are

v=-5 and v=3

8 0

3 years ago

Please help!!! Tony wants to build a fence around his rectangular garden that has

scZoUnD [109]

Answer:

22 posts

Step-by-step explanation:

on the side of 20 feet you have 5 fence posts each in including one on each of the corners. On the 35 feet side you would have 8 including the corners, but since the corner posts are done its 6 on each 35 feet side. So you get 5+5+6+6=22

6 0

3 years ago

A pet store has 2 gray rabbits one eight of the rabbits at the pet store are gray. How many rabbits does the pet store have?

Eduardwww [97]

2 = 1/8
2 x 8 = 16
16 = 8/8
Glad to Help!

3 0

3 years ago