Ask question

Ask question

All categories

Alenkasestr [34]

3 years ago

10

I bet no one can solve this (+)^=∑24_(=0)^▒〖(¦) ^ ^(−) 〗 HAHAHAAAAAAAAAAAAA

2 answers:

umka21 [38]3 years ago

8 0

What's this? Cute. Thnx for the freee pointttssss <3

puteri [66]3 years ago

5 0

Answer:

free points nice

Step-by-step explanation:

You might be interested in

Please help due in 5 minutes !

Elenna [48]

It is horizontal I believe because if you were to flip it horizontally then it would be symmetrical

8 0

3 years ago

Read 2 more answers

Graph the linear equation.find three points that solve the equation,then plot on the graph -3x++2y=11

Katarina [22]

$-3x+2y=11\qquad\text{add 3x to both sides}\\\\2y=3x+11\qquad\text{divide both sides by 2}\\\\y=\dfrac{3}{2}x+\dfrac{11}{2}\\\\for\ x=1\to y=\dfrac{3}{2}(1)+\dfrac{11}{2}=\dfrac{3}{2}+\dfrac{11}{2}=\dfrac{14}{2}=7\to(1,\ 7)\\\\for\ x=-3\to y=\dfrac{3}{2}(-3)+\dfrac{11}{2}=-\dfrac{9}{2}+\dfrac{11}{2}=\dfrac{2}{2}=1\to(-3,\ 1)\\\\for\ x=-5\to y=\dfrac{3}{2}(-5)+\dfrac{11}{2}=-\dfrac{15}{2}+\dfrac{11}{2}=-\dfrac{4}{2}=-2\to(-5,\ -2)$

3 0

3 years ago

What is the 6th term of this geometric sequence? 1,7,49,343

ludmilkaskok [199]

Since it is times 7 each time it will be 16807

3 0

3 years ago

Read 2 more answers

Jill got 36 out of 40 questions correct on her test. What percent of her test did she get correct?

ale4655 [162]

Answer:

36 percent I think ya

Step-by-step explanation:

I think I should know this

5 0

3 years ago

Read 2 more answers

<img src="https://tex.z-dn.net/?f=%5Cfrac%7B%5Csec%5Cleft%28x%5Cright%29%7D%7B%5Ccos%5Cleft%28x%5Cright%29%7D-%5Cfrac%7B%5Csin%5

DanielleElmas [232]

Answer:

1

Step-by-step explanation:

First, convert all the secants and cosecants to cosine and sine, respectively. Recall that $csc(x)=1/sin(x)$ and $sec(x)=1/cos(x)$ .

Thus:

$\frac{sec(x)}{cos(x)} -\frac{sin(x)}{csc(x)cos^2(x)}$

$=\frac{\frac{1}{cos(x)} }{cos(x)} -\frac{sin(x)}{\frac{1}{sin(x)}cos^2(x) }$

Let's do the first part first: (Recall how to divide fractions)

$\frac{\frac{1}{cos(x)} }{cos(x)}=\frac{1}{cos(x)} \cdot \frac{1}{cos(x)}=\frac{1}{cos^2(x)}$

For the second term:

$\frac{sin(x)}{\frac{cos^2(x)}{sin(x)} } =\frac{sin(x)}{1} \cdot\frac{sin(x)}{cos^2(x)}=\frac{sin^2(x)}{cos^2(x)}$

So, all together: (same denominator; combine terms)

$\frac{1}{cos^2(x)}-\frac{sin^2(x)}{cos^2(x)}=\frac{1-sin^2(x)}{cos^2(x)}$

Note the numerator; it can be derived from the Pythagorean Identity:

$sin^2(x)+cos^2(x)=1; cos^2(x)=1-sin^2(x)$

Thus, we can substitute the numerator:

$\frac{1-sin^2(x)}{cos^2(x)}=\frac{cos^2(x)}{cos^2(x)}=1$

Everything simplifies to 1.

7 0

3 years ago