How do I do this? I am having a hard time doing this

Mathematics

2 answers:

Alenkasestr [34]3 years ago

7 0

Answer:

you put ur account nuber where it says savings account number, where it says account in words you will put what you wish to withdraw, and put your signature.

Step-by-step explanation:

you will have to write the number in words and in numbers wheres it says dollars and cents x- i hope this helps

kvv77 [185]3 years ago

7 0

Answer:

they formed

Step-by-step explanation:

really easy question if you ask me

You might be interested in

How can you tell that x and y are NOT directly proportional? A) x is 0 when y is 0. B) y is getting farther away from x. C) x is

Vikki [24]

Answer:

D) y is increasing by different amounts.

7 0

2 years ago

Read 2 more answers

Write the slope-intercept form of the equation of each line given the slope and y-intercept

amid [387]

The slope-intercept form of the equation for a line is ...
y = (Slope)x + (y-intercept)

Put the given information in the appropriate spot in the equation. Simplify as required.
y = (-1/3)x + 0
can be simplified to

y = (-1/3)x

6 0

3 years ago

Please help math questions put number with answers please

natita [175]

Answer:

See below for answers

Step-by-step explanation:

Problem 1

$\frac{1}{sin\theta}=2cos\theta\: ; \: 0\leq\theta < 2\pi\\\\1=2sin\theta cos\theta\\\\1=sin2\theta\\\\\frac{\pi}{2}+2\pi n=2\theta\\ \\\frac{\pi}{4}+\pi n=\theta\\ \\\theta=\bigr\{\frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}\bigr\}$

Therefore, the second option is correct.

Problem 2

$cos(2x)+1=sin(x)+2\:;\: 0\leq x < 2\pi\\\\1-2sin^2x+1=sin(x)+2\\\\-2sin^2x=sin(x)\\\\0=2sin^2x+sinx\\\\0=sinx(2sinx+1)$

$0=sinx\\\\x=\pi n\\\\x=0,\pi$

$0=2sinx+1\\\\-1=2sinx\\\\-\frac{1}{2}=sinx\\ \\x=\frac{7\pi}{6}+2\pi n,\frac{11\pi}{6}+2\pi n\\\\x=\frac{7\pi}{6},\frac{11\pi}{6}$

Therefore, the solution set is $x=\bigr\{0,\pi,\frac{7\pi}{6},\frac{11\pi}{6}\bigr\}$ , making the second option correct.

Problem 3

$2cos^2x=-cosx\: ;\: 0^\circ\leq x < 360^\circ\\\\2cos^2x+cosx=0\\\\cosx(2cosx+1)=0$

$cosx=0\\\\x=\frac{\pi}{2}+\pi n\\ \\x=90^\circ+180n^\circ\\\\x=90^\circ,270^\circ$

$2cosx+1=0\\\\2cosx=-1\\\\cosx=-\frac{1}{2}\\\\x=\frac{2\pi}{3}+2\pi n,\frac{4\pi}{3}+2\pi n\\ \\x=120^\circ+360n^\circ,240^\circ+360n^\circ\\\\x=120^\circ,240^\circ$