So the price of the bike is increased by 30%. If we assume the original price of the bike is b, 125= b+0.3*b = 1.3(b). We rearrange this, dividing 1.3 on both sides of the equation, and we get 125/1.3 = (roughly) 96.15$. Checking, we add (0.3)(96.15) = 28.85, 96.15+28.85 = 125$.
Reach out if you need any help.
I have 54 coins in my pocket; some of them are quarters and the rest are nickels. In total, the coins are worth $6.10. How many of each coin do I have?
d+n=54
10d+5n=610
now solve simultaneously.
Since Bankok is 2 hours ahead Karachi, then it left Karachi at 1530 + 2 =1730 Bankok time.
3840 / 720 = 5 1/3 or 5 hours and 20 minutes it took the airplane to fly to Bankok.
So, 1730 + 520 =2250 or 2250 - 1200 =10:50 pm Bankok time when the plane arrived.
The period is the distance from from one crest (top) to the next crest ;)
p/2 = 16 - 0
p/2 = 16
p = 32
1 - 7^4 x 4 =9604 Seven digits that can be repeated and which end in one of 4 odd digits [1, 3, 5, 7]
2 - [7^5 x 4] /7 =9604 Five out of 7 digits that be can be repeated that can only end in [1, 3, 5, 7] out of 7.
This graph should help
I see the midline is at 20 and the amplitude is 16 but i'm confused about the period
Here is your graph.
If you know what those terms mean then the answers are easy.
If you do not then google them.
If you want you can post your answers and I (or someone else) will comment on their accuracy.
Please no one else finish this question.
17 quarters and 37 nickels
Substitute y for cosx and then solve as a quadratic.
Solve the equation on the interval [0,2π) .
4cos^2x−2cosx=0
ABCD is a square with side length 4 and center X. AB is the diameter of a semicircle that lies outside ABCD and Y is a point on that semicircle so that XY=12. Find the area of CDY
The length of XY is not 12!!!
AB = 4
XY = 4
Area of CDY = 12 u²
Rectangle area is 40 u2
Area MBCN = 3/8 * 40
You are doing lots of great research Cal.
Keep it up!
This site has the same question with a reasonable explanation:
https://www.algebra.com/algebra/homework/coordinate/word/Linear_Equations_And_Systems_Word_Problems.faq.question.1075719.html
Thanks Cal, I am very glad that you found and posted the original post.
Attn Guest:
Am I right in thinking that you are mathtoo? (the original poster)
Why do you think the answer is wrong and why didn't you say so on the original post?
Here is the original post as a link.
https://web2.0calc.com/questions/helppppp_26
sorry i forgot the interval of the function. the function is over the interval [-2pi, 2pi]. what is the answer for this range
Seriously ? TWENTY questions posted today? Go away.....we are not here to do your homework !!!!
So I inputted this on desmos here
https://www.desmos.com/calculator/qxikraoayu
and looking at the graph the normal domain is a real #s
UwU
and sorry if I am wrong but the implicit domain should be all real #s > 0
so here's my work UwU https://www.desmos.com/calculator/hs9c2pcmfd
how I got it is :
1. you find m (rise/run)
2. find b
helps if you put them on a graph
ans= y=-(9/6)x+2
Let A be the first "theta"
A is in the 3Q ......so the sine is negative and is given by -√ [ 3^2 - (√2)^2] / 3 =
- √7 / 3
Let B be the second "theta"
B is in the 1Q and r = 5
sin B = 4/5 cos B = 3/5
sin ( A + B) = sin A cos B + sin Bcos A =
(-√7/3) ( 3/5) + (4/5) ( -√2/3) =
[ -3√7 - 4√2 ] / 15
+118733 
