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Original angles are  45º  and  60º     (Ratio  3 : 4)

The ratio of the complements is      3 : 2 = (90 - 45) : (90 - 60) = 45 : 30

The ratio of the supplements is        9 : 8 = (180 - 45) : (180 - 60) = 135 : 120

16.71 is still wrong?

Sorry, misread your question !!!.

If the quarters have a total value of 6.60, then:

6.60 / 0.25 =26.4 - ???!!!! 

Well, there is something wrong! Your quarters must have a value that is a multiple of 25. Right?

Could be 6.50? Or other amount?

Let the number of quarters = Q

Then, the number of dimes =3Q

0.25Q  +  0.10 * 3Q = 6.60, solve for Q

Q = 12 quarters

6.60 - [12 x 0.25] = 3.60 dollars, which are ALL in dimes.

[3.60 x 100] / 10 = 36 dimes.

Do you understand  your question and what is being asked of you to find out? I'm not quite certain but here is my attempt. Let me know your thinking.

1 - How many "positive odd integers" are there between 1 and 2010? There are: 2010 / 2 =1005. right?

2 -Of those 1005 "positive odd integers", how many do you think have their reciprocals end in a terminating decimal?

3 - My understanding, correct me if I'm wrong, is that only: 5^1 =5,  5^2 =25, 5^3 =125, and 5^4 =625, or a total of 4 integers have that property.

4 - What is the probability of randomly picking one of these 4 integers out of a total of 1005 "positive odd integers"?

5 -  Well, It is: 4 / 1005, which are "relatively prime", that is, their GCD = 1

6 - Therefore, a =4 and b=1005 and a + b =4 + 1005 =1009 .

7 - Did you understand all of the above and what do you think? Let me know if I went astray somewhere !.

I will, tyty

AQ * BQ = 16 * 12 = 192

CQ * DQ = 6 * 32 = 192

CD = CQ + DQ = 6 + 32 = 38

The minimum length of CQ = 6

that is incorrect please more help!

\(\displaystyle \text{If the equation }\: x^{3}+bx^{2}+cx+d=0,\: \text{has roots}\: \alpha, \beta, \gamma,\\ \text{then} \\ \alpha + \beta + \gamma = -b, \\ \alpha \beta + \beta \gamma + \gamma \alpha = c, \text{ and}\\ \alpha \beta \gamma = -d. \)

\(\displaystyle \text{So, for the equation} \: x^{3}-Mx +15=0,\\ \alpha + \beta + \gamma = 0 \dots \dots(1)\\ \alpha \beta + \beta \gamma +\gamma \alpha = -M \dots \dots(2)\\ \alpha\beta \gamma=-15 \dots \dots(3).\)

\(\displaystyle \text{If now } \: \alpha + \beta = 5, \text{then from (1)}, \gamma = -5.\)

\(\displaystyle \text{Substituting into (2)},\: \alpha \beta -5\beta-5\alpha=\alpha \beta -5(\alpha + \beta)=\alpha \beta -25=-M, \\ \text{from which} \: M =25-\alpha\beta \dots \dots (4).\)

\(\displaystyle \text{From (3)} \: \alpha \beta(-5)=-15\quad\text{so } \: \alpha \beta = 3 \: \: \text{and so from (4)}, \\ M=25-3 = 22.\)

Disregard, solved it myself. 

Minimum length of CQ was 2. 

https://web2.0calc.com/questions/a-circle-has-an-arc-of-length-32-that-is-intercepted

Answer seen here. ^^^

There are 154 sweets.

WX/XY = 5/4.

Does AC look like it is only .21 longer than AB ?

Cosine rule

AC^2 = 3^2 + 7^2 - 2 (3)(7)cos 50=.......

The expression (4k + 8)/4 can be written as 4k/4 + 8/4 = 1k + 2 = k + 2. Therefore, a = 1 and b = 2, so a/b = 1/2.  

alright, thank you.

It looks like this forum was a big part of many users when they were growing up. I found this while trying to find answers for my MobyMax. I would love to hear people's stories about how they found this forum and they grew with it.

thanks

that looks wrong

1. How many numbers between 0 and 120 are a multiple of 7? [17]

2. How many multiples of 7 are between -120 and 0? [17]

3. How many multiples of 7 are in 0? [1]

4. Add them all together: [35]

Do you know the number of hours in a day? Good. It's 24. We want to do modular arithmetic, so here it is:

1. How many times does 2025 go into 24? Do the calculation and tell me the remainder. Highlight this for the answer to this part: [84 R9]

2. Take the remainder and add it to the current time (5pm). This gives us: [9+5=14=2am]

3. Thus it will be [2 am] 2025 hours from now.

(Highlight the area between the brackets to see the answer to each step.)

Complementary angles add to 90 degrees.

Supplementary angles add to 180 degrees.

Now that you know this, try to solve the problem. If you want a step-by-step, reply to this thread and I will happily walk you through it.

I just looked at it and saw that Zarina is 10 years old. 

Steps please?

P =0.5283208.........

Q =1.4649735.........

P*Q =0.773976......., so:

8^0.773976...... = 5, so:

Log_10(5) = Log_10(8^0.773976...). Remove the "Log"

5 = 8^0.773976 =8^PQ

It follows that log_{10} 5 = (P + Q)/(8PQ).