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[120 + C] / [60 + C] ==[150 + C] / [120 + C], solve for C

C == - 180

- 180 + 60==- 120,     - 180 + 120 == - 60,    - 180 + 150 == - 30

The new GS will be:  -120,  - 60,  - 30

The common ratio is: - 60 / - 120 == 1 / 2

Acute isosceles triangle.

Sides: a = 40   b = 40   c = 29

Area: T = 540.551
Perimeter: p = 109
Semi-perimeter: s = 54.5

T=rs 
r=T / s
​= 540.55 / 54.5
​=9.92 - the Inradius.

R=[abc] / [4rs] ==[40 * 40 * 29] / [4 * 9.918 * 54.5]
R==21.46 - the Circumradius.

See the answer here:  https://web2.0calc.com/questions/algebra_79619

You sure you can't just do this yourself...

Alright, so here's the deal, I think I might be able to do this problem, but no guarantee that it is going to be right. I advise you to check my work afterward, to make sure that it is right:

So, we first know that there are 10 dogs, and we have to split them up into a 3 group, a 5 group, and a 2 group. However, one dog is already set in stone for the 2 dog group, as well as another dog set in stone for the 5 dog group.

We first pick the 3 dog group. We can do this by:

C(8,3) which means (choose 3 of 8). C(8,3) is 56.

Then, for the 5 dog group, we so far have 1 dog already established, then from the remaining 5 dogs left, we can pick 4 of them.

C(5,4) = 5 ways here.

The last group is already decided because we already picked the other two.

When we do 56 * 5 = 280, so I believe there are 280 ways to do this. (However, please REMEMBER, I might be incorrect [ and probably am] )

...........

Do you have a diagram or something else that can help me better understand it?

69420

F(x) = (x^2 + 1)/(x^3 - 3x + 1)

The area of triangle ABC is 12*sqrt(7).

(a)

Let the amount of 75% red potion needed==R

0.75R  +  0.15(200) ==0.25[R + 200], solve for R

R ==40 mL of red potion needed

(b)

0.75R  +  0.15[400 - R] ==0.30[400], solve for R

R ==100 mL of red potion needed

400 - 100 ==300 mL of blue potion needed.

(c)

Yes, there is. Example:

0.75[40]  +  0.15[80] ==0.35[40 + 80]

    30         +     12         == 42

                42  ==  42

1 - [(3/12) (2/11) (1/10)]

 2 - [+ 3 (3/12) (2/11) (9/10)]

 3 - [+ (1/8 - (1/8) (3/12) (2/11) (1/10))]

 4 - [(1/8) 3 (3/12) (2/11) (9/10)]

Overall probability =[(3/12) (2/11) (1/10)] + [ 3 (3/12) (2/11) (9/10)] + [ (1/8 - (1/8) (3/12) (2/11) (1/10))] - [(1/8) 3 (3/12) (2/11) (9/10)]

=13 / 55 =0.23636363..........

Here's number 1:

Try a similar approach for the others.

I'll give you some initial hints. First of all, angle ACB is 90 degrees because C lies on the circumference. So triangle ACB is a right triangle, thus CM makes a 45 degree angle with AC and BC, the radius of the circle is sqrt(65). How can you use trig or smarter methods to obtain CM?

I'm so sorry, but those are both incorrect... could i also see some work with the answers so I can understand how to do the problems?

The expected value is 17/50.

(a) The probability is 4/25.

(b) The probability is 8/125.

Probably a typo, but the triangle BDI cant exist with the given dimensions.