Rachel, Nancy, and Diego were in fishing competition. Rachel's fish was 7/8 foot long, Nancy's fish was 1/4 foot long, and Diego's fish was 1/2 foot long. What are the lengths of the fish in order from least to greatest

Answers

Answer 1

Answer: It is 1/4 (.25), 1/2 (.5)and 7/8 (.875) from least to greatest

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Twelve less than the product of three and a number is less than 21.

16/4 x 11/5

what is this answer ?

Answers

Answer:

44x/5

Step-by-step explanation:

Answer: what method are you using to solve this?

Step-by-step explanation:

Which fraction is greater than 2/3?

A. 1/12
B. 2/6
C. 5/12
D. 6/8

Answers

The answer is D. 6/8

D. 6/8
2/3=.66 and 6/8=.75 and .66<.75
:)

If A, B,C are the angles of a triangle then prove: (the following in picture)Please help me to prove this.

Answers

Answer: see proof below

Step-by-step explanation:

Given: A + B + C = π → A + B = π - C

→ C = π - (A + B)

Use Sum to Product Identity: cos A + cos B = 2 cos [(A + B)/2] · cos [(A - B)/2]

Use Product to Sum Identity: 2 sin A · sin B = cos [(A + B)/2] - cos [(A - B)/2]

Use the Double Angle Identity: cos 2A = 1 - 2 sin² A

Use the Cofunction Identity: cos (π/2 - A) = sin A

Proof LHS → RHS:

LHS: cos A + cos B + cos C

= (cos A + cos B) + cos C

$\text{Sum to Product:}\qquad 2\cos \bigg((A+B)/(2)\bigg)\cdot \cos \bigg((A-B)/(2)\bigg)+\cos C$

$\text{Given:}\qquad 2\cos \bigg((\pi -C)/(2)\bigg)\cdot \cos \bigg((A-B)/(2)\bigg)+\cos C\n\n\n.\qquad \qquad =2\cos \bigg((\pi)/(2) -(C)/(2)\bigg)\cdot \cos \bigg((A-B)/(2)\bigg)+\cos C$

$\text{Cofunction:}\qquad 2\sin \bigg((C)/(2)\bigg)\cdot \cos \bigg((A-B)/(2)\bigg)+\cos C$

$\text{Double Angle:}\qquad 2\sin \bigg((C)/(2)\bigg)\cdot \cos \bigg((A-B)/(2)\bigg)+\cos\bigg(2\cdot (C)/(2)\bigg)\n\n\n.\qquad \qquad \qquad =2\sin \bigg((C)/(2)\bigg)\cdot \cos \bigg((A-B)/(2)\bigg)+1-2\sin^2 \bigg((C)/(2)\bigg)\n\n\n.\qquad \qquad \qquad =1+2\sin \bigg((C)/(2)\bigg)\cdot \cos \bigg((A-B)/(2)\bigg)-2\sin^2\bigg((C)/(2)\bigg)$

$\text{Factor:}\qquad 1+2\sin \bigg((C)/(2)\bigg)\bigg[\cos \bigg((A-B)/(2)\bigg)-\sin\bigg((C)/(2)\bigg)\bigg]$

$\text{Given:}\qquad 1+2\sin \bigg((C)/(2)\bigg)\bigg[\cos \bigg((A-B)/(2)\bigg)-\sin\bigg((\pi-(A+B))/(2)\bigg)\bigg]\n\n\n.\qquad \qquad 1+2\sin \bigg((C)/(2)\bigg)\bigg[\cos \bigg((A-B)/(2)\bigg)-\sin\bigg((\pi)/(2)-(A+B)/(2)\bigg)\bigg]$

$\text{Cofunction:}\qquad 1+2\sin \bigg((C)/(2)\bigg)\bigg[\cos \bigg((A-B)/(2)\bigg)-\cos\bigg((A+B)/(2)\bigg)\bigg]$

$\text{Product to Sum:}\qquad 1+2\sin \bigg((C)/(2)\bigg)\bigg[2\sin \bigg((A)/(2)\bigg)\cdot \sin\bigg((B)/(2)\bigg)\bigg]\n\n\n.\qquad \qquad \qquad \qquad =1+4\sin \bigg((C)/(2)\bigg)\bigg[\sin \bigg((A)/(2)\bigg)\cdot \sin\bigg((B)/(2)\bigg)\bigg]\n\n\n.\qquad \qquad \qquad \qquad =1+4\sin \bigg((A)/(2)\bigg)\sin \bigg((B)/(2)\bigg) \sin\bigg((C)/(2)\bigg)$

$\text{LHS = RHS:}\ 1+4\sin \bigg((A)/(2)\bigg)\sin \bigg((B)/(2)\bigg) \sin\bigg((C)/(2)\bigg)=1+4\sin \bigg((A)/(2)\bigg)\sin \bigg((B)/(2)\bigg) \sin\bigg((C)/(2)\bigg)\quad \checkmark$

The proof for this is simple. Let's say that A + B + C = π. From here on we require several trigonometric identities that must be applied.

$\cos \left(A\right)+\cos \left(B\right)+\cos \left(C\right) \n= 2 * cos((A + B) / 2) * cos((A - B) / 2) + \cos C \n= 2 * cos((\pi /2) - (C/2)) * cos((A - B) / 2) +\cos C \n= 2 * sin(C/2) * cos((A - B) / 2) + (1 - 2 * sin^2 (C/2)) \n= 1 + 2 sin (C/2) * cos((A - B) / 2) - sin (C/2) \n= 1 + 2 sin (C/2) * cos((A - B) / 2) - sin((\pi /2) - (A + B)/2 ))\n= 1 + 2 sin (C/2) * cos((A - B) / 2) - cos((A + B)/ 2)\n= 1 + 2 sin (C/2) * 2 sin (A/2) * sin(B/2) \n= 1 + 4 sin(A/2) sin(B/2) sin(C/2)$

Hope that helps!

Chromium-51 has a half-life of 28 days. If Georgia leaves a bottle of 10,000-milligram capsules in her medicine cabinet for a year, what will be the strength in milligrams of each capsule? Round to the nearest milligram

Answers

For finding the strength of the capsules after one year, we will use half-life formula. The formula is:

A = A₀ $((1)/(2))^(t)/(h)$

where, A= Final amount

A₀ = Initial amount

t= time elapsed

h= half-life

Here, in this problem A₀ = 10000 milligram, t= 1 year or 365 days

and h= 28 days

So, A = 10000 $((1)/(2) )^(365)/(28)$

⇒ A = 10000 $((1)/(2))^1^3^.^0^4$

⇒ A = 1.187

So, the strength of the capsules after one year will be 1.187 milligrams.

How many fourths are equivalent to 2/8