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Marta_Voda [28]

4 years ago

13

Virgil has one sixth of a birthday cake left over he wants to share the leftover cake with 3 friends what fraction of the origin

al Cake will each of the four people receive

1 answer:

marta [7]4 years ago

8 0

1/10 hope it helps you out

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<img src="https://tex.z-dn.net/?f=12%20%5Cfrac%7B1%7D%7B3%7D%20%5Cdiv%202" id="TexFormula1" title="12 \frac{1}{3} \div 2" alt="1

aev [14]

Answer:

37/6

Step-by-step explanation:

Simplify the following:

(12 + 1/3)/2

Put 12 + 1/3 over the common denominator 3. 12 + 1/3 = (3×12)/3 + 1/3:

((3×12)/3 + 1/3)/2

3×12 = 36:

(36/3 + 1/3)/2

36/3 + 1/3 = (36 + 1)/3:

((36 + 1)/3)/2

36 + 1 = 37:

(37/3)/2

37/3×1/2 = 37/(3×2):

37/(3×2)

3×2 = 6:

Answer: 37/6

3 0

3 years ago

Help please please please please please please please please

Cerrena [4.2K]

Question 1= $7,620

Question 2 (a) $3,211

Question 2 (b) $9,828

5 0

3 years ago

HURRY PLS!! TEST!!!!

stich3 [128]

Answer:

y=-5x+19

Step-by-step explanation:

yeah -ya...... right?

5 0

1 year ago

-412-(-223) worked out

shutvik [7]

Answer:

-189

Step-by-step explanation:

Remember to follow the rules of signs:

- If there are two positive signs next to each other, you add:

- If there is one negative sign and one positive sign next to each other, you subtract.

- If there is two negative signs next to each other, you add:

- 412 - (-223) = -412 + 223

Subtract:

223 - 412 = -189

-189 is your answer.

~

5 0

4 years ago

Read 2 more answers

a toy rocket is launched from the top of a 48 foot hill. The rockets initial upward velocity is 32 feet per second and its heigh

sdas [7]

Answer:

t = 2.11 seconds

Step-by-step explanation:

A toy rocket is launched from the top of a 48 foot hill. The rockets initial upward velocity is 32 feet per second and its height h at any given second t is modeled by the equation:

$h=-16t^2+32t+48$

Let us assume that we need to find the time by it to reach the ground. It means h = 0

$-16t^2+32t+48=0$

The above is a quadratic equation. The value of t is given by :

$t=\dfrac{-b\pm \sqrt{b^2-2ac} }{2a}\\\\t=\dfrac{-b+ \sqrt{b^2-2ac} }{2a},\dfrac{-b- \sqrt{b^2-2ac} }{2a}\\\\t=\dfrac{-32+ \sqrt{(32)^2-2\times (-16)(8)} }{2\times (-16)},\dfrac{-32-\sqrt{(32)^{2}-2\times(-16)(8)}}{2\times(-16)}\\\\t=-0.11\ s, 2.11\ s$

So, it will take 2.11 seconds to reach the ground.

6 0

3 years ago