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Trava [24]
3 years ago
6

Karl is making a pot of chili. The recipe calls for StartFraction 3 Over 8 EndFraction cup of chili powder, but Karl only wants

to use half as much so it won’t be so spicy. How much chili powder should Karl use?
Mathematics
1 answer:
muminat3 years ago
3 0

Answer:

3/16 or 3 over 16

Step-by-step explanation:

Find half of 3/8:

1/2x3/8 = 3/16

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Flura [38]


8-3x2

= 8 - (3x2)

= 8-6

= 2

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Help me out pleasee!!!
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T=sin⁻¹(36/85)=25.06
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(1 point) A tank contains 2640 L of pure water. A solution that contains 0.09 kg of sugar per liter enters the tank at the rate
OverLord2011 [107]

Answer:

t=0      Sugar = 0 Kg

t=1min Sugar=0.27 Kg

Step-by-step explanation:

Data

Tank = 2640 L (pure water)

Sol=0.09kg Sugar per liter

Vin = Vout = 3L/min

Sugar in the beginning = ?

if beginning is t = 0min Amount of sugar = 0, this is due to the fact that at the moment of entering the tank the content is only water , but if beginning is t= 1min then;

\frac{3L}{min}*\frac{0.09Kg}{L}=0.27kg/min

7 0
3 years ago
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What is the area of the triangle in square yards?<br> 6 yd.<br> 15 3/4 yd.<br><br> hurry plss
Lera25 [3.4K]

Answer:

im guessing 6 is your base and your height is 15 3/4.

Formula of a triangle we must remember:

BH x 1/2 (basically dividing by 2)

94.5 (fraction form will be: 94 1/2)

94.5 divided by 2 = 47.25 <---------- decimal form

94 1/2 divided by 2 = 47 1/4 <------- fraction form

8 0
3 years ago
One Sunday, 120 days before Christmas, Aldsworth store publishes an advertisement saying ‘120 shopping days until Christmas'. Al
Lena [83]

Answer:

(a)18

(b)1089

(c)Sunday

Step-by-step explanation:

The problem presented is an arithmetic sequence where:

  • First Sunday, a=1
  • Common Difference (Every subsequent Sunday), d=7

We want to determine the number of Sundays in the 120 days before Christmas.

(a)In an arithmetic sequence:

\text{The nth term}, T_n=a+(n-1)d\\T_n \leq 120\\$Therefore:$\\1+7(n-1) \leq 120\\1+7n-7\leq 120\\7n-6\leq 120\\7n\leq 120+6\\7n\leq 126\\$Divide both sides by 7$\\n\leq 18

Since the result is a whole number, there are 18 Sundays in which Aldsworth advertises.

Therefore, Aldsworth advertised 18 times.

(b)Next, we want to determine the sum of the first 18 terms of the sequence

1,8,15,...

\text{Sum of a sequence}, S_n=\frac{n}{2}( 2a+(n-1)d)\\S_{18}=\frac{18}{2}( 2*1+(18-1)*7)\\=9(2+17*7)\\=9(2+119)\\=9*121\\S_{18}=1089

The sum of the numbers of days published in all the advertisements is 1089.

(c)SInce the 120th day is the 18th Sunday, Christmas is on Sunday.

6 0
3 years ago
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