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grigory [225]
3 years ago
8

1/2h - 12 = -5 How can you isolate the term 1/2h

Mathematics
1 answer:
Ronch [10]3 years ago
3 0
Divide by “h-12” on both sides
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How much of 30% HCI and 45% HCI would be needed to mix and produce 50 liters of 35% HCI
VladimirAG [237]

Answer:

  • 33 1/3 liters of 30%
  • 16 2/3 liters of 45%

Step-by-step explanation:

Let x represent the liters of 45% solution needed. Then the amount of HCl in the mix is ...

  0.45x +0.30(50 -x) = 0.35(50)

  0.15x = 0.05(50) . . . . . simplify, subtract 0.30(50)

  x = (0.05/0.15)(50) = 50/3 = 16 2/3 . . . liters of 45% HCl

33  1/3 liters of 30% and 16 2/3 liters of 45% HCl are needed.

_____

<em>Comment on the solution</em>

You may notice that the general solution to a mixture problem of this sort is that the fraction of the mix that is the highest contributor is ...

  (mix % - low %) / (high % - low %) = (.35 -.30) / (.45 -.30) = .05/.15 = 1/3

5 0
3 years ago
Prove or disprove that the point lies on the circle centered at the origin
tigry1 [53]

Answer:

Step-by-step explanation:

5 0
2 years ago
Pls help me with this problem:( and I’ll be happy if you write steps for this
andreev551 [17]

Answer:

b/(b+a)

Step-by-step explanation:

(1/a)-(1/b) :[ (b²-a²)/ab²]

first  solve :

common denominator ab

(1/a)-(1/b)  = (b-a)/ab

[b-a/ab] : [(b²-a²)/ab²]

when divide fraction ( division sign turn to (×) and flip the second fraction(reciprocal):

[b-a/ab] × [ab²/ (b²-a²)]

then  simplify : ab²/ab = b

(b-a)×(b/b²-a²)    

factorize : b²-a² = (b-a)(b+a)

(b-a)×(b/(b-a)(b+a))  simplify : (b-a)/b-a = 1

[(b-a)(b)]/[(b-a)(b+a)

b/b+a

4 0
3 years ago
Mr. Anderson had 4/5 of a pizza left over and split it among his 4 children. What fraction of the pizza did each child get?
Soloha48 [4]

Answer:

1

Step-by-step explanation:

there were  5 pieces now there is 4. so 4 slices of pizza divided by 4 is 1

3 0
3 years ago
How do you evaluate sin(13π/12)?
Alekssandra [29.7K]

13pi/12 lies between pi and 2pi, which means sin(13pi/12) < 0

Recall the double angle identity,

sin^2(x) = (1 - cos(2x))/2

If we let x = 13pi/12, then

sin(13pi/12) = - sqrt[(1 - cos(13pi/6))/2]

where we took the negative square root because we expect a negative value.

Now, because cosine has a period of 2pi, we have

cos(13pi/6) = cos(2pi + pi/6) = cos(pi/6) = sqrt[3]/2

Then

sin(13pi/12) = - sqrt[(1 - sqrt[3]/2)/2]

sin(13pi/12) = - sqrt[2 - sqrt[3]]/2

7 0
3 years ago
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