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Olenka [21]
3 years ago
12

a+solution+of+salt+and+water+contains+75+grams+of+water+per+150+mililiters+of+the+solution.+if+1+mole+of+water+weighs+16+grams,+

how+many+moles+of+water+would+be+present+in+30+mililiters+of+the+solution
Mathematics
1 answer:
bazaltina [42]3 years ago
5 0
A solution of salt and water contain 75 grams of water per 150 millimeters of the solution. If 1 mole of water weight 16 grams, how many moles of water would be present in 30 mil liter of the solution.
Kindly mark as brainliest
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. One-fifth of the plants in a garden are grape tomato plants. Two-ninths of the plants in the garden are cherry tomato plants.
vodomira [7]

Answer:52

Step-by-step explanation:

Given

one-fifth of the plants in a garden are grape tomato plant

two-ninth of the plants in the garden are cherry tomato plants

Also there are 18 grape tomato and 20 cherry tomato plants.

Let Total plants be x

Then Grape tomato plants =\frac{x}{5}=18

x=90

thus there are total of 90 plants

also grape tomato+cherry tomato=18+20=38

Thus there are 90-38=52 remaining plants.

6 0
3 years ago
A band expects to put 12 songs on their next CD. The band writes and records 25​% more songs than they expect to put on the CD.
sweet-ann [11.9K]
12 songs. 25% of 12 is 3, so you add 3 to 12 and get 15. You then find that 20% of 15 is 3 so you subtract 3 from 15 and get 12 again.
7 0
3 years ago
Read 2 more answers
Please help me with these, oh sweet jesus
Lelechka [254]

Answer:

77.  \cot^{6} x = \cot^{4} x \csc^{2}x - \cot^{4} xProved

78.  \sec^{4}x \tan^{2} x = \sec^{2}x [\tan^{2}x + \tan^{4}x ] Proved

79. \cos^{3} x\sin^{2} x = [\sin^{2}x - \sin^{4}x] \cos x Proved.

80. \sin^{4}x - \cos^{4}x = 1 - 2\cos^{2}x + 2 \cos^{4} x Proved.

Step-by-step explanation:

77. Left hand side

= \cot^{6} x

= \cot^{4} x \times \cot^{2} x

= \cot^{4}x [\csc^{2}x - 1]  

{Since we know, \csc^{2} x - \cot^{2}x = 1}

= \cot^{4} x \csc^{2}x - \cot^{4} x  

= Right hand side (Proved)

78. Left hand side

= \sec^{4}x \tan^{2} x

= \sec^{2} x [1 + \tan^{2}x] \tan^{2} x  

{Since \sec^{2}x - \tan^{2}x = 1}

= \sec^{2}x [\tan^{2}x + \tan^{4}x ]

= Right hand side (Proved)

79. Left hand side  

= \cos^{3} x\sin^{2} x

= \cos x[1 - \sin^{2} x] \sin^{2} x

{Since \sin^{2}x + \cos^{2} x = 1}

= [\sin^{2}x - \sin^{4}x] \cos x

= Right hand side

80. Left hand side  

= \sin^{4}x - \cos^{4}x

= [\sin^{2}x + \cos^{2}x]^{2} - 2\sin^{2} x \cos^{2}x

{Since \sin^{2}x + \cos^{2} x = 1}

= 1 - 2\cos^{2} x[1 - \cos^{2}x ]

= 1 - 2\cos^{2}x + 2 \cos^{4} x

= Right hand side. (Proved)

7 0
3 years ago
7(5-4x) in distributive property
kati45 [8]

Answer:

Distributive Property : <em><u>a(b+c)= ab+ac</u></em>

  • 7(5-4x) = (35-28x) ★
<h3><u>(35-28x)</u> is the right answer.</h3>
3 0
3 years ago
Braden logged 3 2/3 hours last week in his reading journal and 2 3/6 hours last week how many fewer hours did Brayden log in his
lubasha [3.4K]

Answer:

1 1/6

Step-by-step explanation:

im guessing you need this fast so i wont explain it

5 0
3 years ago
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