Answer:
Pure water =74.1 mL
Step-by-step explanation:
You need 420 mL of a 70% alcohol solution.
420 mL of solution and 70% of it should be alcohol.
So,
420 * 70/100 mL
= 420 * 0.7
= 294 mL of alcohol and the rest should be water.
On hand, you have a 85% alcohol mixture.
The alcohol can only come from the 85% mixture. So, how much of 85% alcohol do we need to get 294 mL of alcohol:
Let
y = alcohol solution needed
85% of y = 294
85/100 * y = 294
y = 294 ÷ 85/100
y = 294 × 100/85
y = 29,400/85
y = 345.88235294117
Approximately, 345.9mL
The rest of 420 mL solution has to be pure water
Pure water = 420 mL - 345.9mL
= 74.1 mL
Pure water =74.1 mL
Short Answer: Tony 40 Cleo 30
Givens
T = Tony
C = Cleo
Equations
1/2 T + 1/3 C = 30 (1)
2/5 T + 1/2 C = 31 (2)
Adjustments
Multiply (1) by 3
Multiply (2) by 2
New Equations
3/2 T + C = 90 (1a)
<u>4/5 T + C = 62</u> (2a) Subtract (2a) from (1a)
(3/2 - 4/5)T = 28 the common multiple of 2 and 5 = 10.
(15/10 - 8/10)T = 28
7/10 T = 28 Multiply by 10
7T = 28*10
7T = 280 Divide by 7
T = 280/7 = 40
Put T = 40 into (1)
1/2 T + 1/3 C = 30
1/2 (40) + 1/3 C = 30
20 + 1/3 C = 30 Subtract 20 from both sides.
1/3 C = 30 - 20
1/3 C = 10 Multiply through by 3
C = 10 * 3
C = 30
Answer:
Cleo has 30 books.
Tony has 40 books.
Check
<em>Use Equation 2</em>
2/5 T + 1/2 C = 31
2/5*40 + 1/2 * 30 = ? 31
16 + 15 =?3`
31 = 31 They are equal.
Let width = w
Let length = l
Let area = A
3w+2l=1200
2l=1200-3w
l=1200-3/2
A=w*l
A=w*(1200-3w)/2
A=600w-(3/2)*w^2
If I set A=0 to find the roots, the maximum will be at wmax=-b/2a which is exactly 1/2 way between the roots-(3/2)*w^2+600w=0
-b=-600
2a=-3
-b/2a=-600/-3
-600/-3=200
w=200
And, since 3w+2l=1200
3*200+2l=1200
2l = 600
l = 300
The dimensions of the largest enclosure willbe when width = 200 ft and length = 300 ft
check answer:
3w+2l=1200
3*200+2*300=1200
600+600=1200
1200=1200
and A=w*l
A=200*300
A=60000 ft2
To see if this is max area change w and l slightly but still make 3w+2l=1200 true, like
w=200.1
l=299.85
A=299.85*200.1
A=59999.985
Answer is D
After dilation the image remain same but it is stretched or shrinked to the original size.
That means the ratio of the sides and the angles between the sides remain same .
Here we did dilation of ABCD which made it to EFGH.
Hence the ratio of the corresponding sides of the original rectangle ABCD should remain same even after dilation.
the corresponding sides are : AB and EF
BC and FG
CD and GH
DA and HE
* Let us find ratio of the sides AB and BC
given that AB= 10 and BC= 14
AB/BC= 10/14 = 5 /7 ( 10 and 14 both are multiple of 2 so we reduced them by a factor of 2 )
* the raio of the corresponding sides EF and FG should be same ( 5/7)
in the option D EF= 25 and FG= 35
so EF/FG= 25/35 = 5 /7 ( both are multiple of 5 so we reduced them by the factor of 5 )
Since ratio of the corresponding sides are coming out to be same for the EFGH given in option D it should be the dilation of the ABCD
