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Vince need 12 quarts of a 60% anti-freeze solution. He will continue an amount of 100% anti-freeze with an amount of a 50% anti-freeze solution. How many quarts of each solution should be mixed to make the required amount of the 60% anti-freeze solution?
.
Let x = amount (quarts) of 100% solution
then
12-x = amount (quarts) of 50% solution
.
x + .50(12-x) = .60(12)
x + 6-.50x = 7.2
.50x + 6 = 7.2
.50x = 1.2
x = 2.4 quarts of 100% solution
.
50% solution:
12-x = 12-2.4 = 9.6 quarts of 50% solution

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2 log 5 + log x=2
log 5^2 + log x=2
log 25 + log x=2
log 25x = 2
25x = 10^2
25x = 100
x = 4

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what is the answer to log 3z equal 2 ?





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what is the answer to ln x equal 2 ?



.
whats the answer to 3 ln 5x equal 10 ?





.
whats the answer to ln square root x-8 equal 5 ?

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The area of the piece is 30 ft². The length is 1 less than 3 times the width. How wide should the metal piece be? Round to the nearest hundredth of a foot.
.
Let w = width
then
3w-1 = length
.
w(3w-1) = 30
3w^2-w = 30
3w^2-w-30 = 0
replace middle term:
3w^2 +9w-10w -30 = 0
group terms:
(3w^2+9w) - (10w+30) = 0
3w(w+3) - 10(w+3) = 0
(w+3)(3w-10) = 0
w = {-3, 10/3}
throw out the negative solution (extraneous) leaving:
w = 10/3
w = 3.33 feet

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what is the length of the hypotenuse of a right triangle that has legs with lengths of 5 and 4?
apply the Pythagorean Theorem:
Let h = hypotenuse
then
h^2 = 5^2 + 4^2
h^2 = 25 + 16
h^2 = 41
h = sqrt(41)
h = 6.40

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Li's car gets 23 miles per gallon on the highway. If 1 gallon of gas cost #3.20, and Li has $50.00 worth of gas in her car, how many miles can she drive on the highway? Give the exact answer.
50/3.20 *23
500/32 *23
125/8 *23
2875/8
359 and 3/8 miles (exact answer)

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It takes 9 hours for pump A to fill a tank alone. Pump B takes 15 hours to fill the same tank alone. If pumps A,B, and C are used, the tank fills in 5 hours. How long does it take pump C to fill the tank?
.
Let c = time (hours) it takes for pump C to fill tank
then
5(1/9 + 1/15 + 1/c) = 1
5/9 + 5/15 + 5/c = 1
5/9 + 1/3 + 5/c = 1
multiplying both sides by 9c:
5c + 3c + 45 = 9c
8c + 45 = 9c
45 hours = c

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-3(y+7)
distribute the -3 to each term inside the parenthesis:
-3y-21
answer, then, is B

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8m - 6
has 2 terms: 8m and 6
factoring using prime factors we get:
8m = 2*2*2*m
6 = 2*3
a "common factor" is 2, so:
2(4m - 3) (this is what they're looking for)

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10,11, and 20
let's assign:
10(a),11(b), and 20(c)
.
The triangle inequality rule says that:
c – b < a < c + b
20 – 11 < 10 < 20 + 11
9 < 10 < 31
Since the above is TRUE:
answer: YES, it is possible

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If the length of a rectangle is 15 cm and the width is 12 cm, what is the area?
For any rectangle:
area = width * length
area = 12*15
area = 180 square cm

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A biologist has a 30% solution and a 5% solution of the same plant nutrient. How many cubic centimeters (cc) of each solution should be mixed to obtain a 25 cc of a 25% solution? (Round your answers to 2 decimal places if needed)
.
Let x = amount (cc) of 30% solution
then
25-x = amount (cc) of 5% solution
.
.30x + .05(25-x) = .25(25)
.30x + 1.25-.05x = 6.25
.25x + 1.25 = 6.25
.25x = 5
x = 20 cc (of 30% solution)
.
amount of 5% solution:
25-x = 25-20 = 5 cc

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A 20% iodine solution is mixed with a 70% iodine solution to produce 5 gallons of 50% Iodine solution. How many gallons of each solution are needed?
.
Let x = amount (gallons) of 20% solution
then
5-x = amount (gallons) of 70% solution
.
.20x + .70(5-x) = .50(5)
.20x + 3.5-.70x = 2.5
3.5-.50x = 2.5
-.50x = -1
x = -1/(-.50)
x = 2 gallons (of 20% solution)
.
70% solution:
5-x = 5-2 = 3 gallons

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The perimeter of square is 40m. The perimeter of rectangle of length 12m is as the same as the perimeter of square. Find which encloses the greater area square or rectangle?
Square:
since all four sides are equal:
40/4 = 10 m (length of one side)
area = 10*10 = 100 sq m
.
Rectangle:
Let w = width
then
2(w+12) = 40
w+12 = 20
w = 8 m
Area = 8*12 = 96 sq m
.
Answer: the square

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log5x+log10=0
log5x(10)=0
log50x=0
50x=10^0
50x=1
x = 1/50
x = 0.02

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AMY CAN CLEAN THE ELEPHANT ENCLOSURE IN 4 HOURS, BUT IT TAKE JOEL 6 HOURS. HOW LONG WOULD IT TAKE THEM WORKING TOGETHER.....PLEASE HELP
.
Let x = time (hours) it takes together
then
x(1/4 + 1/6) = 1
multiplying both sides by 12:
x(3 + 2) = 12
x(5) = 12
x = 12/5
x = 2.4 hours
or
x = 2 hours and 24 minutes

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Translate the following statement into an equation. Seven less then twice a number is 4 more than the number
.
Let n = the number
then
2n-7 = n+4

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1/2(2n-5)=4n-1
multiplying both sides by 2:
(2n-5)=8n-2
2n-5=8n-2
subtracting 2n from both sides:
-5=6n-2
adding two to both sides:
-3=6n
dividing both sides by 6:
-3/6 = n
-1/2 = n

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Liam thinks of a number.
He multiplies the number by 5 and then subtracts 60 from the result .
His answer equals the number he started with .
.
Let x = number Liam started with
then
5x-60 = x
5x = x+60
4x = 60
x = 60/4
x = 15
.
What was the number Liam started with ? 15

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what is the dimension of a rectangle whose length is 5 less than twice its width and whose area is 63 square units?
.
Let w = width
then from: "length is 5 less than twice its width" we get
2w-5 = length
.
since:
area = width * length
63 = w(2w-5)
63 = 2w^2-5w
0 = 2w^2-5w-63
rewrite the middle term as:
0 = 2w^2-14w+9w-63
group:
0 = (2w^2-14w) + (9w-63)
0 = 2w(w-7) + 9(w-7)
0 = (w-7)(2w+9)
w = {-9/2, 7}
toss out the negative term (extraneous) leaving:
w = 7 units (width)
.
Length:
2w-5 = 2(7)-5 = 14-5 = 9 units
.
Dimension: 7 units by 9 units

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After 10% discount has been given the discounted cost of a coat is $765. What was the original cost of the coat?
.
Let x = original cost of coat
then
x - .10x = 765
x(1 - .10) = 765
x(.90) = 765
x = 765/.90
x = $850

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log(base 5) 125^1/2 + log(base 11) 121^1/3= 13/6
log(base 5) (5^3)^1/2 + log(base 11) (11^2)^1/3= 13/6
log(base 5) (5^3/2) + log(base 11) (11^2/3)= 13/6
3/2 + 2/3= 13/6
9/6 + 4/6= 13/6
13/6 = 13/6

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log2(x - 2) + log2(x + 1) = log2 10
log2(x - 2)(x + 1) = log2 10
(x - 2)(x + 1) = 10
x^2+x-2x-2 = 10
x^2-x-2 = 10
x^2-x-12 = 0
(x-4)(x+3) = 0
x = {-3, 4}
The -3 is an "extraneous" solution so we throw it out leaving:
x = 4

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log x^3 - 2log 1/x - 1= 0
log x^3 - 2log 1/x = 1
log x^3 - log (1/x)^2 = 1
log x^3 - log (1/x^2) = 1
log x^3 - log (x^-2) = 1
log x^3/(x^-2) = 1
log x^3x^2 = 1
log x^5 = 1
x^5 = 10^1
x^5 = 10
x = 10^(1/5)
x = 1.5849

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1)In(2m+3)=8
2m+3 = e^8
2m = e^8 - 3
m = (e^8 - 3)/2
m = 1488.98
.
2)1.1+In x^2=6
Ln x^2 = 6 - 1.1
x^2 = e^(6 - 1.1)
x = sqrt(e^(6 - 1.1))
x = 11.59
.
3)e^x+1=30
e^x=30-1
x = ln(30-1)
x = 3.37

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Apply "log rules":


.


.

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Okay I'm stumped on this problem. I have tried to figure out the equation but I keep geting stuck. The problem is: "A chemist has a 90% solution of sulfuric acid and a 70% solution of the same acid. How many gallons of each must be mixed in order to make 200 gallons of 75% solution?"
Here's what I know: I know that x is the amount of the 90% solution and that y would be the amount of the 70% solution. I think that the equation would be .90x + .70y = .75 and that the other equation is x+y=200. but I am not sure I did that right and I keep getting fractions as my answer. Help please.
.
You almost had it right. It should have been:
Let x = amount (gallons) of 90% solution
and y = amount (gallons) of 70% solution
.
x+y=200 (equation 1)
.90x + .70y = .75(200) (equation 2)
.
Solving equation 1 for y
x+y=200
y=200-x
Substitute above into equation 2:
.90x + .70y = .75(200)
.90x + .70(200-x) = .75(200)
.90x + 140-.70x = 150
.20x + 140 = 150
.20x = 10
x = 50 gallons (of 90% solution)
.
70% solution:
x+y=200
50+y=200
y = 150 gallons

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h=-16t^2+vt+s, where h is the height (in feet), t is the time in motion (in seconds), v is the initial velocity (in feet per second), and s is the initial height(in feet)
A gymnast dismounts the uneven parallel bars at a height of 8 feet with an initial upward velocity of 8 feet per second. Find the time (in seconds) it takes for the gymnast to reach the ground.
.
from the statement of the problem we have:
v is 8 ft per sec
s is 8 ft
set h to zero (ground) and solve for t:
h=-16t^2+vt+s
0=-16t^2+8t+8
dividing both sides by 8:
0=-2t^2+t+1
multiplying both sides by -1:
0=2t^2-t-1
0=2t^2-2t+t-1
0=(2t^2-2t)+(t-1)
0=2t(t-1)+(t-1)
0=(t-1)(2t+1)
t = {-1/2, 1}
throw out the negative solution leaving:
t = 1 second