The denominator of a fraction is 1 more than the numerator. If both the numerator and denominator are decreased by 3, the resulting fraction is equal to 4/5. Find the fraction
Answers
Related Questions
Express each repeating decimal as a fraction.
1.
1.8[4]
2.
2.1[26]
Answers
1.8444444....=1.8+1/10*0.44444....
=1.8+1/10*4/9
=1.8+4/90
=162/90+4/90
=166/90
=83/45
2.126262626...=2.1+1/10*0.262626...
=2.1+1/10*26/99
=2.1+26/990
=2079/990+26/990
=2105/990
=421/198
Answer:
1.8444444....=1.8+1/10*0.44444....
=1.8+1/10*4/9
=1.8+4/90
=162/90+4/90
=166/90
=83/45
2.126262626...=2.1+1/10*0.262626...
=2.1+1/10*26/99
=2.1+26/990
=2079/990+26/990
=2105/990
=421/198
which
of the following is the least?
A.0.105
B.0.501
c.o.015
D.0.15
Answers
Solve For L: A = L ⋅ W ⋅ H
L=aw/h
L=wh/a
L=a/wh
L=w/ha
Answers
A=L*WH
[tex] \frac{A}{WH} [/tex]=L
So you answer would be: C
Between two tenths of a mile will the second relay exchange occur? Explain.
Answers
2/10
There is one tenth in a mile. I have the same question. But I am just guessing.
Use the function below to find f(4).
Answers
f(4) = 1/3 * 4^4
f(4) = 1/3 * 256
f(4) = 256/3 <===
Answer: The correct option is (C) [tex]\dfrac{256}{3}.[/tex]
Step-by-step explanation: The given function is
[tex]f(x)=\dfrac{1}{3}\times 4^x.[/tex]
We are to find the value of f(4).
To find the value of f(4), we need to substitute the value of x as 4 in the given definition of function f(x).
Substituting x = 4 in f(x), we have
[tex]f(4)=\dfrac{1}{3}\times 4^4=\dfrac{1}{3}\times 256=\dfrac{256}{3}.[/tex]
Therefore, the value of f(4) is [tex]\dfrac{256}{3}.[/tex]
Option (C) is correct.
The following formula relates three quantities force mass and acccelerations
F=ma
Answers
There are 60 minutes in 1 hour. What fraction of an hour is 4 minutes? Write your answer in lowest terms.
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Divide the numerator and denominator by 4
1/15
Final answer: 1/15
The fraction of an hour is 4 minutes in lowest terms is 1/15.
What fraction of an hour is 4 minutes in lowest terms?To find the fraction of an hour that 4 minutes represents, we need to compare the number of minutes to the total number of minutes in an hour, which is 60.
We can set up a fraction where the numerator represents the number of minutes we want to find the fraction for (4 minutes) and the denominator represents the total number of minutes in an hour (60 minutes).
So, the fraction can be written as 4/60.
To simplify the fraction to its lowest terms, we can divide both the numerator and the denominator by their greatest common divisor (GCD). In this case, the GCD of 4 and 60 is 4.
Dividing both the numerator and denominator by 4, we get:
4/60 = (4 ÷ 4) / (60 ÷ 4) = 1/15.
Therefore, 4 minutes in lowest terms is equal to 1/15th of an hour.
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Find the range of the graphed function?
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B is true
B is the correct answer
You want to replace the countertops in a kitchen. One counter measures 24 inches by 6 feet. The other is 30 inches by 4 feet. If the new countertop costs $42.00 per square foot, what will be the cost of the new top?
Answers
2 ft x 6 ft=12 square feet
2.5 ft x 4 ft=10 square feet
Total square feet of the counter is 22 square feet.
To find the total cost, just multiply the area by the cost per square foot.
$42 x 22=$924
Answer:
The cost of the new top $924.
Step-by-step explanation:
You want to replace the counter-tops in a kitchen.
The dimensions of the first counter is 24 inches by 6 feet.
i.e [tex]24 \text{ inch} \times 6 \text{ feet}[/tex]
or [tex]2\text{ feet} \times 6 \text{ feet}[/tex]
As 1 feet =12 inches
Hence the area of the counter is [tex]2\times 6=12\text{ feet}^2[/tex]
The dimensions of the second counter is 30 inches by 4 feet.
i.e [tex]30 \text{ inch} \times 4 \text{ feet}[/tex]
or [tex]2.5\text{ feet} \times 4 \text{ feet}[/tex]
Hence the area of the counter is [tex]2.5\times 4=10\text{ feet}^2[/tex]
Hence, the total area will be [tex]12+10=22 \text{ feet}^2[/tex]
The unit cost of the counter-top is [tex]\$42.00\text{ feet}^2[/tex]
So the net cost will be,
[tex]=42\times 22=\$924[/tex]
write 4 hundred-thousand, 13 thousand, 11 hundreds, 4 ones in standard form
Answers
The standard form of the expression 4 hundred-thousand, 13 thousand, 11 hundreds, 4 ones is [tex]4.14104 \times10^5[/tex]
In order to write 4 hundred-thousand, 13 thousand, 11 hundreds, 4 ones in standard form, let us write each of the expressions in figures
4 hundred-thousand = 400,000
13 thousand = 13,000
11 hundreds = 11 x 100
11 hundreds = 1100
4 ones = 4 x 1
4 ones = 4
The next step is to sum up all the figures together
400000 + 13000 + 1100 + 4 = 414,104
Convert 414104 to standard form
[tex]414104 = 4.14104 \times10^5[/tex]
Therefore, the standard form of the expression 4 hundred-thousand, 13 thousand, 11 hundreds, 4 ones is [tex]4.14104 \times10^5[/tex]
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one positive integer is 5 less than another. the product of the two integers is 36. what are the integers?
Answers
x(x-5)=36
x^2-5x-36=0
solving the equation we get
x=9 and x= -4
now find out the integers
1st number = x
2nd number = x-5
x*(x-5) = 36
x^2-5x =36
x^2 -5x-36 = 0
factor the polynomial:
(x-9) (x+4)
9-9 = 0, -4 +4 = 0
so x = 9 or -4
the question states a positive number
so x = 9
the 2 numbers are 9 and 4
There are six different sixth roots of 64. That is, there are six complex numbers that solve x^6=64
Please help
Answers
1)
When we know that:
[tex]\sqrt[n]{z}=\sqrt[n]{|z|}\Big(\cos\frac{\phi+2k\pi}{n}+i\sin\frac{\phi+2k\pi}{n}\Big)\qquad\text{for}\quad k=0,1,\ldots,n-1[/tex]
when [tex]z=|z|(\cos\phi+i\sin\phi)[/tex]
then:
[tex]x^6=64\quad|\sqrt[6]{(\ldots)}\\\\x=\sqrt[6]{64}[/tex]
For [tex]x=64[/tex] we have:
[tex]64=|64|(\cos0+i\sin0)\quad\Rightarrow\quad \phi=0[/tex]
and:
[tex]\sqrt[6]{64}=\sqrt[6]{64}\Big(\cos\frac{0+2k\pi}{6}+i\sin\frac{0+2k\pi}{6}\Big)\qquad\text{for}\quad k=0,1,\ldots,5\\\\\\ \sqrt[6]{64}=2\Big(\cos\frac{k\pi}{3}+i\sin\frac{k\pi}{3}\Big)\qquad\text{for}\quad k=0,1,\ldots,5 [/tex]
k = 0
[tex]2\Big(\cos\frac{0}{3}+i\sin\frac{0}{3}\Big)=2(1+0i)=2\cdot1=\boxed{2}[/tex]
k = 1
[tex]2\Big(\cos\frac{\pi}{3}+i\sin\frac{\pi}{3}\Big)=2\Big(\frac{1}{2}+i\frac{\sqrt{3}}{2}\Big)=\boxed{1+i\sqrt{3}}[/tex]
k = 2
[tex]2\Big(\cos\frac{2\pi}{3}+i\sin\frac{2\pi}{3}\Big)=2\Big(-\frac{1}{2}+i\frac{\sqrt{3}}{2}\Big)=\boxed{-1+i\sqrt{3}}[/tex]
k = 3
[tex]2\Big(\cos\frac{3\pi}{3}+i\sin\frac{3\pi}{3}\Big)=2\Big(\cos\pi+i\sin\pi\Big)=2(-1+0i)=\boxed{-2}[/tex]
k = 4
[tex]2\Big(\cos\frac{4\pi}{3}+i\sin\frac{4\pi}{3}\Big)=2\Big(-\frac{1}{2}-i\frac{\sqrt{3}}{2}\Big)=\boxed{-1-i\sqrt{3}}[/tex]
k = 5
[tex]2\Big(\cos\frac{5\pi}{3}+i\sin\frac{5\pi}{3}\Big)=2\Big(\frac{1}{2}-i\frac{\sqrt{3}}{2}\Big)=\boxed{1-i\sqrt{3}}[/tex]
So the answer is:
[tex]x=\{2,\,1+i\sqrt{3},\,-1+i\sqrt{3},\,-2,\,-1-i\sqrt{3},\,1-i\sqrt{3}\}[/tex]
2)
We don't know method (1). If so, we could use following identities:
[tex](1)\quad a^2-b^2=(a+b)(a-b)\\\\(2)\quad a^3-b^3=(a-b)(a^2+ab+b^2)\\\\(3)\quad a^3+b^3=(a+b)(a^2-ab+b^2)[/tex]
There will be:
[tex]x^6=64\\\\x^6-64=0\\\\(x^3)^2-8^2=0 \qquad\text{from (1)}\\\\(x^3+8)(x^3-8)=0\\\\(x^3+2^3)(x^3-2^3)=0\qquad\text{from (2) and (3)}\\\\ (x+2)(x^2-2x+4)(x-2)(x^2+2x+4)=0\qquad(\star)[/tex]
Now, we complete the square for:
[tex]x^2-2x+4=x^2-2x+1+3=(x^2-2x+1)+3=(x-1)^2+3=\\\\=(x-1)^2+(\sqrt{3})^2=(x-1)^2-(-1)(\sqrt{3})^2=(x-1)^2-i^2(\sqrt{3})^2=\\\\=(x-1)^2-(i\sqrt{3})^2=\text{from (1)}=\boxed{(x-1-i\sqrt{3})(x-1+i\sqrt{3})}[/tex]
and for:
[tex]x^2+2x+4=x^2-2x+1+3=(x^2+2x+1)+3=(x+1)^2+3=\\\\=(x+1)^2+(\sqrt{3})^2=(x+1)^2-(-1)(\sqrt{3})^2=(x+1)^2-i^2(\sqrt{3})^2=\\\\=(x+1)^2-(i\sqrt{3})^2=\text{from (1)}=\boxed{(x+1-i\sqrt{3})(x+1+i\sqrt{3})}[/tex]
When we return to [tex](\star)[/tex]:
[tex](x+2)(x^2-2x+4)(x-2)(x^2+2x+4)=0\\\\(x+2)(x-1-i\sqrt{3})(x-1+i\sqrt{3})(x-2)(x+1-i\sqrt{3})(x+1+i\sqrt{3})=\\=0[/tex]
And we have answer:
[tex]x=\{-2,\,1+i\sqrt{3},\,1-i\sqrt{3},\,2,\,-1+i\sqrt{3},\,-1-i\sqrt{3}\} [/tex]
To find the six sixth roots of 64, you can use polar form and De Moivre's Theorem to calculate the complex numbers. The roots are 2, 1+i√3, -1+i√3, -2, -1-i√3, and 1-i√3.
[tex]X^6 = 64[/tex]
The sixth roots of 64 are complex numbers and can be found by writing 64 in polar form and applying De Moivre's Theorem. Let's find the six roots:
First root: 2(cos(0) + i sin(0)) = 2
Second root: 2(cos(π/3) + i sin(π/3)) = 2(1/2 + i√3/2) = 1 + i√3
Third root: 2(cos(2π/3) + i sin(2π/3)) = 2(-1/2 + i√3/2) = -1 + i√3
Fourth root: -2
Fifth root: -2(cos(-π/3) + i sin(-π/3)) = -2(1/2 - i√3/2) = -1 - i√3
Sixth root: -2(cos(-2π/3) + i sin(-2π/3)) = -2(-1/2 - i√3/2) = 1 - i√3
Tyquan uses the greatest common factor and the distributive property to rewrite this sum: 75 + 60 What expression does Tyquan write? 3(25 + 20) 5(15 + 12) 25(3 + 60) 15(5 + 4)
Answers
75/15 = 5
60/15 = 4
so ur expression is : 15(5 + 4) <==
Simplify the expression 3 1/3/3 1/3. Explain how the expression and its simplified form show that the set of irrational numbers is not closed under division
Answers
From the context, I assume it to mean
(3^(1/3)) / (3^(1/3)) = 1
Since 3^(1/3) is the cube-root of 3, and is thus irrational.
However, the quotient is exactly 1, which is a rational number.
The example is therefore a counter-example for the statement that "set of irrational numbers is closed under division", or stated differently,
"the set of irrational numbers is not closed under division"
Here is the five-number summary for a group of 100 runners in a 5-kilometer race. the variable is the time to complete the race. five-number summary: minimum: 15 minutes q1: 27 minutes median: 31 minutes q3: 32 minutes maximum: 50 minutes are there any outliers in the runners’ finish times by the 1.5 * iqr definition?
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The interquartile range (IQR) is 5 minutes. Any values below 19.5 minutes or above 39.5 minutes would be considered outliers. In this data set, both the minimum time of 15 minutes and the maximum time of 50 minutes are outliers.
Explanation:
The question is whether there are any outliers in the finish times of a group of runners based on the IQR definition. To determine this, we first calculate the interquartile range (IQR), which is the range of the middle 50 percent of the data. Q1 = 32 min - 2 For this data set, we calculate the IQR by subtracting the first quartile (Q1) from the third quartile (Q3). Here, IQR = Q3 - 7 min = 5 min.
Next, we multiply the IQR by 1.5, which is our test for outliers. This gives 1.5 * IQR = 1.5 * 5 min = 7.5 min.
An outlier is defined as a data point that is below Q1 - 1.5 * IQR or above Q3 + 1.5 * IQR. So, any time less than 27 min - 7.5 min = 19.5 min or greater than 32 min + 7.5 min = 39.5 min would be considered an outlier.
Given the five-number summary: min=15 min, Q1=27 min, median=31 min, Q3=32 min, max=50 min, we can clearly see that the minimum time of 15 min is an outlier because it's less than 19.5 min. Also, the maximum time of 50 min is an outlier because it's greater than 39.5 min.
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The minimum time is 15 minutes and below the lower bound of 19.5 minutes, there is at least one outlier in the given data set.
To determine if there are any outliers using the 1.5 * IQR (interquartile range) definition, we need to calculate the IQR first. The IQR is the difference between the third quartile (Q3) and the first quartile (Q1).
Given:
- Q1 = 27 minutes
- Q3 = 32 minutes
We can calculate the IQR as follows:
[tex]\[ IQR = Q3 - Q1 = 32 - 27 = 5 \text{ minutes} \][/tex]
Now, we can use the 1.5 * IQR rule to find the lower and upper bounds for potential outliers:
Lower Bound: Q1 - 1.5 * IQR
Upper Bound: Q3 + 1.5 * IQR
[tex]\[ \text{Lower Bound} = 27 - 1.5 \times 5 = 27 - 7.5 = 19.5 \text{ minutes} \]\[ \text{Upper Bound} = 32 + 1.5 \times 5 = 32 + 7.5 = 39.5 \text{ minutes} \][/tex]
Any finish times below 19.5 minutes or above 39.5 minutes would be considered potential outliers according to the 1.5 * IQR rule.
Since the minimum time is 15 minutes and below the lower bound of 19.5 minutes, there is at least one outlier in the data set.
Express the area of each square below as a monomial
14g5h9
Answers
So, the resulting expression is found using the properties of the power.
=> (14)^2 (g^5)^2 (h^9)^2 = 196 g^10 h^18.
Answer: 196 g^10 h^18
A picture frame has a total perimeter P of 5 meters. The height of the frame is
2/3 times its width.
Answers
H should equal 1 and the width is 1.5
The height of the frame is 1.5 meters.
The Width of the frame is 1 meter.
What is the perimeter of the rectangle?The perimeter of a rectangle is defined as the addition of the lengths of the rectangle's four sides.
The perimeter of a rectangle = 2(L+W)
Where W is the width of the rectangle and L is the length of the rectangle
Given that,
P = 5 meters
w = 2/3h = 0.67h
We have to determine the height of the frame.
The perimeter of the frame = 2h + 2W
We know all the other values So we can plug in values to solve for h:
5 = 2h + 2(.67h)
5 = 2h+ 1.33h
5 = 3.33h
5/3.33 = h
h = 1.5
So the height of the frame is 1.5 meters
We need to find the width and we know that W = 0.67h
W = 0.67(1.5)
W = 1
So the Width of the frame is 1 meter.
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what is (9x^4-x^2y^2-5y^4)-(3x^4+2x^2y^2-7y^4)
Answers
9x⁴ - x²y² - 5y⁴ - 3x⁴ - 2x²y² + 7y⁴
6x⁴ - 3x²y² + 2y⁴
if f(x)=x/2-2 and g(x)=2x^2+x-3, find (f+g)(x)
Answers
If f(x)=x/2-2 and g(x)=2x^2+x-3, find (f+g)(x)
(f+g)(x)= f(x) +g(x) = x/2-2 +2x²+x-3 = x/2 +2x²+x-5
(f+g)(x)= (x+4x²-2x-10) /2
(f+g)(x)= (4x²+3x-10)/2 = 2x²+3/2 x -5
Answer:
B. [tex]2x^2+\frac{3x}{2}-5[/tex]
Step-by-step explanation:
Given functions,
[tex]f(x)=\frac{x}{2}-2[/tex]
[tex]g(x)=2x^2+x-3[/tex]
Since, (f+g)(x) = f(x) + g(x)
[tex]=\frac{x}{2}-2+2x^2+x-3[/tex]
[tex]=2x^2+\frac{x+2x}{2}-5[/tex] ( combine like terms )
[tex]=2x^2+\frac{3x}{2}-5[/tex]
Option 'B' is correct.
Distributive property of 68/4
Answers
The trick here is to rewrite 68 as 4(17).
Marcus walked 5 km due east and then he turned around and walk 5 km to West how many kilometers is Marcus now from his starting point
Answers
I
I
W------------------E
I
I
S o km
Find the equation in standard form of the line with slope
[tex] \frac{7}{2} [/tex] that passes through the point (5,7)
Please explain as well
Answers
Rational numbers are _____ natural numbers.
always
sometimes
never
Answers
What is the value of x?
6 + 8x
-------------- = 5x
2
A) -2
B) -1
C) 1/3
D) 3
Answers
domain of the function
Answers
y exist : x+6 ≥ 0
x ≥ -6
pls I help I need answers and work shown
Answers
0.30 as a fraction is 30/100 = 3/10
0.30 as a percent = 0.30 * 100 = 30%
answer is D. 3/10 and 30%
.30 is .3
do .30 divided by 1 and you get 3/10
Percentage= move the decimal 2 places to the right and you get 30%
So your answer is D.
How many pairs of whole numbers have a sum of 99
Answers
Jacks has 18 marbles,he loses 6 of them.
What fraction of the marbles does he lose?
Answers
1/3
Hope this helps!
I have 5 digits. My 4s are worth 4 [10,000s] and 4 * 10. One of my 3s is worth 3,000. The other is worth 1/10 as much. My other digit is a 2. What number am I?
Answers
4(10,000) + 4(10) + 3,000 + 3,000(1/10) + 2 = 43,342
Thus, the 5-digit number is 43,342.
Which of the following equations is of a parabola with a vertex at (0, 3)?
y = (x - 3) 2
y = (x + 3) 2
y = x 2 - 3
y = x 2 + 3
Answers
Answer:
Option D.
Step-by-step explanation:
We have to find the equation of the parabola which has vertex at ( 0,3 )
If we write the equation of a parabola in the vertex from then we will write
y = ( x - h )² + (k)²
In this equation ( h, k) will be the vertex.
(a) y = (x-3)² + 0
so ( 3,0) will be the vertex
(b) y = ( x+3)² + 0
or y = [x - (-3)]² + 0
then ( -3,0) will be the vertex
(c) y = (x² - 3)
y = ( x-0 )² + (-3)
then ( 0, -3) is the vertex.
(d) y = x² + 3
y = ( x-0 )² + 3
Then ( 0, 3 ) will be the vertex of this parabola.
Option D. is the answer.