Answer:
Lol yu lateee das "who i smoke by yung ace"
Step-by-step explanation:
Answer:
woogle said woo hoo by rock a teens
Step-by-step explanation:
Define the operation a∇b = 2+b^a What is the value of (1∇2)∇3?
Answer:
83
Step-by-step explanation:
1∇2= (2+2^1)
=2+2=4
(4)∇3= (2+3^4)
=2+81
=83
I have know idea how to do this problem and teach my child
k(t)=13t-2
Answer:
K(t)=37
Step-by-step explanation:
k(t) = 13t - 2
k(3) = 13(3) - 2
k(3) = 39 - 2
k(3) = 37 <===
8 meters for every 2inches what is the area of 144 meters squared
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Answer:
9 square inches
Step-by-step explanation:
The area is proportional to the square of the linear scale factor. We can use this to write the proportion ...
A/(144 m²) = ((2 in)/(8 m))²
A = (144·4/64) in² = 9 in²
The area representing 144 square meters is 9 square inches.
Determine x&y
(2+i) (x+yi) = -7+3i
Answer:
x = -11/5 or -2.2
y = 13/5 or 2.6
Step-by-step explanation:
well, start by doing the multiplication. then we will see better.
2x + 2yi + xi + yii = -7 + 3i
2x + 2yi + xi - y = -7 + 3i
this is because, remember, i = sqrt(-1), and ii = -1.
now we group the i-factors and the terms without i and compare it to the corresponding parts on the right side.
2x - y = -7
2yi + xi = 3i
=> 2y + x = 3
x = 3 - 2y
and that we use ihr the first equation again
2×(3-2y) - y = -7
6 - 4y - y = -7
-5y = -13
y = 13/5
x = 3 - 2×13/5 = 3 - 26/5 = 15/5 - 26/5 = -11/5
a bag contains 7 red chips and 11 blue chips. two chips are selected randomly without replacement from the bag. what is the probability that the two chips are NOT the same coler
Answer:
77/306 or around 25.2%
Step-by-step explanation:
[tex]\frac{7}{18} *\frac{11}{17}[/tex] section 1/total * section 2/(total-1) since there is no replacement
just solve and you get 77/306
What is the value of x that makes l1||l2
A. 35
B. 25
C. 37
D. 18
Answer:
B
Step-by-step explanation:
For l1 and l2 to be parallel, these two angles need to be equal. 3x-15=2x+10, x=25
time in months it would take for a $1500 dollar investment in a TFSA to grow to $1545 if the simple interest at a rate paid was 2% per annum.
It would take 17 months and 14 days for the investment to grow to $1545.
To determine the time in months it would take for a $ 1500 dollar investment in a TFSA to grow to $ 1545 if the simple interest at a rate paid was 2% per annum, the following calculation must be performed:
First, you must obtain 2% of 1545 to determine the interest generated per year.
1545 x 2/100 = X 30.9 = XThen, a cross multiplication must be carried out considering the number of months it took to generate said interest, and compare it with the interest that arises from the subtraction of 1545 - 1500, that is, 45.
30.9 = 1245 = X45 x 12 / 30.9 = X540 / 30.9 = X17.47 = X 1 = 300.47 = X14 = XTherefore, it would take 17 months and 14 days for a $ 1500 dollar investment in a TFSA to grow to $ 1545.
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Suppose y varies inversely with x, and y = 32 when x = 4. What is the value of y when x = 8?
a. 1/8
b. 64
c. 16
d. 8
NO LINKS OR ANSWERING QUESTIONS YOU DON'T KNOW!!!
Answer:
16
Step-by-step explanation:
Inverse variation is of the form
xy = k where k is a constant
x=4 and y = 32
4*32 = k
128 = k
xy = 128
Let x = 8
8y = 128
Divide each side by 8
8y/8 = 128/8
y =16
A vector has a magnitude of 43
meters at an angle (0). If the
Y-component of the vector is 30
meters, what is the X-component?
[?]m
30.8 m
Step-by-step explanation:
Given: [tex]V = 43\:\text{m}[/tex], [tex]V_y = 30\:\text{m}[/tex]
The x-component of vector [tex]\vec{\text{V}}[/tex] is
[tex]V_x = \sqrt{V^2 - V_y^2} = \sqrt{(43)^2 - (30)^2} = 30.8\:\text{m}[/tex]
The x- component of the vector is 30.8meters.
What is the magnitude and direction of vector?If [tex]v = < a. b >[/tex] be a position vector then the magnitude of vector v is found by [tex]|v| =\sqrt{a^{2}+b^{2} } }[/tex] , where a and b are the x and y component respectively.
And the direction is equals to the angle formed x- axis or y axis.
According to the given question
We have
Magnitude of the vector, |v| = 43meters
Y- component of the vector, b = 30meters
Since, we know that
[tex]|v| =\sqrt{a^{2} +b^{2} }[/tex]
Substitute the value of |v| = 43 and b = 30 in the above formula of magnitude.
⇒ [tex]43 = \sqrt{a^{2}+30^{2} }[/tex]
⇒ [tex]43 = \sqrt{a^{2}+900 }[/tex]
⇒ [tex]43^{2} =a^{2} + 900[/tex]
⇒ [tex]1849 = a^{2} + 900[/tex]
⇒ [tex]1849-900=a^{2}[/tex]
⇒ [tex]949=a^{2}[/tex]
⇒ [tex]a =\sqrt{949}[/tex]
⇒ [tex]a = 30.8[/tex]
Hence, the x- component of the vector is 30.8meters.
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2. Mandla spent one quarter of his pocket money on sweets. a. What fraction does he have left? b. If he had R40 pocket money, how much did he spend?
Answer:
a. 3/4 of pocket money left
b. R10
Step-by-step explanation:
a. 4/4 - 1/4 = 3/4
b. 40/4 = 10 = 1/4 of pocket money
(a). The fraction of pocket money left is 3/4
(b). If he had R40 pocket money, he spend R10.
What is the Ratio?The ratio is defined as a relationship between two quantities, it is expressed one divided by the other.
What are Arithmetic operations?Arithmetic operations can also be specified by the subtract, divide, and multiply built-in functions.
The operator that perform arithmetic operation are called arithmetic operators .
Operators which let do basic mathematical calculation
+ Addition operation : Adds values on either side of the operator.
For example 4 + 2 = 6
- Subtraction operation : Subtracts right hand operand from left hand operand.
for example 4 -2 = 2
* Multiplication operation : Multiplies values on either side of the operator
For example 4*2 = 8
Mandla spent one-quarter of his pocket money on sweets
Let x be the total amount of pocket money Mandla had originally
Solution of (a).
⇒ x - (1/4)x
⇒ (3/4)x
The fraction of pocket money left = 3/4
Solution of (b).
1/4 of pocket money
⇒ (1/4)x
⇒ (1/4)40
⇒ 10
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What is the value of X? HELP
with no further informations, just go by looking at it.
it's 90°, all other options are too far off
.......... is a factor of every even number.
Answer:
2 is the factor of every even number hope this help you
You found the prime factorization of the number 73 explain how you can check your answer
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Explanation:
You can check your answer by making sure that each of the primes you found is actually a prime. (Compare to a list of known primes, for example.) After you have determined your factors are primes, multiply them together to see if the result is 73. If so, you have found the correct prime factorization.
__
Additional comment
73 is prime, so its prime factor is 73.
73 = 73
A shoe store carries one brand of shoe in 4
different styles, 5 sizes, and 5 colors. How many
different shoes are available of this one brand?
Answer:
100 different shoes
Step-by-step explanation:
4 styles * 5 sizes * 5 colors
4*5*5 = 100
Complete the function table.
Input (n) Output (n-2)
Answer: Choice C
This is because the input n = 2 leads to the output n-2 = 2-2 = 0
As another example: the input n = 4 leads to the output n-2 = 4-2 = 2
Whatever the input is, subtract 2 from it to get the output.
What is the shape of a sorbital
Answer:
Spherical-Like Shape
Step-by-step explanation:
An s-orbital is spherical with the nucleus at its center.
Abigail plans to repaint some classroom bookcases. She has 6/25
gallons of paint. All of the bookcases are the same size and each requires 2/3
gallon of paint. How many bookcases will the custodian be able to repaint with that amount of paint?
Answer:
Step-by-step explanation: Hello! Do
The percent of data between z=0.23 and z = 1.27 is
(Round to two decimal places as needed.)
Answer:
0.40905 - 0.10204 = .30701 = 30.7 %
Step-by-step explanation:
0.23 0.40905
1.27 0.10204
Find the differential coefficient of
[tex]e^{2x}(1+Lnx)[/tex]
Answer:
[tex] \rm \displaystyle y' = 2 {e}^{2x} + \frac{1}{x} {e}^{2x} + 2 \ln(x) {e}^{2x} [/tex]
Step-by-step explanation:
we would like to figure out the differential coefficient of [tex]e^{2x}(1+\ln(x))[/tex]
remember that,
the differential coefficient of a function y is what is now called its derivative y', therefore let,
[tex] \displaystyle y = {e}^{2x} \cdot (1 + \ln(x) )[/tex]
to do so distribute:
[tex] \displaystyle y = {e}^{2x} + \ln(x) \cdot {e}^{2x} [/tex]
take derivative in both sides which yields:
[tex] \displaystyle y' = \frac{d}{dx} ( {e}^{2x} + \ln(x) \cdot {e}^{2x} )[/tex]
by sum derivation rule we acquire:
[tex] \rm \displaystyle y' = \frac{d}{dx} {e}^{2x} + \frac{d}{dx} \ln(x) \cdot {e}^{2x} [/tex]
Part-A: differentiating $e^{2x}$
[tex] \displaystyle \frac{d}{dx} {e}^{2x} [/tex]
the rule of composite function derivation is given by:
[tex] \rm\displaystyle \frac{d}{dx} f(g(x)) = \frac{d}{dg} f(g(x)) \times \frac{d}{dx} g(x)[/tex]
so let g(x) [2x] be u and transform it:
[tex] \displaystyle \frac{d}{du} {e}^{u} \cdot \frac{d}{dx} 2x[/tex]
differentiate:
[tex] \displaystyle {e}^{u} \cdot 2[/tex]
substitute back:
[tex] \displaystyle \boxed{2{e}^{2x} }[/tex]
Part-B: differentiating ln(x)•e^2x
Product rule of differentiating is given by:
[tex] \displaystyle \frac{d}{dx} f(x) \cdot g(x) = f'(x)g(x) + f(x)g'(x)[/tex]
let
[tex]f(x) \implies \ln(x) [/tex][tex]g(x) \implies {e}^{2x} [/tex]substitute
[tex] \rm\displaystyle \frac{d}{dx} \ln(x) \cdot {e}^{2x} = \frac{d}{dx}( \ln(x) ) {e}^{2x} + \ln(x) \frac{d}{dx} {e}^{2x} [/tex]
differentiate:
[tex] \rm\displaystyle \frac{d}{dx} \ln(x) \cdot {e}^{2x} = \boxed{\frac{1}{x} {e}^{2x} + 2\ln(x) {e}^{2x} }[/tex]
Final part:
substitute what we got:
[tex] \rm \displaystyle y' = \boxed{2 {e}^{2x} + \frac{1}{x} {e}^{2x} + 2 \ln(x) {e}^{2x} }[/tex]
and we're done!
Answer:
Product Rule for Differentiation
[tex]\textsf{If }y=uv[/tex]
[tex]\dfrac{dy}{dx}=u\dfrac{dv}{dx}+v\dfrac{du}{dx}[/tex]
Given equation:
[tex]y=e^{2x}(1+\ln x)[/tex]
Define the variables:
[tex]\textsf{Let }u=e^{2x} \implies \dfrac{du}{dx}=2e^{2x}[/tex]
[tex]\textsf{Let }v=1+\ln x \implies \dfrac{dv}{dx}=\dfrac{1}{x}[/tex]
Therefore:
[tex]\begin{aligned}\dfrac{dy}{dx} & =u\dfrac{dv}{dx}+v\dfrac{du}{dx}\\\\\implies \dfrac{dy}{dx} & =e^{2x} \cdot \dfrac{1}{x}+(1+\ln x) \cdot 2e^{2x}\\\\& = \dfrac{e^{2x}}{x}+2e^{2x}(1+\ln x)\\\\ & = \dfrac{e^{2x}}{x}+2e^{2x}+2e^{2x} \ln x\\\\& = e^{2x}\left(\dfrac{1}{x}+2+2 \ln x \right)\end{aligned}[/tex]
Given that V=4/3πrcube , make r the subject of formula
Answer:
[tex]V = \frac{4}{3} \pi {r}^{3} \\ \\ 3V = 4\pi {r}^{3} \\ {r}^{3} = \frac{3V}{4\pi} \\ \\ r = \sqrt[3]{ \frac{3V}{4\pi} } [/tex]
A triangle has side lengths of (4.6x-4.4y)(4.6x−4.4y) centimeters, (7.5x-8.8z)(7.5x−8.8z) centimeters, and (7.7z-9.2y)(7.7z−9.2y) centimeters. Which expression represents the perimeter, in centimeters, of the triangle?
The expression that represents the perimeter, in centimeters, of the triangle is [tex]77.41x^{2}+104y^{2}+ 136.73z^{2} -40.48xy - 132xz - 141.64yz[/tex] cm OR 77.41x²+104y²+136.73z²-40.48xy-132xz-141.64yz centimeters
From the question, the side lengths are
(4.6x-4.4y)(4.6x−4.4y) cm, (7.5x-8.8z)(7.5x−8.8z) cm, and (7.7z-9.2y)(7.7z−9.2y) cm.
First, we will clear the brackets one after the other
For (4.6x-4.4y)(4.6x−4.4y) cm
[tex]4.6x(4.6x-4.4y) -4.4y(4.6x-4.4y)[/tex]
[tex]21.16x^{2} -20.24xy -20.24xy+19.36y^{2}[/tex]
[tex]21.16x^{2} -40.48xy+19.36y^{2}[/tex]
∴ (4.6x-4.4y)(4.6x−4.4y) cm = [tex]21.16x^{2} -40.48xy+19.36y^{2}[/tex] cm
For (7.5x-8.8z)(7.5x−8.8z) cm
[tex]7.5x(7.5x-8.8z) -8.8z(7.5x-8.8z)[/tex]
[tex]56.25x^{2} - 66xz -66xz + 77.44z^{2}[/tex]
[tex]56.25x^{2} - 132xz + 77.44z^{2}[/tex]
∴ (7.5x-8.8z)(7.5x−8.8z) cm = [tex]56.25x^{2} - 132xz + 77.44z^{2}[/tex] cm
For (7.7z-9.2y)(7.7z−9.2y) cm
[tex]7.7z(7.7z-9.2y)-9.2y(7.7z-9.2y)[/tex]
[tex]59.29z^{2} - 70.84yz-70.84yz+84.64y^{2}[/tex]
[tex]59.29z^{2} - 141.64yz+84.64y^{2}[/tex]
∴ (7.7z-9.2y)(7.7z−9.2y) cm = [tex]59.29z^{2} - 141.64yz+84.64y^{2}[/tex] cm
Now, for the expression that represents the perimeter of the triangle,
Perimeter of a triangle can be calculated by determining the sum of all its sides
That is,
Perimeter of the triangle = [tex]21.16x^{2} -40.48xy+19.36y^{2}[/tex] cm + [tex]56.25x^{2} - 132xz + 77.44z^{2}[/tex] cm + [tex]59.29z^{2} - 141.64yz+84.64y^{2}[/tex] cm
[tex]=21.16x^{2} -40.48xy+19.36y^{2} + 56.25x^{2} - 132xz + 77.44z^{2} + 59.29z^{2} - 141.64yz+84.64y^{2}[/tex]
Collect like terms
[tex]= 21.16x^{2}+ 56.25x^{2} -40.48xy+19.36y^{2}+84.64y^{2} - 132xz + 77.44z^{2} + 59.29z^{2} - 141.64yz[/tex]
[tex]= 77.41x^{2} -40.48xy+104y^{2} - 132xz + 136.73z^{2} - 141.64yz[/tex]
[tex]= 77.41x^{2}+104y^{2}+ 136.73z^{2} -40.48xy - 132xz - 141.64yz[/tex] cm
Hence, the expression that represents the perimeter, in centimeters, of the triangle is [tex]77.41x^{2}+104y^{2}+ 136.73z^{2} -40.48xy - 132xz - 141.64yz[/tex] cm OR 77.41x²+104y²+136.73z²-40.48xy-132xz-141.64yz centimeters
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Solve the formula for t
V = 4(3.14)ct + 6(3.14)c^2
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Answer:
t = (V -6(3.14)c^2)/(4(3.14)c)
Step-by-step explanation:
Isolate the term containing t, then divide by the coefficient of t
[tex]V=4(3.14)ct+6(3.14)c^2\\\\V-6(3.14)c^2=t(4(3.14)c)\\\\\boxed{t=\dfrac{V-6(3.14)c^2}{4(3.14)c}}[/tex]
convert the following to decimal fractions 99 by 5
Answer:
divide 99 by 5
99/5= 19.8
Use the power series method to solve the given initial-value problem. (Format your final answer as an elementary function.)
(x − 1)y'' − xy' + y = 0, y(0) = −7, y'(0) = 3
You're looking for a solution of the form
[tex]\displaystyle y = \sum_{n=0}^\infty a_n x^n[/tex]
Differentiating twice yields
[tex]\displaystyle y' = \sum_{n=0}^\infty n a_n x^{n-1} = \sum_{n=0}^\infty (n+1) a_{n+1} x^n[/tex]
[tex]\displaystyle y'' = \sum_{n=0}^\infty n(n-1) a_n x^{n-2} = \sum_{n=0}^\infty (n+1)(n+2) a_{n+2} x^n[/tex]
Substitute these series into the DE:
[tex]\displaystyle (x-1) \sum_{n=0}^\infty (n+1)(n+2) a_{n+2} x^n - x \sum_{n=0}^\infty (n+1) a_{n+1} x^n + \sum_{n=0}^\infty a_n x^n = 0[/tex]
[tex]\displaystyle \sum_{n=0}^\infty (n+1)(n+2) a_{n+2} x^{n+1} - \sum_{n=0}^\infty (n+1)(n+2) a_{n+2} x^n \\\\ \ldots \ldots \ldots - \sum_{n=0}^\infty (n+1) a_{n+1} x^{n+1} + \sum_{n=0}^\infty a_n x^n = 0[/tex]
[tex]\displaystyle \sum_{n=1}^\infty n(n+1) a_{n+1} x^n - \sum_{n=0}^\infty (n+1)(n+2) a_{n+2} x^n \\\\ \ldots \ldots \ldots - \sum_{n=1}^\infty n a_n x^n + \sum_{n=0}^\infty a_n x^n = 0[/tex]
Two of these series start with a linear term, while the other two start with a constant. Remove the constant terms of the latter two series, then condense the remaining series into one:
[tex]\displaystyle a_0-2a_2 + \sum_{n=1}^\infty \bigg(n(n+1)a_{n+1}-(n+1)(n+2)a_{n+2}-na_n+a_n\bigg) x^n = 0[/tex]
which indicates that the coefficients in the series solution are governed by the recurrence,
[tex]\begin{cases}y(0)=a_0 = -7\\y'(0)=a_1 = 3\\(n+1)(n+2)a_{n+2}-n(n+1)a_{n+1}+(n-1)a_n=0&\text{for }n\ge0\end{cases}[/tex]
Use the recurrence to get the first few coefficients:
[tex]\{a_n\}_{n\ge0} = \left\{-7,3,-\dfrac72,-\dfrac76,-\dfrac7{24},-\dfrac7{120},\ldots\right\}[/tex]
You might recognize that each coefficient in the n-th position of the list (starting at n = 0) involving a factor of -7 has a denominator resembling a factorial. Indeed,
-7 = -7/0!
-7/2 = -7/2!
-7/6 = -7/3!
and so on, with only the coefficient in the n = 1 position being the odd one out. So we have
[tex]\displaystyle y = \sum_{n=0}^\infty a_n x^n \\\\ y = -\frac7{0!} + 3x - \frac7{2!}x^2 - \frac7{3!}x^3 - \frac7{4!}x^4 + \cdots[/tex]
which looks a lot like the power series expansion for -7eˣ.
Fortunately, we can rewrite the linear term as
3x = 10x - 7x = 10x - 7/1! x
and in doing so, we can condense this solution to
[tex]\displaystyle y = 10x -\frac7{0!} - \frac7{1!}x - \frac7{2!}x^2 - \frac7{3!}x^3 - \frac7{4!}x^4 + \cdots \\\\ \boxed{y = 10x - 7e^x}[/tex]
Just to confirm this solution is valid: we have
y = 10x - 7eˣ ==> y (0) = 0 - 7 = -7
y' = 10 - 7eˣ ==> y' (0) = 10 - 7 = 3
y'' = -7eˣ
and substituting into the DE gives
-7eˣ (x - 1) - x (10 - 7eˣ ) + (10x - 7eˣ ) = 0
as required.
A right triangle has side 14 and hypotenuse 50. Use the Pythagorean Theorem to find the length of the third side.
Answer:
48
Step-by-step explanation:
Pythagorean Theorem = h^2=p^2+b^2
We have,
(Hypotenuse)h=50
Let 14 be p, i.e (Perpendicular ,Known side)p=14
(Remaining side ,base)b=?(
Now,
h^2=p^2+b^2
or, 50^2=14^2+b^2
or, 2500-196=b^2
or, √2304=b
b=48
if the area of the triangle is 5cm^2. find the angle.
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Answer:
the angle is 30°
Step-by-step explanation:
The area of a triangle given two sides and the included angle is ...
A = 1/2ab·sin(C)
5 cm² = 1/2(4 cm)(5 cm)sin(θ)
0.5 = sin(θ) . . . . . . divide by 10 cm²
θ = arcsin(.5) ≈ 30°
_____
Additional comment
An obtuse angle of 150° in that location will give a triangle with the same area.
use quadratic formula to solve the following equation
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Answer:
x = 2 or x = 9
Step-by-step explanation:
To use the quadratic formula, we first need the equation in standard form for a quadratic. We can get there by multiplying the equation by 3(x -3).
2(3) +4(3(x -3)) = (x +4)(x -3)
6 +12x -36 = x² +x -12
x² -11x +18 = 0
Using the quadratic formula to find the solutions, we have ...
[tex]x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}=\dfrac{-(-11)\pm\sqrt{(-11)^2-4(1)(18)}}{2(1)}\\\\x=\dfrac{11\pm\sqrt{49}}{2}=\{2,9\}[/tex]
The solutions are x=2 and x=9.
Evaluate the following expression
Write in a shorter form:7m -7 +7m +7
Answer:
14m
Step-by-step explanation:
[tex]7m-7+7m-7\\[/tex]
First, we need to eliminate the like term and collect the like term.
[tex]-7+-7=0[/tex]
Now, we have 7m +7m, sum them up and you will get the answer.[tex]7m+7m=14m[/tex]
So, the answer is 14m.
Answer:
14m
Step-by-step explanation:
7m -7 +7m +7
7m + 7m - 7 + 7
14m
They are 10 ice cream flavors, 5 different toppings and it could be either in cup or in cone, how many 2-scoop combinations are possible?
Using the fundamental counting principle, it is found that: 50 2-scoop combinations are possible.
----------------------------------
The flavors and the toppings are independent, which means that the fundamental counting principle is used to solve this question, which states that if there are p ways to do a thing, and q ways to do another thing, and these two things are independent, there are p*q ways to do both things.
----------------------------------
In this question:
10 ice cream flavors.5 toppings.So,
[tex]10 \times 5 = 50[/tex]
50 2-scoop combinations are possible.
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