Answer:
15x+6u−3
Step-by-step explanation:
This means -3 times -5x, -3 times -2u, and -3 times 1.
Do this and you have, 15x+6u-3.
The size of a television is the length of the diagonal of its screen in inches. The aspect ratio of the screens of older televisions is 4:3, while the aspect ratio of newer wide-screen televisions is 16:9. Find the width and height of an older 35-inch television whose screen has an aspect ratio of 4:3.
Answer:
The Width = 28 inches
The Height = 21 inches
Step-by-step explanation:
We are told in the question that:
The width and height of an older 35-inch television whose screen has an aspect ratio of 4:3
Using Pythagoras Theorem
Width² + Height² = Diagonal²
Since we known that the size of a television is the length of the diagonal of its screen in inches.
Hence, for this new TV
Width² + Height² = 35²
We are given ratio: 4:3 as aspect ratio
Width = 4x
Height = 3x
(4x)² +(3x)² = 35²
= 16x² + 9x² = 35²
25x² = 1225
x² = 1225/25
x² = 49
x = √49
x = 7
Hence, for the 35 inch tv set
The Width = 4x
= 4 × 7
= 28 inches.
The Height = 3x
= 3 × 7
= 21 inches
Each student in a school was asked, "What is your favorite color?" The circle graph below shows how they answered
Which color was chosen by approximately one fourth of the students?
Approximately what percentage of the students chose purple or green?
Answer:
a). BLUE color
b). 20%
Step-by-step explanation:
a). "Which color was chosen by approximately one fourth of the students?"
Since one fourth of the students will be represented by one fourth area of the circle given.
That means color of choice represented by the quarter of the circle will be the color liked by one fourth students.
In the figure attached, BLUE color is the choice of one fourth students in the class.
b). Area represented by purple, green and other colors is a quarter of the circle.
If we divide this quarter into five equal sections, then the total of purple and green will be [tex]4\times \frac{1}{5}[/tex] of the the quarter of the circle.
Measure of the angle defined by purple or green sections = [tex]\frac{4}{5}\times 90[/tex]
= 72°
Percentage of the students who preferred purple or green = [tex]\frac{72}{360}\times 100[/tex]
= 20%
Answer:
blue
20%
Step-by-step explanation:
what is the distance between the first and third quartiles of a data set called?
Answer:
Interquartile range is the distance between the first and third of a data.
Step-by-step explanation:
Hope it will help you :)
the fourth term of an AP is 5 while the sum of the first 6 terms is 10. Find the sum of the first 19 terms
Answer: S₁₉ = 855
Step-by-step explanation:
T₄ = a + ( n - 1 )d = 5 , from the statement above , but n = 4
a + 3d = 5 -------------------------1
S₆ = ⁿ/₂[(2a + ( n - 1 )d] = 10, where n = 6
= ⁶/₂( 2a + 5d ) = 10
= 3( 2a + 5d ) = 10
= 6a + 15d = 10 -----------------2
Now solve the two equation together simultaneously to get the values of a and d
a + 3d = 5
6a + 15d = 10
from 1,
a = 5 - 3d -------------------------------3
Now put (3) in equation 2 and open the brackets
6( 5 - 3d ) + 15d = 10
30 - 18d + 15d = 10
30 - 3d = 10
3d = 30 - 10
3d = 20
d = ²⁰/₃.
Now substitute for d to get a in equation 3
a = 5 - 3( ²⁰/₃)
a = 5 - 3 ₓ ²⁰/₃
= 5 - 20
a = -15.
Now to find the sum of the first 19 terms,
we use the formula
S₁₉ = ⁿ/₂( 2a + ( n - 1 )d )
= ¹⁹/₂( 2 x -15 + 18 x ²⁰/₃ )
= ¹⁹/₂( -30 + 6 x 20 )
= ¹⁹/₂( -30 + 120 )
= ¹⁹/₂( 90 )
= ¹⁹/₂ x 90
= 19 x 45
= 855
Therefore,
S₁₉ = 855
Identify each x-value at which the slope of the tangent line to the function f(x) = 0.2x^2 + 5x − 12 belongs to the interval (-1, 1).
Answer:
Step-by-step explanation:
Hello, the slope of the tangent is the value of the derivative.
f'(x) = 2*0.2x + 5 = 0.4x + 5
So we are looking for
[tex]-1\leq f'(x) \leq 1 \\ \\<=> -1\leq 0.4x+5 \leq 1 \\ \\<=> -1-5=-6\leq 0.4x \leq 1-5=-4 \\ \\<=> \dfrac{-6}{0.4}\leq 0.4x \leq \dfrac{-4}{0.4} \\\\<=> \boxed{-15 \leq x\leq -10}[/tex]
Hope this helps.
Do not hesitate if you need further explanation.
Thank you
Using derivatives, it is found that the x-values in which the slope belong to the interval (-1,1) are in the following interval is (-15,-10).
What is the slope of the tangent line to a function f(x) at point x = x_0?It is given by the derivative at x = x_0, that is:
m = f'(x_0)
In this problem, the function is:
f(x) = 0.2x^2 + 5x − 12
Hence the derivative is:
f'(x) = 0.4x + 5
For a slope of -1, we have that,
0.4x + 5 = -1
0.4x = -6
x = -15.
For a slope of 1, we have that,
0.4x + 5 = 1.
0.4x = -4
x = -10
Hence it is found that the x-values in which the slope belong to the interval (-1,1) are in the following interval is (-15,-10).
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Question 2 Rewrite in simplest radical form 1 x −3 6 . Show each step of your process.
Answer:
√(x)
Step-by-step explanation:
(1)/(x^-(1/2)) that's 3 goes into -3 leaving 1 and goes into 6 leaving 2
1/2 is same as 2^-1
so therefore we can simplify the above as
x^-(-1/2)
x^(1/2)
and 4^(1/2)
is same as √(4)
so we conclude as
√(x)
What is the value of 20 + 3 (7 + 4) + 5 + 2 (7 + 9)?
Answer:
90
Step-by-step explanation:
Answer:
90
Step-by-step explanation:
Here is the equation
[tex]20+3\times(7+4)+5+2\times(7+9)[/tex]
In the order of operations parentheses go first so we get
[tex]20+3\times11+5+2\times16[/tex]
Next we do the multiplication
[tex]20+33+5+32\\[/tex]
And finally we add them all up
[tex]20+33+5+32=90\\[/tex]
Thus, 90 is the answer of [tex]20+3\times(7+4)+5+2\times(7+9)[/tex] or [tex]20+3(7+4)+5+2(7+9)[/tex]
Simplify to create an equivalent expression.
-k-(-8k+7)
a=7k−7
b=-7k-7
c=7k+7
d=-7k+7
choose one
Answer:
a. 7k - 7
Step-by-step explanation:
Step 1: Write out expression
-k - (-8k + 7)
Step 2: Distribute negative
-k + 8k - 7
Step 3: Combine like terms
7k - 7
And we have our answer!
what is the domain of f(x)=(1/4)^x
Answer:
B All real numbers
hope you wil understand
Answer:
[tex]\boxed{\sf B. \ All \ real \ numbers}[/tex]
Step-by-step explanation:
The domain is all possible values for x.
[tex]f(x)=(\frac{1}{4} )^x[/tex]
There are no restrictions on the value of x.
The domain is all real numbers.
The double number lines show the ratio of cups to gallons. How many cups are in 333 gallons? _____ cups
Answer:
5328 cups.
Step-by-step explanation:
Given that 333 gallons
We know that
1 gallons = 16 cups
1 cups = 0.0625 gallons
Therefore,from the above conversion we can say that
Now by putting the values in the above conversion
333 gallons = 16 x 333 cups
333 gallons = 5328 cups
So , we can say that 333 gallons is equal to 5328 cups.
Thus the answer will be 5328 cups.
Answer:
48 cups(BTW he meant 33 galons, IVE had this before). lol you need to put the double number line image. first u have to divide 64/4 to get 16, Then it says "How many cups are in 3 gallons". There fore, U multiply 16 to 3 to get ur answer "48".
Suppose the radius of a circle is 5 units. What is its circumference?
Answer:
C≈31.42
Step-by-step explanation:
C=2πr
C=2xπx5
C≈31.42
pls mark as brainliest
Tom is afraid of heights above 9 feet. He is asked to repair a side of a high deck. The bottom of the ladder must be placed 6 feet from a deck. The ladder is 10 feet long. How far above the ground does the ladder touch the deck? Is Tom afraid of the height?
Answer:
8 ftnoStep-by-step explanation:
The height on the side of the deck (h) can be found using the Pythagorean theorem. It tells you ...
6^2 + h^2 = 10^2
h = √(10^2 -6^2) = √64 = 8
The ladder touches the deck 8 feet above the ground. Tom is not afraid of that height.
Can somebody please solve this problem for me!
Answer:
x = 200.674
Step-by-step explanation:
tan∅ = opposite/adjacent
Step 1: Find length of z
tan70° = 119/z
ztan70° = 119
z = 119/tan70°
z = 43.3125
Step 2: Find length z + x (denoted as y)
tan26° = 119/y
ytan26° = 119
y = 119/tan26°
y = 243.986
Step 3: Find x
y - z = x
243.986 - 43.3125 = x
x = 200.674
Of the three properties, reflexive, symmetric, and transitive that define the relation "is equal to," which one could also apply to "is less than" and "is greater than?" transitive reflexive symmetric
Answer: Transitive property.
Step-by-step explanation:
First, for the equality we have:
Reflexive:
For all real numbers x, x = x.
Symmetric:
For all real numbers x, y
if x= y, then y = x.
Transitive:
For reals x, y and z.
if x = y, and y = z, then x = z.
Now, let's talk about inequalities.
first, the reflexive property will say that:
x > x.
This has no sense, so this property does not work for inequalities.
Now, the reflexive.
If x > y, then y > x.
Again, this has no sense, if x is larger than y, then we can never have that y is larger than x. This property does not work for inequalities.
Not, the transitive property.
if x > y, and y > z, then x > z.
This is true.
x is bigger than y, and y is bigger than z, then x should also be bigger than z.
x > y > z.
And this also works for the inverse case:
x < y and y < z, then x < z.
So the correct option is transitive property.
Express the quotient of z1 and z2 in standard form given that [tex]z_{1} = -3[cos(\frac{-\pi }{4} )+isin(\frac{-\pi }{4} )][/tex] and [tex]z_{2} = 2\sqrt{2} [cos(\frac{-\pi }{2} )+isin(\frac{-\pi }{2} )][/tex]
Answer:
Solution : [tex]-\frac{3}{4}-\frac{3}{4}i[/tex]
Step-by-step explanation:
[tex]-3\left[\cos \left(\frac{-\pi }{4}\right)+i\sin \left(\frac{-\pi \:}{4}\right)\right]\:\div \:2\sqrt{2}\left[\cos \left(\frac{-\pi \:\:}{2}\right)+i\sin \left(\frac{-\pi \:\:\:}{2}\right)\right][/tex]
Let's apply trivial identities here. We know that cos(-π / 4) = √2 / 2, sin(-π / 4) = - √2 / 2, cos(-π / 2) = 0, sin(-π / 2) = - 1. Let's substitute those values,
[tex]\frac{-3\left(\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i\right)}{2\sqrt{2}\left(0-1\right)i}[/tex]
=[tex]-3\left(\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i\right)[/tex] ÷ [tex]2\sqrt{2}\left(0-1\right)i[/tex]
= [tex]3\left(-\frac{\sqrt{2}i}{2}+\frac{\sqrt{2}}{2}\right)[/tex] ÷ [tex]-2\sqrt{2}i[/tex]
= [tex]\frac{3\left(1-i\right)}{\sqrt{2}}[/tex]÷ [tex]2\sqrt{2}i[/tex] = [tex]-3-3i[/tex] ÷ [tex]4[/tex] = [tex]-\frac{3}{4}-\frac{3}{4}i[/tex]
As you can see your solution is the last option.
I need help on this question, can someone please answer it correctly?
Answer:the one area < with line underneath then -4
St-by-step explanation: I’m pretty sure this is correct
Answer:
[tex] \boxed{x \leqslant - 4}[/tex]Step-by-step explanation:
[tex] \mathrm{16x - 7 \leqslant - 71}[/tex]
Move constant to Right hand side and change its sign
[tex] \mathrm{16x \leqslant - 71 + 7}[/tex]
Calculate
[tex] \mathrm{16x \leqslant - 64}[/tex]
Divide both sides of the equation by 16
[tex] \mathrm{ \frac{16x}{16} \leqslant \frac{ - 64}{16} }[/tex]
Calculate
[tex] \mathrm{x \leqslant - 4}[/tex]
Hope I helped!
Best regards!
5 STARS IF CORRECT! In general, Can you translate a phrase or sentence into symbols? Explain the answer.
Answer:
Step-by-step explanation:
I answered this already a few minutes ago.
Answer:
yes you can
Step-by-step explanation:
you can write algebraic expressions and use variables for the unknown
When proving a statement using mathematical induction, part of the process is assuming that the statement is true for the nth case. (True or False).
Answer:
True
Step-by-step explanation:
We assume that is true for the nth case and prove it for the n+1 case
and show that it is true for the case when n=1
For a certain casino slot machine, the odds in favor of a win are given as 17 to 83. Express the indicated degree of likelihood as a probability value between 0 and 1 inclusive.
Step-by-step explanation:
83P (E)=17-17P (E),
P (E)=17/100=0.17
please help with this
Answer:
[tex]\sin \left(\theta \right)-\frac{1}{2}\cos \left(2\theta \rightt)+C[/tex]
Step-by-step explanation:
We are given the graph of r = cos( θ ) + sin( 2θ ) so that we are being asked to determine the integral. Remember that [tex]\:r=cos\left(\theta \right)+sin\left(2\theta \right)[/tex] can also be rewritten as [tex]\int \cos \left(\theta \right)+\sin \left(2\theta \right)d\theta \right[/tex].
Let's apply the functional rule [tex]\int f\left(x\right)\pm g\left(x\right)dx=\int f\left(x\right)dx\pm \int g\left(x\right)dx[/tex],
[tex]\int \cos \left(\theta \right)+\sin \left(2\theta \right)d\theta \right[/tex] = [tex]\int \cos \left(\theta \right)d\theta \right+\int \sin \left(2\theta \right)d\theta \right[/tex]
At the same time [tex]\int \cos \left(\theta \right)d\theta \right=\sin \left(\theta \right)[/tex] = [tex]sin( \theta \right ))[/tex], and [tex]\int \sin \left(2\theta \right)d\theta \right[/tex] = [tex]-\frac{1}{2}\cos \left(2\theta \right)[/tex]. Let's substitute,
[tex]\int \cos \left(\theta \right)d\theta \right+\int \sin \left(2\theta \right)d\theta \right[/tex] = [tex]\sin \left(\theta \right)-\frac{1}{2}\cos \left(2\theta \right)[/tex]
And adding a constant C, we receive our final solution.
[tex]\sin \left(\theta \right)-\frac{1}{2}\cos \left(2\theta \rightt)+C[/tex] - this is our integral
Emily made a pot of cream of pumpkin soup for thanksgiving dinner. She put 5
cups of cream in the soup. She poured the soup into 24 small soup bowls. How
much cream (measured in oz.) is used for each small bowl of soup?
Answer:
1 2/3 ounces in each bowl
Step-by-step explanation:
We need to convert 5 cups to ounces
1 cup = 8 ounces
5 cups = 5*8 = 40 ounces
We divide the 40 ounces into 24 bowls
40 ounces / 24 bowl
5/3 ounces per bowl
1 2/3 ounces in each bowl
Answer:
each bowl can contain 5/3 oz. of soup.
Step-by-step explanation:
1 cup = 8 oz.
8 oz.
5 cups x -------------- = 40 oz.
1 cup
to get the measurement of each bowl,
40 oz. divided into 24 bowls.
therefore, each bowl can contain 5/3 oz. of soup.
The balances in two separate bank accounts that grow each month at different rales are represented by the functions f(x) and gix) In what month do the funds in the f(x) bank account exceed those in the glx)
bank account?
Month (x) f(x) = 2* g(x) = 4x + 12
1
2
16
2.
4
20
O Month 3
O Month 4
O Month 5
O Month 6
Answer:
The balance in two separate bank accounts grows each month at different rates. the growth rates for both accounts are represented by the functions f(x) = 2x and g(x) = 4x 12. in what month is the f(x) balance greater than the g(x) balance?
Answer:
6 months
A function is a relationship between inputs where each input is related to exactly one output.
x = 5,
f(5) = [tex]2^5\\[/tex] = 32
g(5) = 4 x 5 + 12 = 20 + 12 = 32
x = 6,
f(6) = [tex]2^6[/tex] = 64
g(6) = 4 x 6 + 12 = 24 + 12 = 36
At month 6 the funds in the f(x) bank account exceed those in the g(x) bank account.
What is a function?A function is a relationship between inputs where each input is related to exactly one output.
Example:
f(x) = 2x + 1
f(1) = 2 + 1 = 3
f(2) = 2 x 2 + 1 = 4 + 1 = 5
The outputs of the functions are 3 and 5
The inputs of the function are 1 and 2.
We have,
f(x) = [tex]2^{x}[/tex]
g(x) = 4x + 12
x = number of months
Now,
x = 3,
f(3) = 2³ = 8
g(3) = 4 x 3 + 12 = 12 + 12 = 24
x = 4,
f(4) = [tex]2^4[/tex] = 16
g(4) = 4 x 4 + 12 = 16 + 12 = 28
x = 5,
f(5) = [tex]2^5\\[/tex] = 32
g(5) = 4 x 5 + 12 = 20 + 12 = 32
x = 6,
f(6) = [tex]2^6[/tex] = 64
g(6) = 4 x 6 + 12 = 24 + 12 = 36
We see that,
At x = 6,
f(5) = 64
g(5) = 36
Thus,
At month 6 the funds in the f(x) bank account exceed those in the g(x) bank account.
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20 points!
Please help.
Write an expression to represent the given statement. Use n for the variable. Three times the absolute value of the sum of a number and 6
Answer:
3 · |x+6|
Step-by-step explanation:
Write out what you see. "Three times" is 3 · something; "the absolute value of the sum of a number and 6" is |number + 6|. We'll use x for our number. Put it all together and you get 3 · |x+6|
The expression of the statement, Three times the absolute value of the sum of a number and 6 is [tex]\[3\left| n+6 \right|\][/tex] .
Representation of statement:Let n be the number.The sum of the numbers n and 6 is n+6.The absolute value of the sum of the numbers n and 6 is [tex]\[\left| n+6 \right|\][/tex].Hence, three times the absolute value of the sum of a number and 6 is [tex]\[3\left| n+6 \right|\][/tex].
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Use spherical coordinates. Evaluate e x2 + y2 + z2 dV, E where E is enclosed by the sphere x2 + y2 + z2 = 25 in the first octant.
Answer:
[tex]\displaystyle \iiint_E {e^{\sqrt{x^2 + y^2 + z^2}}} \, dV = \frac{\pi (17e^5 - 2)}{2}[/tex]
General Formulas and Concepts:
Calculus
Integration
IntegralsIntegration Rule [Reverse Power Rule]:
[tex]\displaystyle \int {x^n} \, dx = \frac{x^{n + 1}}{n + 1} + C[/tex]
Integration Rule [Fundamental Theorem of Calculus 1]:
[tex]\displaystyle \int\limits^b_a {f(x)} \, dx = F(b) - F(a)[/tex]
Integration Property [Multiplied Constant]:
[tex]\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx[/tex]
Integration Property [Addition/Subtraction]:
[tex]\displaystyle \int {[f(x) \pm g(x)]} \, dx = \int {f(x)} \, dx \pm \int {g(x)} \, dx[/tex]
Integration Method [Integration by Parts]:
[tex]\displaystyle \int {u} \, dv = uv - \int {v} \, du[/tex]
Multivariable Calculus
Triple Integrals
Cylindrical Coordinate Conversions:
[tex]\displaystyle x = r \cos \theta[/tex][tex]\displaystyle y = r \sin \theta[/tex][tex]\displaystyle z = z[/tex][tex]\displaystyle r^2 = x^2 + y^2[/tex][tex]\displaystyle \tan \theta = \frac{y}{x}[/tex]Spherical Coordinate Conversions:
[tex]\displaystyle r = \rho \sin \phi[/tex][tex]\displaystyle x = \rho \sin \phi \cos \theta[/tex][tex]\displaystyle z = \rho \cos \phi[/tex][tex]\displaystyle y = \rho \sin \phi \sin \theta[/tex][tex]\displaystyle \rho = \sqrt{x^2 + y^2 + z^2}[/tex]Integral Conversion [Spherical Coordinates]:
[tex]\displaystyle \iiint_T {f( \rho, \phi, \theta )} \, dV = \iiint_T {\rho^2 \sin \phi} \, d\rho \, d\phi \, d\theta[/tex]
Step-by-step explanation:
*Note:
Recall that φ is bounded by 0 ≤ φ ≤ 0.5π from the z-axis to the x-axis.
I will not show/explain any intermediate calculus steps as there isn't enough space.
Step 1: Define
Identify given.
[tex]\displaystyle \iiint_E {e^{\sqrt{x^2 + y^2 + z^2}}} \, dV[/tex]
[tex]\displaystyle \text{Region E:} \ x^2 + y^2 + z^2 = 25 \ \text{bounded by first octant}[/tex]
Step 2: Integrate Pt. 1
Find ρ bounds.
[Sphere] Substitute in Spherical Coordinate Conversions:Find θ bounds.
[Sphere] Substitute in z = 0:Find φ bounds.
[Circle] Substitute in Cylindrical Coordinate Conversions:Step 3: Integrate Pt. 2
[Integrals] Convert [Integral Conversion - Spherical Coordinates]:We evaluate this spherical integral by using the integration rules, properties, and methods listed above:
[tex]\displaystyle \begin{aligned} \iiint_E {e^{\sqrt{x^2 + y^2 + z^2}}} \, dV & = \int\limits^{\frac{\pi}{2}}_0 \int\limits^{\frac{\pi}{2}}_0 \int\limits^5_0 {e^{\rho} \rho^2 \sin \phi} \, d\rho \, d\phi \, d\theta \\ & = \int\limits^{\frac{\pi}{2}}_0 \int\limits^{\frac{\pi}{2}}_0 {\bigg[ (\rho^2 - 2 \rho + 2) e^{\rho} \sin \phi \bigg] \bigg| \limits^{\rho = 5}_{\rho = 0}} \, d\phi \, d\theta\end{aligned}[/tex]
[tex]\displaystyle \begin{aligned}\iiint_E {e^{\sqrt{x^2 + y^2 + z^2}}} \, dV & = \int\limits^{\frac{\pi}{2}}_0 \int\limits^{\frac{\pi}{2}}_0 {(17e^5 - 2) \sin \phi} \, d\phi \, d\theta \\& = \int\limits^{\frac{\pi}{2}}_0 {\bigg[ -(17e^5 - 2) \cos \phi \bigg] \bigg| \limits^{\phi = \frac{\pi}{2}}_{\phi = 0}} \, d\theta \\& = \int\limits^{\frac{\pi}{2}}_0 {17e^5 - 2} \, d\theta \\& = (17e^5 - 2) \theta \bigg| \limits^{\theta = \frac{\pi}{2}}_{\theta = 0} \\& = \frac{\pi (17e^5 - 2)}{2}\end{aligned}[/tex]
∴ the given integral equals [tex]\displaystyle \bold{\frac{\pi (17e^5 - 2)}{2}}[/tex].
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Topic: Multivariable Calculus
Unit: Triple Integrals Applications
What word phrase can you use to represent the algebraic expression 7x?
A. 7 more than a number x
B. the product of 7 and a number x
C. the quotient of 7 and a number x
D. 7 less than a number x
Answer:
B. the product of 7 and a number x
Step-by-step explanation:
7x is 7 multiplied by x.
Answer:
b is the product
Step-by-step explanation:
A hot metal bar is submerged in a large reservoir of water whose temperature is 60°F. The temperature of the bar 20 s after submersion is 120°F. After 1 min submerged, the temperature has cooled to 100°F. (Let y be measured in degrees Fahrenheit, and t be measured in seconds.) (a) Determine the cooling constant k. k = s−1 (b) What is the differential equation satisfied by the temperature y(t)? (Use y for y(t).) y'(t) = (c) What is the formula for y(t)? y(t) = (d) Determine the temperature of the bar at the moment it is submerged. (Round your answer to one decimal place.)
Answer:
a. k = -0.01014 s⁻¹
b. [tex]\mathbf{\dfrac{dy}{dt} = - \dfrac{In(\dfrac{3}{2})}{40}(y-60)}[/tex]
c. [tex]\mathbf{y(t) = 60+ \dfrac{60 \sqrt{3}}{\sqrt{2}} \ e^{\dfrac{-In(\dfrac{3}{2})\ t}{40}}}[/tex]
d. y(t) = 130.485°F
Step-by-step explanation:
A hot metal bar is submerged in a large reservoir of water whose temperature is 60°F. The temperature of the bar 20 s after submersion is 120°F. After 1 min submerged, the temperature has cooled to 100°F.
(Let y be measured in degrees Fahrenheit, and t be measured in seconds.)
We are to determine :
a. Determine the cooling constant k. k = s−1
By applying the new law of cooling
[tex]\dfrac{dT}{dt} = k \Delta T[/tex]
[tex]\dfrac{dT}{dt} = k(T_1-T_2)[/tex]
[tex]\dfrac{dT}{dt} = k (T - 60)[/tex]
Taking the integral.
[tex]\int \dfrac{dT}{T-60} = \int kdt[/tex]
㏑ (T -60) = kt + C
T - 60 = [tex]e^{kt+C}[/tex]
[tex]T = 60+ C_1 e^{kt} ---- (1)[/tex]
After 20 seconds, the temperature of the bar submersion is 120°F
T(20) = 120
From equation (1) ,replace t = 20s and T = 120
[tex]120 = 60 + C_1 e^{20 \ k}[/tex]
[tex]120 - 60 = C_1 e^{20 \ k}[/tex]
[tex]60 = C_1 e^{20 \ k} --- (2)[/tex]
After 1 min i.e 60 sec , the temperature = 100
T(60) = 100
From equation (1) ; replace t = 60 s and T = 100
[tex]100 = 60 + c_1 e^{60 \ t}[/tex]
[tex]100 - 60 =c_1 e^{60 \ t}[/tex]
[tex]40 =c_1 e^{60 \ t} --- (3)[/tex]
Dividing equation (2) by (3) , we have:
[tex]\dfrac{60}{40} = \dfrac{C_1e^{20 \ k } }{C_1 e^{60 \ k}}[/tex]
[tex]\dfrac{3}{2} = e^{-40 \ k}[/tex]
[tex]-40 \ k = In (\dfrac{3}{2})[/tex]
- 40 k = 0.4054651
[tex]k = - \dfrac{0.4054651}{ 40}[/tex]
k = -0.01014 s⁻¹
b. What is the differential equation satisfied by the temperature y(t)?
Recall that :
[tex]\dfrac{dT}{dt} = k \Delta T[/tex]
[tex]\dfrac{dT}{dt} = \dfrac{- In (\dfrac{3}{2})}{40}(T-60)[/tex]
Since y is the temperature of the body , then :
[tex]\mathbf{\dfrac{dy}{dt} = - \dfrac{In(\dfrac{3}{2})}{40}(y-60)}[/tex]
(c) What is the formula for y(t)?
From equation (1) ;
where;
[tex]T = 60+ C_1 e^{kt} ---- (1)[/tex]
Let y be measured in degrees Fahrenheit
[tex]y(t) = 60 + C_1 e^{-\dfrac{In (\dfrac{3}{2})}{40}t}[/tex]
From equation (2)
[tex]C_1 = \dfrac{60}{e^{20 \times \dfrac{-In(\dfrac{3}{2})}{40}}}[/tex]
[tex]C_1 = \dfrac{60}{e^{-\dfrac{1}{2} {In(\dfrac{3}{2})}}}[/tex]
[tex]C_1 = \dfrac{60}{e^ {In(\dfrac{3}{2})^{-1/2}}}}[/tex]
[tex]C_1 = \dfrac{60}{\sqrt{\dfrac{2}{3}}}[/tex]
[tex]C_1 = \dfrac{60 \times \sqrt{3}}{\sqrt{2}}}[/tex]
[tex]\mathbf{y(t) = 60+ \dfrac{60 \sqrt{3}}{\sqrt{2}} \ e^{\dfrac{-In(\dfrac{3}{2})\ t}{40}}}[/tex]
(d) Determine the temperature of the bar at the moment it is submerged.
At the moment it is submerged t = 0
[tex]\mathbf{y(0) = 60+ \dfrac{60 \sqrt{3}}{\sqrt{2}} \ e^{\dfrac{-In(\dfrac{3}{2})\ 0}{40}}}[/tex]
[tex]\mathbf{y(t) = 60+ \dfrac{60 \sqrt{3}}{\sqrt{2}} }[/tex]
y(t) = 60 + 70.485
y(t) = 130.485°F
You drive 15 miles in 0.1hours . How fast did you travel if 8=d/t
Answer:
150Step-by-step explanation:
[tex]distance = 15 miles\\time = 0.1 hours\\\\Speed = \frac{Distance}{time}\\ Speed = \frac{15}{0.1}\\ Speed =150[/tex]
Answer:
[tex]150mph[/tex]
Step-by-step explanation:
Given:
s=15miles
t=0.1hours
Required:
v=?
Formula:
[tex]v = \frac{s}{t} [/tex]
Solution:
[tex]v = \frac{s}{t} = \frac{15m}{0.1h} = \frac{150m}{1h} = 150mph[/tex]
Hope this helps ;) ❤❤❤
Determine the number of degrees of freedom for the two-sample t test or CI in each of the following situations. (Round your answers down to the nearest whole number.)a. m = 12, n = 15, s1 = 4.0, s2 = 6.0b. m = 12, n = 21, s1 = 4.0, s2 = 6.0c. m = 12, n = 21, s1 = 3.0, s2 = 6.0d. m = 10, n = 24, s1 = 4.0, s2 = 6.0
Answer:
Part a ) The degrees of freedom for the given two sample non-pooled t test is 24
Part b ) The degrees of freedom for the given two sample non-pooled t test is 30
Part c ) The degrees of freedom for the given two sample non-pooled t test is 30
Part d ) The degrees of freedom for the given two sample non-pooled t test is 25
Step-by-step explanation:
Degrees of freedom for a non-pooled two sample t-test is given by;
Δf = {[ s₁²/m + s₂²/n ]²} / {[( s₁²/m)²/m-1] + [(s₂²/n)²/n-1]}
Now given the information;
a) :- m = 12, n = 15, s₁ = 4.0, s₂ = 6.0
we substitute
Δf = {[ 4²/12 + 6²/15 ]²} / {[( 4²/12)²/12-1] + [(6²/15)²/15-1]}
Δf = 30184 / 1241
Δf = 24.3223 ≈ 24 (down to the nearest whole number)
b) :- m = 12, n = 21, s₁ = 4.0, s₂ = 6.0
we substitute using same formula
Δf = {[ s₁²/m + s₂²/n ]²} / {[( s₁²/m)²/m-1] + [(s₂²/n)²/n-1]}
Δf = {[ 4²/12 + 6²/21 ]²} / {[( 4²/12)²/12-1] + [(6²/21)²/21-1]}
Δf = 56320 / 1871
Δf = 30.1015 ≈ 30 (down to the nearest whole number)
c) :- m = 12, n = 21, s₁ = 3.0, s₂ = 6.0
we substitute using same formula
Δf = {[ s₁²/m + s₂²/n ]²} / {[( s₁²/m)²/m-1] + [(s₂²/n)²/n-1]}
Δf = {[ 3²/12 + 6²/21 ]²} / {[( 3²/12)²/12-1] + [(6²/21)²/21-1]}
Δf = 29095 / 949
Δf = 30.6585 ≈ 30 (down to the nearest whole number)
d) :- m = 10, n = 24, s₁ = 4.0, s₂ = 6.0
we substitute using same formula
Δf = {[ s₁²/m + s₂²/n ]²} / {[( s₁²/m)²/m-1] + [(s₂²/n)²/n-1]}
Δf = {[ 4²/10 + 6²/24 ]²} / {[( 4²/10)²/10-1] + [(6²/24)²/24-1]}
Δf = 1044 / 41
Δf = 25.4634 ≈ 25 (down to the nearest whole number).
The areas of two similar octagons are 4 m² and 9 m². What is the scale factor of their side lengths? PLZ PLZ HELP PLZ
Answer:
[tex] \frac{2}{3} [/tex]
Step-by-step explanation:
Area of Octagon A = 4 m²
Side length of Octagon A = a
Area of Octagon B = 9 m²
Side length of Octagon B = b
The scale factor of their side lengths = [tex] \frac{a}{b} [/tex]
According to the area of similar polygons theorem, [tex] \frac{4}{9} = (\frac{a}{b})^2 [/tex]
Thus,
[tex] \sqrt{\frac{4}{9}} = \frac{a}{b} [/tex]
[tex] \frac{\sqrt{4}}{\sqrt{9}} = \frac{a}{b} [/tex]
[tex] \frac{2}{3} = \frac{a}{b} [/tex]
Scale factor of their sides = [tex] \frac{2}{3} [/tex]
Answer:
3:5
Step-by-step explanation:
square root of 9 is 3.
square root if 25 is 5.
therefore, 3:5.