1 hour = 60 minutes
45 min = (45 min)/(60 min) = 45/60 = 3/4 of an hour.
To go from 45/60 to 3/4, I divided both top and bottom by the GCF 15.
Answer:
45 minutes equal to 3/4 of an hour because, if you convert the one hour into minutes (60 minutes) it is divisible by 15 four times.
3 · 15 = 45.
Please Help me. Thanks!
Multiply.
52 × 4
Enter your answers in the boxes to correctly complete the equations.
Answer:
52 x 4 = (50 + 2) x 4
= 50 x 4 + 2 x 4
=200 + 8
=208
this should be right!
brainliest?
The correct the correct multiple of ( 52 x 4 ) is 208.
What is multiple?The number you obtain when you multiple a certain number by an integer. Multiple as like consisting of, including, or involving more than one.
Assume the multiplication of 52 x 4 is A.
To find the value of A
52 x 4 = ( 50 + 2 ) x 4
52 x 4 = 50 x ( 4 ) + ( 2 ) x 4
52 x 4 = ( 200 ) + 8
52 x 4 = ( 208 ).
A = 208.
Therefore, the correct multiple of ( 52 x 4 ) is 208.
Learn more about multiple here:
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find the smallest possible value of n for which 99n is multiple of 24
Answer:
99 is not multiple of 24 you will get it wrong if you think it is
Step-by-step explanation:
Answer:
The answer is 8.
7. 20 is what percent of 50?
A. 250%
B. 10%
C. 40%
D. 30%
Answer:
I think the answer is B or C
help plssssssssssssssssss
Answer:
35x + 14
Step-by-step explanation:
multiply in the brackets
5x × 7 = 35x
(+)2 × 7 = (+)14
HELP PLS - ITS DUE TMR
Using the net below, find the surface area
of the rectangular prism.
7 cm
3 cm
7 cm
5 cm
5 cm
3 cm
3 cm
3 cm
Surface Area
Answer:
the surface area of the rectangular prism is 142cm³
Step-by-step explanation:
35+15+21+21+35+15
PLEASE HELP ASAP! I am so confused by this for some reason.
=====================================================
Explanation:
You can use your calculator to make short work of this problem. Simply plot the parabola and use the "minimum" feature on the calculator to spot the lowest point (specifically the coordinates of that point).
However, I have a feeling your teacher wants you to use a bit of math here. So I'll focus on another approach instead.
---------
The parabolic equation is of the form y = ax^2+bx+c
In this case, we have
a = 0.3b = -192c = 40357Use the values of 'a' and b to find the x coordinate of the vertex h
h = -b/(2a)
h = -(-192)/(2*0.3)
h = 320
The x coordinate of the vertex is 320.
Plug this into the original equation to find the y coordinate of the vertex.
y = 0.3x^2 - 192x + 40357
y = 0.3(320)^2 - 192(320) + 40357
y = 9637 which is the minimum unit cost, in dollars.
The vertex is (h,k) = (320, 9637). It is the lowest point on this parabola.
Interpretation: If the company made 320 cars, then the unit cost (aka cost per car) is the smallest at $9637 per car.
This shows two functions.
f(x)=5x-7
g(x)=x+4
which function represents
h(x) = f(x) ⋅ g(x)
Answer:
its c just so u know
Step-by-step explanation:
please answer with explanation
Answer:
The length of the other diagonal is 42 cm.
Step-by-step explanation:
Area of Rhombus = 966 cm^2 (squared)
one length of diagonal(d2) =46cm
The are of rhombus= d1 x d2/2
*substitution from equation above* = 966= 46 x d2/2 *plug answer in* : 966=23 x d2 *divide next*
d2=966/23
d2=42cm
I don't know if that makes sense hopefully it helps.
Find the difference between 3x+5 and 10x-4.
Answer:
7x+9
Step-by-step explanation:
10x- 3x=7x
5--4=9
so you just bring them together so 7x+9
Charlotte earns $1675 every month of which she spends $85 on cosmetics. Identify the percent of earnings that Charlotte spends on cosmetics to the nearest tenth of a percent?
O 5.1%
O 2.5%
O 3.9%
O 7.4%
Answer:
5.1%
Step-by-step explanation:
percent = part/whole × 100%
percent = 85/1675 × 100%
percent = 5.1%
The percentage of earnings that Charlotte spends on cosmetics to the nearest tenth of a per cent is 5.1%.
What is the percentage?The percentage is defined as representing any number with respect to 100. It is denoted by the sign %. The percentage stands for "out of 100." Imagine any measurement or object being divided into 100 equal bits.
Numbers (constants), variables, operations, functions, brackets, punctuation, and grouping can all be represented by mathematical symbols, which can also be used to indicate the logical syntax's order of operations and other features.
Charlotte earns $1675 every month of which she spends $85 on cosmetics.
The percentage will be calculated as,
Percent = part/whole × 100%
Percent = 85/1675 × 100%
Percent = 5.1%
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Section 8.1 Introduction to the Laplace Transforms
Problem 1.
Find the Laplace transforms of the following functions by evaluating the integral
[tex]F(s) = {∫}^{ \infty } _{0} {e}^{ - st} f(t)dt[/tex]
[tex](a)t[/tex]
[tex](b) {te}^{ - t} [/tex]
[tex](c) {sinh} \: bt[/tex]
[tex](d) {e}^{2t} - {3e}^{t} [/tex]
[tex](e) {t}^{2} [/tex]
For the integrals in (a), (b), and (e), you'll end up integrating by parts.
(a)
[tex]\displaystyle \int_0^\infty t e^{-st} \, dt[/tex]
Let
[tex]u = t \implies du = dt[/tex]
[tex]dv = e^{-st} \, dt \implies v = -\dfrac1s e^{-st}[/tex]
Then
[tex]\displaystyle \int_0^\infty t e^{-st} \, dt = uv\bigg|_{t=0}^{t\to\infty} - \int_0^\infty v\, du[/tex]
[tex]\displaystyle \int_0^\infty t e^{-st} \, dt = \left(-\frac1s te^{-st}\right)\bigg|_0^\infty + \frac1s \int_0^\infty e^{-st} \, dt[/tex]
[tex]\displaystyle \int_0^\infty t e^{-st} \, dt = -\frac1s \left(\lim_{t\to\infty} te^{-st} - 0\right) + \frac1s \int_0^\infty e^{-st} \, dt[/tex]
[tex]\displaystyle \int_0^\infty t e^{-st} \, dt = \frac1s \int_0^\infty e^{-st} \, dt[/tex]
[tex]\displaystyle \int_0^\infty t e^{-st} \, dt = -\frac1{s^2} e^{-st} \bigg|_0^\infty e^{-st} \, dt[/tex]
[tex]\displaystyle \int_0^\infty t e^{-st} \, dt = -\frac1{s^2} \left(\lim_{t\to\infty}e^{-st} - 1\right) = \boxed{\frac1{s^2}}[/tex]
(b)
[tex]\displaystyle \int_0^\infty t e^{-t} e^{-st} \, dt = \int_0^\infty t e^{-(s+1)t} \, dt[/tex]
Let
[tex]u = t \implies du = dt[/tex]
[tex]dv = e^{-(s+1)t} \, dt \implies v = -\dfrac1{s+1} e^{-(s+1)}t[/tex]
Then
[tex]\displaystyle \int_0^\infty t e^{-(s+1)t} \, dt = -\dfrac1{s+1} te^{-(s+1)t} \bigg|_0^\infty + \frac1{s+1} \int_0^\infty e^{-(s+1)t} \, dt[/tex]
[tex]\displaystyle \int_0^\infty t e^{-(s+1)t} \, dt = -\dfrac1{s+1} \left(\lim_{t\to\infty}te^{-(s+1)t} - 0\right) + \frac1{s+1} \int_0^\infty e^{-(s+1)t} \, dt[/tex]
[tex]\displaystyle \int_0^\infty t e^{-(s+1)t} \, dt = \frac1{s+1} \int_0^\infty e^{-(s+1)t} \, dt[/tex]
[tex]\displaystyle \int_0^\infty t e^{-(s+1)t} \, dt = -\frac1{(s+1)^2} e^{-(s+1)t} \bigg|_0^\infty[/tex]
[tex]\displaystyle \int_0^\infty t e^{-(s+1)t} \, dt = -\frac1{(s+1)^2} \left(\lim_{t\to\infty}e^{-(s+1)t} - 1\right) = \boxed{\frac1{(s+1)^2}}[/tex]
(e)
[tex]\displaystyle \int_0^\infty t^2 e^{-st} \, dt[/tex]
Let
[tex]u = t^2 \implies du = 2t \, dt[/tex]
[tex]dv = e^{-st} \, dt \implies v = -\dfrac1s e^{-st}[/tex]
Then
[tex]\displaystyle \int_0^\infty t^2 e^{-st} \, dt = -\frac1s t^2 e^{-st} \bigg|_0^\infty + \frac2s \int_0^\infty t e^{-st} \, dt[/tex]
[tex]\displaystyle \int_0^\infty t^2 e^{-st} \, dt = -\frac1s \left(\lim_{t\to\infty} t^2 e^{-st} - 0\right) + \frac2s \int_0^\infty t e^{-st} \, dt[/tex]
[tex]\displaystyle \int_0^\infty t^2 e^{-st} \, dt = \frac2s \int_0^\infty t e^{-st} \, dt[/tex]
The remaining integral is the transform we found in (a), so
[tex]\displaystyle \int_0^\infty t^2 e^{-st} \, dt = \frac2s \times \frac1{s^2} = \boxed{\frac2{s^3}}[/tex]
Computing the integrals in (c) and (d) is much more immediate.
(c)
[tex]\displaystyle \int_0^\infty \sinh(bt) e^{-st} \, dt = \int_0^\infty \frac{e^{bt}-e^{-bt}}2 \times e^{-st} \, dt[/tex]
[tex]\displaystyle \int_0^\infty \sinh(bt) e^{-st} \, dt = \frac12 \int_0^\infty \left(e^{(b-s)t} - e^{(b+s)t}\right) \, dt[/tex]
[tex]\displaystyle \int_0^\infty \sinh(bt) e^{-st} \, dt = \frac12 \left(\frac1{b-s} e^{(b-s)t} - \frac1{b+s} e^{(b+s)t}\right) \bigg|_0^\infty[/tex]
[tex]\displaystyle \int_0^\infty \sinh(bt) e^{-st} \, dt = \frac12 \left[\lim_{t\to\infty}\left(\frac1{b-s} e^{(b-s)t} - \frac1{b+s} e^{(b+s)t}\right) - \left(\frac1{b-s} - \frac1{b+s}\right)\right][/tex]
[tex]\displaystyle \int_0^\infty \sinh(bt) e^{-st} \, dt = \frac12 \left(\frac1{b+s} - \frac1{b-s}\right) = \boxed{\frac{s}{s^2-b^2}}[/tex]
(d)
[tex]\displaystyle \int_0^\infty (e^{2t} - 3e^t) e^{-st} \, dt = \int_0^\infty \left(e^{(2-s)t} - 3e^{(1-s)t}\right) \, dt[/tex]
[tex]\displaystyle \int_0^\infty (e^{2t} - 3e^t) e^{-st} \, dt = \left( \frac1{2-s} e^{(2-s)t} - \frac3{1-s} e^{(1-s)t} \right) \bigg|_0^\infty[/tex]
[tex]\displaystyle \int_0^\infty (e^{2t} - 3e^t) e^{-st} \, dt = \lim_{t\to\infty} \left( \frac1{2-s} e^{(2-s)t} - \frac3{1-s} e^{(1-s)t} \right) - \left( \frac1{2-s} - \frac3{1-s} \right)[/tex]
[tex]\displaystyle \int_0^\infty (e^{2t} - 3e^t) e^{-st} \, dt = \frac3{1-s} - \frac1{2-s} = \boxed{-\frac{2s-5}{s^2-3s+2}}[/tex]
(1−x21)5)=?
can someone help me
Answer:
answer is 105 hope u like it
Make this Answer as complex as possible. 5 + 5
Answer:
Tennnnnnnnnnnnn.............
Help with math?Please? ANYONE
Answer:
(-1, 1)
Step-by-step explanation:
Hi there!
We want to solve the system of equations given as:
2x-3y=-5
3x+y=-2
Let's solve this equation by substitution, where we will set one variable equal to an expression containing the other variable, substitute the expression as the variable that it equals, solve for the other variable (the variable that the expression contains), and then use the value of the solved variable to find the value of the first variable
In the second equation, we have y by itself; therefore, if we subtract 3x from both sides, then we will get an expression that y is equal to.
So subtract 3x from both sides
y=-3x-2
Now substitute -3x-2 as y in the first equation.
It will look something like this:
2x - 3(-3x-2)=-5
Now do the distributive property.
2x+9x+6=-5
Combine like terms
11x+6=-5
Subtract 6 from both sides
11x=-11
Divide both sides by 11
x=-1
Now substitute -1 as x in the equation y=-3x-2 to solve for y:
y=-3(-1)-2
multiply
y=3-2
Subtract
y=1
The answer is x=-1, y=1; this can also be written as an ordered pair, which would be (-1, 1)
Hope this helps!
If you would like to see another problem for additional practice, take a look here: https://brainly.com/question/19212538
[tex]\begin{cases} 2x-3y=-5\\ 3x+y=-2 \end{cases} \\\\[-0.35em] ~\dotfill\\\\ 2x-3y=-5\implies 2x=3y-5\implies x=\cfrac{3y-5}{2} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{substituting on the 2nd equation}}{3\left( \cfrac{3y-5}{2} \right)+y=-2}\implies \cfrac{3(3y-5)}{2}+y=-2 \\\\\\ \stackrel{\textit{multiplying both sides by }\stackrel{LCD}{2}}{2\left( \cfrac{3(3y-5)}{2}+y \right)}=2(-2)\implies 3(3y-5)+2y=-4 \\\\\\ 9y-15+2y=-4\implies 11y-15=-4\implies 11y=11[/tex]
[tex]y=\cfrac{11}{11}\implies \blacktriangleright y=1 \blacktriangleleft \\\\\\ \stackrel{\textit{since we know that}}{x=\cfrac{3y-5}{2}}\implies x=\cfrac{3(1)-5}{2}\implies x=\cfrac{-2}{2}\implies \blacktriangleright x=-1 \blacktriangleleft \\\\[-0.35em] ~\dotfill\\\\ ~\hfill (-1~~,~~1)~\hfill[/tex]
Kiran wrote the expression x−10 for this number puzzle: “Pick a number, add -2, and multiply by 5.”
Lin thinks Kiran made a mistake.
How can she convince Kiran he made a mistake?
What would be a correct expression for this number puzzle?
Step-by-step explanation:
the number is represented by x. you need to add -2 first then multiply by 5.
thus, the correct expression will be:
( x + -2) *5 = (x-2) *5 = 5(x-2)
for e.g, let x be 3.
acc to the puzzle,
3+-2 = 3-2=1
1*5= 5
using Kiran's expression:
x-10; 3-10= -7
here you can see the answer is wrong.
using the "correct expression" :
5(x-2) = 5( 3-2)
= 5(1)
= 5.
hope this helped you!
-s.
what is the wavelength of a wave that has a frequency of 15 Hz and a speed of 2 m/s?
Wavelength=speed/frequency
=2/15
=0.1333 m
Find x. Round to the nearest tenth if necessary.
Answer:
Step-by-step explanation:
14 times 14 then do 15 times 15 then add both answers up and take that answer and square route it
6.Name two streets that intersect.
7.Name two streets that are parallel
Answer:
6.Elm and Oak intersect
7.Birch and Maple are parallel
Step-by-step explanation:
If Elm and Oak continue, they will intersect with each other
Birch and Maple have same distance consistently between them
The radius of a circle is 4 inches. How would you calculate the circumference?
A. π · 4² in.²
B. 2 · π· 4 in.
C. π · 8² in.²
D. 2· π · 8 in.
Step-by-step explanation:
the formula for the circumference of a circle is
2×pi×r
so, B is correct : 2×pi×4 in
Calculus AB Homework, does anyone know how to do this...
(a) f(x) is continuous at x = 1 if the limits of f(x) from either side of x = 1 both exist and are equal:
[tex]\displaystyle \lim_{x\to1^-}f(x) = \lim_{x\to1} (2x-x^2) = 1[/tex]
[tex]\displaystyle \lim_{x\to1^+}f(x) = \lim_{x\to1} (x^2+kx+p) = 1 + k + p[/tex]
So we must have 1 + k + p = 1, or k + p = 0.
f(x) is differentiable at x = 1 if the derivative at x = 1 exists; in order for the derivative to exist, the following one-sided limits must also exist and be equal:
[tex]\displaystyle \lim_{x\to1^-}f'(x) = \lim_{x\to1^+}f'(x)[/tex]
Note that the derivative of each piece computed here only exists on the given open-ended domain - we don't know for sure that the derivative *does* exist at x = 1 just yet:
[tex]f(x) = \begin{cases}2x-x^2 & \text{for }x\le1 \\ x^2+kx+p & \text{for }x>1\end{cases} \implies f'(x) = \begin{cases}2 - 2x & \text{for }x < 1 \\ ? & \text{for }x = 1 \\ 2x + k & \text{for }x > 1 \end{cases}[/tex]
Compute the one-sided limits of f '(x) :
[tex]\displaystyle \lim_{x\to1^-}f'(x) = \lim_{x\to1} (2 - 2x) = 0[/tex]
[tex]\displaystyle \lim_{x\to1^+}f'(x) = \lim_{x\to1} (2x+k) = 2 + k[/tex]
So if f '(1) exists, we must have 2 + k = 0, or k = -2, which in turn means p = 2, and these values tell us that we have f '(1) = 0.
(b) Find the critical points of f(x), where its derivative vanishes. We know that f '(1) = 0. To assess whether this is a turning point of f(x), we check the sign of f '(x) to the left and right of x = 1.
• When e.g. x = 0, we have f '(0) = 2 - 2•0 = 2 > 0
• When e.g. x = 2, we have f '(2) = 2•2 - 2 = 2 > 0
The sign of f '(x) doesn't change as we pass over x = 1, so this critical point is not a turning point. However, since f '(x) is positive to the left and right of x = 1, this means that f(x) is increasing on (-∞, 1) and (1, ∞).
(c) The graph of f(x) has possible inflection points wherever f ''(x) = 0 or is non-existent. Differentiating f '(x), we get
[tex]f'(x) = \begin{cases}2-2x & \text{for }x<1 \\ 0 & \text{for }x=1 \\ 2x+k & \text{for }x>1\end{cases} \implies f''(x) = \begin{cases}- 2 & \text{for }x < 1 \\ ? & \text{for }x = 1 \\ 2 & \text{for }x > 1 \end{cases}[/tex]
Clearly f ''(x) ≠ 0 if x < 1 or if x > 1.
It is also impossible to choose a value of f ''(1) that makes f ''(x) continuous, or equivalently that makes f(x) twice-differentiable. In short, f ''(1) does not exist, so we have a single potential inflection point at x = 1.
From the above, we know that f ''(x) < 0 for x < 1, and f ''(x) > 0 for x > 1. This indicates a change in the concavity of f(x), which means x = 1 is the only inflection point.
9. Suppose you are comparing frequency data for two different
groups, 25 managers and 150 blue collar workers. Why would a
relative frequency distribution be better than a frequency
distribution?
ents
Answer:
A relative frequency distribution is better for comparison between groups whose numbers are different, since ratios are readily comparable.
Step-by-step explanation:
PLEASE HELP I'LL DO ANYTHING
Simplify the expression
1/5 (5x + 9) + 4/5 (1 - 9x)
Step-by-step explanation:
Use the Distributive Property:
[tex]1/5(5x+9)+4/5(1-9x)[/tex]
[tex]x+1.8+4/5-7.2x[/tex]
Combine Like-Terms:
[tex]-6.2x+2.6[/tex]
Jeff is buying books at a used bookstore he wants to approximate the total cost of his purchase before checking out which amount is the most reasonable approximation of the total price of the five books
Answer:
$35.00
Step-by-step explanation:
I just did it rn
Is anybody able to help me with this?
Answer:
Step-by-step explanation:
PLEASE HELP ANYONE HELP ME W THIS
Answer:
a. 1
b. 5
Step-by-step explanation:
a.Ordered pairs are (x, f(x)). You want f(2), so you look for an ordered pair that lists 2 as the first number. That one is (2, 1). This means f(2) = 1.
__
b.f(2) is plotted as the point (2, f(2)). That is, the x-coordinate of the point will be 2. The point that has x-coordinate 2 is (2, 5).
As above, this means f(2) = 5.
How do you solve -8>v-3?
Answer:
v < -5
Step-by-step explanation:
Add 3 to both sides. This gives -8+(-3)>v-3+3, or -5>v. Rewriting the inequality gives v < -5.
which systems have infinite solution? check all that apply
A. y = 0.5x + 2.75 and 2y = x + 2.75
B. y = 0.5x + 2.75 and y - 0.5x = 2.75
C. y = 0.5x + 2.75 and 0.5x + y = 2.75
D. y = 0.5x + 2.75 and y = 0.5(x + 5.5)
E. y = 0.5x + 2.75 and y = -2(-0.25x) + 2.75
Answer:
B
Step-by-step explanation:
Answer:
Ok got it
Step-by-step explanation:
the answers are y = 0.5x + 2.75 and y – 0.5x = 2.75
y = 0.5x + 2.75 and y = 0.5(x + 5.5)
y = 0.5x + 2.75 and y = –2(–0.25x )+ 2.75
The perimeter of a regular octagon is 96 mm. How long each side?
Answer:
12
Step-by-step explanation:
regular implies that all sides are equal
8 sides
total of 96
96/8
12
Does anyone know the answer to this question?
Answer:
4
Step-by-step explanation:
You can set up the equation as 5x-3=2x+9. Subtract 2x from both sides, then add three to both sides which gives you 3x=12. Divide both sides by 3, and you get x=4.