Water is flowing into the tank at a rate of 5 litres per minute. At the same time water flows out of the tank at a rate of k litres per minute. The volume (in litres) of water in the tank at time t (in minutes) is given by the formula V(t) = 100-4t. What is initial volume of water in the tank?


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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%.

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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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.

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.

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.

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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]

Answer:

I think the answer is B or C

Answer:

ok so we need to first find the amount of black squares vs white squares

top row: 2 black 3 white

2nd top row: 1 black 4 white

middle row: 3 black 2 white

2nd bottom row: 1 black 4 white

bottom row: 2 black 3 white

so lets add that up

9 black 16 white

so then each one is 5 meters so

9*5=45

45m^2

Hope This Helps!!!

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.

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.

"This is non response bias because many customers may choose to not fill out the cards"
Not everyone will chose to leave a review when leaving the store, and those that do were probably in some way inconvenienced, and felt inclined to leave one, while the others left and went about their day.

Answer:

12

Step-by-step explanation:

regular implies that all sides are equal
8 sides
total of 96
96/8
12

Answer:

Step-by-step explanation:

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.

(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.

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.

Answer:

Tennnnnnnnnnnnn.............

Answer:

$35.00

Step-by-step explanation:

I just did it rn

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

Answer:

the surface area of the rectangular prism is 142cm³

Step-by-step explanation:

35+15+21+21+35+15

Step-by-step explanation:

the formula for the circumference of a circle is

2×pi×r

so, B is correct : 2×pi×4 in

Answer:

The answer is 4x² + 2x + 11

Step-by-step explanation:

4x⁴ – 2x³ + 5x² + 2x ÷ x² – x – 1

Now,

[tex]4 {x}^{2} + 2x + 11\\ x² - x - 1 \sqrt{4 x⁴ - 2x³ + 5x² + 2x} \\ \: \: \: \: \: \: 4 {x}^{4} - 4 {x}^{3} - {4x}^{2} \\ \frac{ \: \: \: - \: \: \: \: \: \: + \: \: \: \: \: \: \: \: + \: \: }{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: 2x {}^{3} + 9x {}^{2} + 2x } \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: 2 {x}^{3} - {2x}^{2} - 2x \\ \frac{ \: \: \: \: \: - \: \: \: \: \: \: + \: \: \: \: + }{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 11 {x}^{2} + 4x } \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 11 {x}^{2} - 11x - 11 \\ \frac{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: - \: \: \: \: \: \: + \: \: \: \: \: \: \: \: + }{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 15x + 11} [/tex]

Thus, The answer is 4x² + 2x + 11

-TheUnknownScientist 72

Answer:

4x² + 2x + 11

Step-by-step explanation:

hope it's helpful

Answer:

answer is 105 hope u like it

Answer:

35x + 14

Step-by-step explanation:

multiply in the brackets

5x × 7 = 35x

(+)2 × 7 = (+)14

Answer:

  a. 1

  b. 5

Step-by-step explanation:

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.

__

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.

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

Answer:

7x+9

Step-by-step explanation:

10x- 3x=7x

5--4=9

so you just bring them together so 7x+9

=====================================================

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

Use 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.

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]

Answer:

its c just so u know

Step-by-step explanation:

Answer:

5 x - 4

Step-by-step explanation:

Since g(f(x))=x g ( f ( x ) ) = x , f−1(x)=x5+45 f - 1 ( x ) = x 5 + 4 5 is the inverse of f(x)=5x−4 f ( x ) = 5 x - 4 .