Simplify the expression 6h +(-7.7d) – 16 + 5d - 2.6h.

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.

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:

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.

Learn more about multiple here:
https://brainly.com/question/29276732

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Answer:

7x+9

Step-by-step explanation:

10x- 3x=7x

5--4=9

so you just bring them together so 7x+9

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:

(-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]

(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.

Answer:

A relative frequency distribution is better for comparison between groups whose numbers are different, since ratios are readily comparable.

Step-by-step explanation:

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.

Answer:

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

Answer:

Step-by-step explanation:

Answer:

$35.00

Step-by-step explanation:

I just did it rn

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:

12

Step-by-step explanation:

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

Answer:

the area of the triangle is 36

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.

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:

I think the answer is B or C

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:

  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:

its c just so u know

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.

Answer:

the surface area of the rectangular prism is 142cm³

Step-by-step explanation:

35+15+21+21+35+15

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

Answer:

35x + 14

Step-by-step explanation:

multiply in the brackets

5x × 7 = 35x

(+)2 × 7 = (+)14

Wavelength=speed/frequency

=2/15

=0.1333 m

Answer:

answer is 105 hope u like it

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.

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

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.