Need the answer been stuck on this

Answer:

4/420=105

Step-by-step explanation:

decompose 42 and divide by 4 is 10 and is 20 divided by 4 is 5

Answer:

f(x) = 2x^3 - 6x^2 - 20x + 48

Step-by-step explanation:

A polynomial with zeros a, b, c, etc., is the product of (x - a)(x - b)(x - c)...

f(x) = (x + 3)(x - 2)(x - 4)

f(x) = (x^2 + x - 6)(x - 4)

f(x) = x^3 - 4x^2 + x^2 - 4x - 6x + 24

f(x) = x^3 - 3x^2 - 10x + 24

This polynomial function has the zeros listed in the problem.

Now we need to make sure it includes the point (6, 144).

f(x) = (x + 3)(x - 2)(x - 4)

f(6) = (6 + 3)(6 - 2)(6 - 4)

f(6) = 9 * 4 * 2

f(6) = 72

The polynomial function has point (6, 72). We want it to include the point (6, 144). We multiply the function by 2.

f(x) = 2(x^3 - 3x^2 - 10x + 24)

f(x) = 2x^3 - 6x^2 - 20x + 48

Answer:

[tex]y = x-7[/tex]    

Step-by-step explanation:

To write the equation of the line in slope-intercept form, use the slope-intercept formula, [tex]y = mx +b[/tex]. Find real values for the [tex]m[/tex] and [tex]b[/tex] and substitute them into the formula.  

1) We know that [tex]m[/tex] represents the slope, so substitute 1 in its place.  However, we still need to find [tex]b[/tex], or the y-intercept. So, along with substituting that 1 in for [tex]m[/tex], substitute the x and y values for (7,0) for the x and y in the formula as well. Then, isolate [tex]b[/tex] to find its value:

[tex]y = mx + b\\0 = (1)(7)+b\\0 = 7 + b\\-7 = b[/tex]

So, [tex]b[/tex] = -7.  

2) Now, just substitute the found values for [tex]m[/tex] and [tex]b[/tex] into the slope-intercept formula. Remember that the slope is [tex]m[/tex], so substitute 1 in, and [tex]b[/tex] is -7, so substitute -7 in its place as well. This gives the following equation in slope-intercept form:

[tex]y = 1x -7[/tex] or [tex]y = x-7[/tex]

Answer:

10/1

Step-by-step explanation:

Answer:

yo sorry I'm dum need points

If the 0's are indeed zeroes, then it would appear that you're using degrees, so that the equation

r(θ) = 2 sin(0°) + 3 sin(90°)

describes a circle r = 3 (since sin(0°) = 0 and sin(90°) = 1). In this case, the area would simply be π r ² = 9π.

But I suspect you meant to use θ, so the curve has the far more interesting equation,

r(θ) = 2 sin(θ) + 3 sin(9θ)

Since sin(9θ) has a period of 2π/9 and sin(θ) has a period of 2π, their sum has a period of 2π. (That is, 2π times the LCM of 1 and 1/9, which is 1.) But we observe r(θ) is odd, since

r(-θ) = 2 sin(-θ) + 3 sin(-9θ)

… = -2 sin(θ) - 3 sin(9θ)

… = - r(θ)

which tells us that the curve closes itself over a half-period. So the area bounded by the curve is given by the integral,

[tex]\displaystyle\int_0^\pi\int_0^{r(\theta)}r\,\mathrm dr\,\mathrm d\theta[/tex]

If you're not familiar with double integrals yet, all you need to know is that this reduces to the area formula you may/should be familiar with,

[tex]\displaystyle\frac12\int_0^\pi r(\theta)^2\,\mathrm d\theta[/tex]

Compute the integral:

[tex]=\displaystyle\frac12\int_0^\pi(2\sin(\theta)+3\sin(9\theta))^2\,\mathrm d\theta[/tex]

[tex]=\displaystyle\frac12\int_0^\pi(4\sin^2(\theta)+12\sin(\theta)\sin(9\theta)+9\sin^2(9\theta))\,\mathrm d\theta[/tex]

Apply some identities:

sin²(θ) = (1 - cos(2θ)) / 2

sin(θ) sin(9θ) = (cos(9θ - θ) - cos(9θ + θ)) / 2 = (cos(8θ) - cos(10θ)) / 2

sin²(9θ) = (1 - cos(18θ)) / 2

[tex]=\displaystyle\frac14\int_0^\pi(4(1-\cos(2\theta))+12(\cos(8\theta)-\cos(10\theta))+9(1-\cos(18\theta))\,\mathrm d\theta[/tex]

[tex]=\displaystyle\frac14\int_0^\pi(13 - 4\cos(2\theta) + 12\cos(8\theta) - 12\cos(10\theta) - 9\cos(18\theta))\,\mathrm d\theta[/tex]

[tex]=\dfrac14 \left(13\theta-2\sin(2\theta)+\dfrac32\sin(8\theta)-\dfrac65\sin(10\theta)-\dfrac12\sin(18\theta)\right)\bigg|_0^\pi[/tex]

[tex]=\dfrac1{40} \left(130\theta-20\sin(2\theta)+15\sin(8\theta)-12\sin(10\theta)-5\sin(18\theta)\right)\bigg|_0^\pi[/tex]

[tex]=\dfrac{130\pi-20\sin(2\pi)+15\sin(8\pi)-12\sin(10\pi)-5\sin(18\pi)}{40}[/tex]

= 13π/4

Answer:

1143

Step-by-step explanation:

Answer:

To maximize profit, the amount of each type of coffees made are;

The number of regular latte made = 52 cups

The number of low-fat latte made = 48 cups

Step-by-step explanation:

The given parameters are;

The number of espresso in a regular latte = 1 shot

The number of cups of 2% milk in regular latte = 1 cup

The amount of whipped cream in a regular latte = 0.5 cup

The number of cups of skim milk in the low-fat latte = 1.25 cup

The number of espresso in a low-fat latte = 1 shot

The amount of whipped cream in a low-fat latte = 0 cup

The number of cups of espresso in the journey = 100 shots

The number of cups of skim milk in the journey = 60 cups

The number of cups of 2% milk in the journey = 60 cups

The amount of profit the airline makes on each regular latte = $1.58

The amount of profit the airline makes on each low-fat latte = $1.65

Let 'x' represent the number regular lattes made and let 'y' represent the number low-fat lattes made, we have;

x ≤ 30/0.5 = 60

x ≤ 60

y ≤ 60

y ≤ 60/1.25 = 48

∴ y ≤ 48

x + y ≤ 100

Therefore, the given that more profit is made from the sale of low-fat latte than for regular latte, the maximum number of low-fat latte should be made

Therefore, the number of low-fat latte made, y = 48

Therefore, the number of regular latte made, x ≤ 100 - 48 = 52

For maximum profit, the maximum number of low-fat latte, 'x', should be made

Therefore, x = 52

The ideal mix of regular and low-fat lattes to maximize profit is 52 cups of regular latte and 48 cups of low-fat latte

Answer:

Options A and C are correct answers

Step-by-step explanation:

A) [tex] \angle 5\: and \:\angle 1[/tex] are complementary.

C) [tex] \angle 3\: and \:\angle 4[/tex] form a linear pair.

Hope it helps you in your learning process.

no xdxdxdxdxdxxdxdxdxdxdxdxdxdxddxd

Answer:

-1 and -5

Step-by-step explanation:

Answer:

The equation of the line that has a slope of 3/2 and a y-intercept of -4 is:

y=3/2x-4.

Step-by-step explanation:

You want to find the equation for a line that has a slope of 3/2 and a y-intercept of -4.  First of all, remember what the equation of a line is:

y = mx+b  Where:  m is the slope, and  b is the y-intercept

All you really have to do here is  replace m with 3/2, which is the slope you gave, and  replace b with -4, the y-intercept you gave,  in the equation y=mx+b.

Question 7

Exponential functions have a variable degree.

Answer is 3^x + 6.

Question 8

Changing 15.3% to a decimal, we get 0.153.

Answer: Choice (4)

Answer:

$671.44

Step-by-step explanation:

$23.98 * 28 (people) = $671.44

Answer:

$671.44

Step-by-step explanation:

Answer:

the slope is 2/3 bcoz u do 2-0/3-0

Answer:

No solutions.

Step-by-step explanation:

No solutions because these two lines are parallel.

You can tell by the slopes. Both slopes are equal -3.

The -2 and [tex]\frac{5}{8}[/tex] vertically shift the lines.

Answer:

it’s just 2x^2

it’s not factorable

Answer:

The answer is 20/3

Step-by-step explanation:

(5 x 1 ) + ( 5 x 1/3)

5 + ( 5 × 1/3 )

5 + 5/3 = 20/3

Answer:

There are 479,001,600 arrangments of 12 distinct items.

Step-by-step explanation:

Arrangements of n items:

The number of possible arrangements of n itens is given by the factorial of n, that is:

[tex]A_{n} = n![/tex]

12 items:

[tex]A_{12} = 12! = 479001600[/tex]

There are 479,001,600 arrangments of 12 distinct items.

Answer: 14*x

Step-by-step explanation: I am 99% sure

Answer:

C

Step-by-step explanation:

Answer:

m = 5/9

Step-by-step explanation:

rise = 5

run = 9

m = rise/run

m = 5/9

Answer:

m = 5/9

Step-by-step explanation:

Use the slope formula to find the slope m.

[tex]m = \frac{5}{9}[/tex]

Answer:

Step-by-step explanation:

[tex]r_1(t) = < t^2 , 8t-15,t^2> \\ \\ r_2(t) = < 11t-30, \ t^2 , \ 13t-40> \\ \\ at \ r_1 = r_2 \\ \\ t^2 = 11t - 30 , \ \ \ \ \ \ \ \ \ \ \ \ 8t - 15 = t^2 , \ \ \ \ \ \ \ \ \ t^2 = 13t -40 \\ \\ t^2 - 11t + 30\ \ \ \ \ \ \ \ \ \ \ \ t^2 -8t + 15 , \ \ \ \ \ \ \ \ \ \ \ \ t^2 -13t + 40 \\ \\ t^2 -8t-3t + 30=0 \ \ \ \ \ \ t^2 - 2t -4t + 15=0 \ \ \ \ \ \ t^2 - 1t - 13t + 40 =0[/tex]

[tex](t-6)(t-5) \ \ \ \ \ \ \ \ \ \ \ (t-5)(t-3) \ \ \ \ \ \ \ \ \ \ \ (t-5)(t-8)[/tex]

[tex]\mathbf{The \ common \ value \ of \ t \ = \ 5}[/tex]