which system of equation does the graph represent

Answer:

 A) y = 3(x -3)^2 -46

 B) (3, -46)

 C) look at the y-coordinate of the vertex

Step-by-step explanation:

A) Factor the leading coefficient from the variable terms.

 y = 3(x^2 -6x) -19

Inside parentheses, add the square of half the x-coefficient. Outside, subtract the same value.

 y = 3(x^2 -6x +9) -19 -3(9)

 y = 3(x -3)^2 -46

__

B) Compared to the vertex form, ...

 y = a(x -h)^2 +k

we find a=3, (h, k) = (3, -46).

The vertex is (3, -46).

__

C) The vertex is an extreme value (as is any vertex). The sign of the leading coefficient tells you whether the parabola opens upward (+) or downward (-). This parabola opens upward, so the vertex is a minimum.

If the leading coefficient is positive, the y-coordinate of the vertex is a minimum. If the leading coefficient is negative, the y-coordinate of the vertex is a maximum.

Step-by-step explanation:

Answer:

im not low on points but thank you <3.

Step-by-step explanation:

Answer:

:) :) :) :) :) :) :) :) :) :)

Answer:

Ramona spent 110 minutes on homework

Step-by-step explanation:

You want to add up all the minutes spent on homework altogether.

20 +20 = 40

40 + 40 = 80

80 + 30 = 110

110 + 0 = 110

So, Ramona spent a total of 110 minutes on homework. Hope this helps!

Answer:the first option

Step-by-step explanation:

The best way to do this is to convert the mixed numbers into improper fractions. Improper fractions are when the numerator is greater than the denominator. An example of an improper fraction would be 5/4. Since 4/4 is one whole that means 5/4 as a mixed number would be 1 1/4. I'll use the expression 3 1/4 + 5 2/9 as an example.

3 1/4 is 3 wholes and 1/4 left over. We know that 4/4 is 1 whole so to find 3 holes we multiply the 1 by 3. 4 x 3 =12 so there are 12/4 in 3 holes. We then add the leftover 1/4 to get 13/4 as our first number.

For the second number, we have 5 2/9. This one is a bit harder. We know that 9/9 is in one whole, but how do we find how many are in 5 wholes? We can use multiplication. 1x 5=5. 5 x 9 =45. That means we have 45/9 in 5 holes. Add the leftover 2/9 to have 47/9 as our 2nd number.

Now that we have these two improper fractions, what do we do with them? The two numbers we are working with are 13/4 and 47/9. We need to find the common denominator. The common denominator is when both fractions have the same bottom number. If two fractions have different denominators we cannot add them.

To make the denominators the same, let's find a common multiple. A common multiple is a number that is a product of all chosen numbers. This is kind of confusing but another way to think of a common multiple is a number that is divisible by all chosen numbers. Take the two denominators and multiply them. 9x4=36. 36 is a common multiple of both 9 and 4 because both of them can go into 36 without being a decimal or a fraction.

To convert the fractions we multiply again. Whatever you do to one side you must always do to the other. In 13/4 we multiplied 4x9 to get 36. So now we multiply the numerator 13x9 to get 117. That means we have 117/36. We multiplied 9x4 to also get 36. So our numerator for our second number is 47x4=188 or 188/36.

Now we have 117/36 and 188/36. We add them together like normal and get 305/36. In some cases you'd need to simplify it, but for the example equation I used this is already the fraction's lowest form. I hope this helped! Comment if you need another example or have any questions! :)

[tex]sin^2(\theta)+cos^2(\theta)=1\qquad \qquad sin(2\theta)=2sin(\theta)cos(\theta) \\\\[-0.35em] \rule{34em}{0.25pt}[/tex]

[tex]tan(x)+cot(x)=\cfrac{2}{sin(2x)} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{doing the left-hand-side}}{\cfrac{sin(x)}{cos(x)}+\cfrac{cos(x)}{sin(x)}\implies \cfrac{sin^2(x)+cos^2(x)}{\underset{\textit{using this LCD}}{sin(x)cos(x)}}} \implies \cfrac{1}{sin(x)cos(x)}[/tex]

now, let's recall that anything times 1 is just itself, namely 5*1 =5, 1,000,000 * 1 = 1,000,000, "meow" * 1 = "meow" and so on, so we can write anything as time 1.

let's recall something else, that same/same = 1, so

[tex]\cfrac{cheese}{cheese}\implies \cfrac{spaghetti}{spaghetti}\implies \cfrac{horse}{horse}\implies \cfrac{butter}{butter}\implies \cfrac{25^7}{25^7}=1[/tex]

therefore

[tex]\cfrac{1}{sin(x)cos(x)}\cdot \cfrac{2}{2}\implies \cfrac{2}{2sin(x)cos(x)}\implies \cfrac{2}{sin(2x)}[/tex]

Answer: It is C

Step-by-step explanation:

Answer:

the slope of the line is -2.

Step-by-step explanation:

Answer:

559

Step-by-step explanation:

[tex]\dfrac{\sin(x+y)}{\cos x \cos y}\\\\\\=\dfrac{\sin x \cos y + \sin y \cos x}{\cos x \cos y}\\\\\\=\dfrac{\tfrac{\sin x \cos y}{\cos x \cos y} + \tfrac{\sin y \cos x}{\cos x \cos y}}{\tfrac{\cos x \cos y}{\cos x \cos y}}\\\\\\=\dfrac{\tan x + \tan y}1\\\\\\= \tan x + \tan y[/tex]

Answer:

The answer is 20,880

Step-by-step explanation:

This won't be a very long explanation. What I did was add 18,000 plus 16% and I got 20,880.

Hope this helps!  

Have a nice day/night/afternoon.  

(Can you mark me brainliest?)

- ❤ 7272033Alt ❤

There's something very off about this question.

In spherical coordinates,

x² + y² + z² = ρ²

so that

f(x, y, z) = 6 - (x² + y² + z²)

transforms to

g(ρ, θ, φ) = 6 - ρ²

When transforming to spherical coordinates, we also introduce the Jacobian determinant, so that

dV = dx dy dz = ρ² sin(φ) dρ dθ dφ

Since we integrate over a sphere with radius 4, the domain of integration is the set

E = {(ρ, θ, φ) : 0 ≤ ρ ≤ 4 and 0 ≤ θ ≤ 2π and 0 ≤ φ ≤ π}

so that the integral is

[tex]\displaystyle \int_{\phi=0}^{\phi=\pi} \int_{\theta=0}^{\theta=2\pi} \int_{\rho=0}^{\rho=4} (6 - \rho^2) \rho^2 \sin(\phi) \, d\rho \, d\theta \, d\phi[/tex]

Computing the integral is simple enough.

[tex]\displaystyle = \int_{\phi=0}^{\phi=\pi} \int_{\theta=0}^{\theta=2\pi} \int_{\rho=0}^{\rho=4} (6 \rho^2 - \rho^4) \sin(\phi) \, d\rho \, d\theta \, d\phi[/tex]

[tex]\displaystyle = 2\pi \int_{\phi=0}^{\phi=\pi} \int_{\rho=0}^{\rho=4} (6 \rho^2 - \rho^4) \sin(\phi) \, d\rho \, d\phi[/tex]

[tex]\displaystyle = 2\pi \left(\int_{\phi=0}^{\phi=\pi} \sin(\phi) \, d\phi\right) \left(\int_{\rho=0}^{\rho=4} (6 \rho^2 - \rho^4) \, d\rho\right)[/tex]

[tex]\displaystyle = 2\pi \cdot 2 \cdot \left(-\frac{384}5\right) = \boxed{-\frac{1536\pi}5}[/tex]

but the mass can't be negative...

Chances are good that this question was recycled without carefully changing all the parameters. Going through the same steps as above, the mass of a spherical body with radius R and mass density given by

[tex]\delta(x, y, z) = k - (x^2 + y^2 + z^2)[/tex]

for some positive number k is

[tex]\dfrac{4\pi r^3}{15} \left(5k - 3r^2\right)[/tex]

so in order for the mass to be positive, we must have

5k - 3r² ≥ 0   ⇒   k ≥ 3r²/5

In this case, k = 6 and r = 4, but 3•4²/5 = 9.6.

Answer:

Answer:

R(x)=2x.

Step-by-step explanation:

We get the total revenue by multiplying the revenue per unit times the numbers of units sold, so the general Revenue function is given by

R(x)=selling price per unit⋅x.

The manufacturer sells each energy drink for $2 so the Revenue function is :R(x)=2x.

Answer:

10.

Step-by-step explanation:

X is often associated with the value of "independent variable"

y=20-2(5)

y=20-10

y=10

Answer:

25

Step-by-step explanation:

x+7 = 32 (similar triangles)

x = 25

Step-by-step explanation:

the answer as shown in the photo

Answer:

$0.10

Step-by-step explanation:

Let X = total cost of game

          (4/5) X = $25

Multiply both sides of equation by 5/3

   (5/3)(3/5) X = (5/3) $25

                   X = $35

Answer:

solution given :

Since opposite side and side of a congruent triangle are equal.

Now

the statement which is true is:

<HDG=~ <HFE

Answer: 3y = 9x -12

Step-by-step explanation:

Answer:

A system of two equations can be classified as follows: If the slopes are the same but the y-intercepts are different, the system has no solution. If the slopes are different, the system has one solution. If the slopes are the same and the y-intercepts are the same, the system has infinitely many solutions.

Mark me as brainlist plz :)

Answer:

c

Step-by-step explanation:

Keep in mind, whatever is inside of the square root, stays inside of the square root, and whatever is outside, stays outside, they don't multiply together. With parenthesis, exponents multiply, if there's no parenthesis, the exponents don't multiply. In this case, the exponents add.

Multiply each one together:

sqrt(6x^2) x sqrt(9x) = sqrt(54x^3)

sqrt(6x^2) x -(x sqrt(5x^5)) = -x sqrt(30x^7)

4 sqrt(8x^3) x sqrt(9x) = 4 sqrt(72x^4)

4 sqrt(8x^3) x -(x sqrt(5x^5)) = -4x sqrt(40x^8)

Then simplify:

sqrt(54x^2) = 3x sqrt(6x)

x sqrt(30x^7) = -x^4 sqrt(30x)

4 sqrt(72x^4) = 24x^2 sqrt(2)

-4x sqrt(40x^8) = -8x^5 sqrt(10)

Answer:

This a vector addition problem and you need to know the law of cosines and law of sines to solve this problem.

Two vector forces of 80lbs and 70 lbs act on an object at 40 degree angle between them.

The magnitude of the resultant force can be found by applying the law of cosines:

c2 = a2 + b2 - 2abcos140°

where a = 80, b = 70 and c is the resultant force vector you are asked to find

Once you find the magnitude of the resultant force, then you can find the angle it makes wrt the 70lb force using the law of sines:

sinα/a = sin140°/c

where α is the angle opposite side a, or the angle between the resultant and the 70lb vector that you are asked to find. so it would be 140lbs

Step-by-step explanation

Answer:

Goofy unlimited lighting 12 pack of light bulbs

Step-by-step explanation:

12 = 8.40/12 = .70 cents

19 = 15.35 = .81 cents

Answer:

6 times or 6.4

Answer:

1003

Step-by-step explanation:

The problem is a classic example of a telescoping series of products, a series in which each term is represented in a certain form such that the multiplication of most of the terms results in a massive cancelation of subsequent terms within the numerators and denominators of the series.

The simplest form of a telescoping product[tex]a_{k} \ = \ \displaystyle\frac{t_{k}}{t_{k+1}}[/tex], in which the products of n terms is

[tex]a_{1} \ \times \ a_{2} \ \times \ a_{3} \ \times \ \cdots \times \ a_{n-1} \ \times \ a_{n} \ = \ \displaystyle\frac{t_{1}}{t_{2}} \ \times \ \displaystyle\frac{t_{2}}{t_{3}} \ \times \ \displaystyle\frac{t_{3}}{t_{4}} \ \times \ \cdots \ \times \ \displaystyle\frac{t_{n-1}}{t_{n}} \ \times \ \displaystyle\frac{t_{n}}{t_{n+1}} \\ \\ \-\hspace{5.55cm} = \ \displaystyle\frac{t_{1}}{t_{n+1}}.[/tex].

In this particular case, [tex]t_{1} \ = \ 2[/tex] , [tex]t_{2} \ = \ 3[/tex], [tex]t_{3} \ = \ 4[/tex], ..... , in which each term follows a recursive formula of [tex]t_{n+1} \ = \ t_{n} \ + \ 1[/tex]. Therefore,

[tex]\displaystyle\frac{t_{2}}{t_{1}} \times \displaystyle\frac{t_{3}}{t_{2}} \times \displaystyle\frac{t_{4}}{t_{3}} \times \cdots \times \displaystyle\frac{t_{n}}{t_{n-1}} \times \displaystyle\frac{t_{n+1}}{t_{n}} \ = \ \displaystyle\frac{3}{2} \times \displaystyle\frac{4}{3} \times \displaystyle\frac{5}{4} \times \cdots \times \displaystyle\frac{2005}{2004} \times \displaystyle\frac{2006}{2005} \\ \\ \-\hspace{5.95cm} = \ \displaystyle\frac{2006}{2} \\ \\ \-\hspace{5.95cm} = 1003[/tex]

Answer:

27

Step-by-step explanation:

compitative property

Answer:

Area = 4x^2 + 12x + 9

Perimeter = 8x + 12

Step-by-step explanation: