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2 years ago

The vertex of the function y=-3(x - 1)² +2 is?

1 answer:

7 0

Formula: f(x) = a(x – h)2 + k

transformation: 1 unit right 2 units up

vertex: (h,k) -> (1,2) it’s positive because it has to be the opposite

maximum/minimum: maximum

range: (-∞, ∞)
domain: y ≥2

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2. Which of the following ordered pairs (x,y) satisfies the system of equations below?

bixtya [17]

Answer:

b(2,-2)

Step-by-step explanation:

x= 2 and y=-2

let me know if you want further eplanation

3 0

3 years ago

If \triangle ABC△ABC is equilateral, solve for x. x=

Gwar [14]

Answer:

x = 13

Step-by-step explanation:

The tirangle ABC is given as equilateral. Every angle is therefore =. Not only that, each angle is 60 degrees.

If you need proof of that, remember that x + x + x = 180. Every triangle has 180 degrees.

if 3x = 180 then 1 x is 60

8x - 44 = 60 Add 44 to both sides.

8x = 60 + 44

8x = 104 Divide by 8

x = 104/8

x = 13

6 0

3 years ago

Read 2 more answers

For the right triangle shown, the lengths of two sides are given. Find the third side. Leave your answer in simplified, radical

Salsk061 [2.6K]

5^2 + 10^2 = 125
c^2 = 125
c = √125 = √(25•5) = = 5√5

7 0

3 years ago

Read 2 more answers

Consider the following region R and the vector field F. a. Compute the two-dimensional curl of the vector field. b. Evaluate bo

Shalnov [3]

Looks like we're given

$\vec F(x,y)=\langle-x,-y\rangle$

which in three dimensions could be expressed as

$\vec F(x,y)=\langle-x,-y,0\rangle$

and this has curl

$\mathrm{curl}\vec F=\langle0_y-(-y)_z,-(0_x-(-x)_z),(-y)_x-(-x)_y\rangle=\langle0,0,0\rangle$

which confirms the two-dimensional curl is 0.

It also looks like the region $R$ is the disk $x^2+y^2\le5$ . Green's theorem says the integral of $\vec F$ along the boundary of $R$ is equal to the integral of the two-dimensional curl of $\vec F$ over the interior of $R$ :

$\displaystyle\int_{\partial R}\vec F\cdot\mathrm d\vec r=\iint_R\mathrm{curl}\vec F\,\mathrm dA$

which we know to be 0, since the curl itself is 0. To verify this, we can parameterize the boundary of $R$ by

$\vec r(t)=\langle\sqrt5\cos t,\sqrt5\sin t\rangle\implies\vec r'(t)=\langle-\sqrt5\sin t,\sqrt5\cos t\rangle$

$\implies\mathrm d\vec r=\vec r'(t)\,\mathrm dt=\sqrt5\langle-\sin t,\cos t\rangle\,\mathrm dt$

with $0\le t\le2\pi$ . Then

$\displaystyle\int_{\partial R}\vec F\cdot\mathrm d\vec r=\int_0^{2\pi}\langle-\sqrt5\cos t,-\sqrt5\sin t\rangle\cdot\langle-\sqrt5\sin t,\sqrt5\cos t\rangle\,\mathrm dt$

$=\displaystyle5\int_0^{2\pi}(\sin t\cos t-\sin t\cos t)\,\mathrm dt=0$

7 0

3 years ago

How do you express your terms in pi

il63 [147K]

Answer:

To express your answer in terms of pi, simply refrain from substituting pi's numerical value for its symbol in the equation. That way, your answer will look like xπ where x is whatever number you come up with, and π is simply a placeholder.

4 0

3 years ago

Read 2 more answers