F(x) = -(x+1)^2+4 What is the vertex

Let y = 2. Solve for t: y = 3t - 4 A. B. c. D. -2 -2/3 2 3

Pavel [41]

Answer:

a - t = 2

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Someone please help!!! Can’t find this one anywhere!!! please and thank you!!!!!!!

Tema [17]

$\dfrac{ 2+ 2i}{5+4i}\\\\\\=\dfrac{(2+2i)(5-4i)}{(5+4i)(5-4i)}\\\\\\=\dfrac{10-8i + 10i -8i^2}{5^2 -(4i)^2}\\\\\\=\dfrac{10+2i +8}{25+16}~~~~;[i^2 =-1]\\\\\\=\dfrac{18+2i}{41}\\\\\\=\dfrac{18}{41} + \dfrac{2}{41}i$

4 0

2 years ago

A floor of a room is in the shape of square. It is to be covered with rectangular tiles that each have an area of 192 square inc

Grace [21]

Find area of room
aera=legnth wtimes width
in a square, legnth=width
aera=legnth^2
legnth=14 feet
convert o inches
14*12=168 inches
area=168^2=28224 square inches

how many tiles will cover all of floor?
aka
how many 192 in^2 is at least 28224 (at least, because we need to cover the whole floor without gaps)
192x>28224
divide both sides by 192
x>147

147 is the number of tiles we need

8 0

3 years ago

Read 2 more answers

Determine whether theses lines are parallel, perpendicular or neither

belka [17]

Answer:

Lines 2 and 3 are parallel with slope (-2/3) and different y intercepts, and both are perpendicular to line 1 (-1) / (3/2) = (-2/3)

Step-by-step explanation:

Parallel lines have the same slope

Perpendicular lines have negative reciprocal slopes

line 1: 6x - 4y = 2

line 1: 4y = 6x - 2

line 1: y = (3/2)x - 0.5

line 2: y = (-2/3)x - 6

line 3:3y = -2x + 4

line 3: y = (-2/3)x + 4/3

8 0

3 years ago

The graphs of the polar curves r = 4 and r = 3 + 2cosθ are shown in the figure above. The curves intersect at θ = π/3 and θ = 5π

Gennadij [26K]

(a)

$\displaystyle \frac{1}{2} \cdot \int_{\frac{\pi}{3}}^{\frac{5\pi}{3}} \left(4^2 - (3 + 2\cos\theta)^2 \right) \, d\theta$

or, via symmetry

$\displaystyle\frac{1}{2} \cdot 2 \int_{\frac{\pi}{3}}^{\pi} \left(4^2 - (3 + 2\cos\theta)^2 \right) \, d\theta$

____________

(b)

By the chain rule:

$\displaystyle \frac{dy}{dx} = \frac{ dy/ d\theta}{ dx/ d\theta}$

For polar coordinates, x = rcosθ and y = rsinθ. Since
r = 3 + 2cosθ, it follows that

$x = (3 + 2\cos\theta) \cos \theta \\ 
y = (3 + 2\cos\theta) \sin \theta$

Differentiating with respect to theta

$\begin{aligned}
\displaystyle \frac{dy}{dx} &= \frac{ dy/ d\theta}{ dx/ d\theta} \\
&= \frac{(3 + 2\cos\theta)(\cos\theta) + (-2\sin\theta)(\sin\theta)}{(3 + 2\cos\theta)(-\sin\theta) + (-2\sin\theta)(\cos\theta)} \\ \\
\left.\frac{dy}{dx}\right_{\theta = \frac{\pi}{2}}
&= 2/3
\end{aligned}$

2/3 is the slope

____________

(c)

"distance between the particle and the origin increases at a constant rate of 3 units per second" implies dr/dt = 3

Angle θ and r are related via r = 3 + 2cosθ, so implicitly differentiating with respect to time

 $\displaystyle\frac{dr}{dt} = -2\sin\theta \frac{d\theta}{dt} \quad \stackrel{\theta = \pi/3}{\implies} \quad 3 = -2\left( \frac{\sqrt{3}}{2}}\right) \frac{d\theta}{dt} \implies \\ \\ \frac{d\theta}{dt} = -\sqrt{3} \text{ radians per second}$

5 0

3 years ago