Write standered form equation for the circle center (-7,4)

154=−4(8+6r) + 24r thanks to anyone who helps

vova2212 [387]

Answer:

When you do the math, you get an incorrect equation.

Step-by-step explanation:

154 = -32 -24r +24r

154 = -32

so you would get 186 = 0

4 0

2 years ago

An airliner carries 50 passengers and has doors with a height of 70 in. Heights of men are normally distributed with a mean of 6

Korolek [52]

Answer:

dfdfdfdfdfd

Step-by-step explanation:

3 0

3 years ago

Graph the following piecewise functions for the specified domain.

Anni [7]

Answer:

Answer is given in the pictures

Step-by-step explanation:

Refer the figure given to see the plots

7 0

3 years ago

Read 2 more answers

HELP???? PLEASE AND FAST IM ON THE BRINK OF DEATH --

aleksandrvk [35]

Answer:

See below.

Step-by-step explanation:

Create a system of equations to represent this scenario.

Lin: 12 - 1/3x = y
Diego: 20 - 2/3x = y

1) A graph of these equations is attached below. Lin is in red; Diego is in blue.

2) The time (seconds) is on the x-axis, while the milkshake (oz) is on the y-axis. The graph shows the rate of change that the volume of the milkshake is decreasing for both Lin and Diego. The intersection point tells us at what time t (s) Lin and Diego have the same amount of milkshake left.

There is only one solution to this system of equations: (24, 4). This tells us that at t = 24 s, Lin and Diego both have 4 oz of milkshake left.

The zeros, aka where the graph touches the x-axis, tell us at what time Lin and Diego finish their milkshakes.

Lin finishes her milkshake later than Diego, at t = 36 s (36, 0), while Diego finishes his milkshake at t = 30 s (30, 0).

3 0

3 years ago

Find the absolute maximum and absolute minimum values of f on the given interval. f(t) = 2 cos(t) + sin(2t),[0, π/2].

castortr0y [4]

Answer:

The absolute maximum is $\frac{3\sqrt 3}2$ and the absolute minimum value is $0.$

Step-by-step explanation:

Differentiate of $f$ both sides w.r.t. $t$ ,

$f(t)=2 \cos t+\sin 2t$

$\Rightarrow f'(t)=-2\sin t+2\cos 2t$

Now take $f'(t)=0$

$\Rightarrow -2\sin t+2\cos 2t=0$

$\Rightarrow 2\cos 2t=2\sin t$

$\Rightarrow \cos 2t=\sin t$

$\Rightarrow 1-2\sin ^2t =\sin t \quad \quad [\because \cos 2t = 1-2\sin ^2t]$

$\Rightarrow 2\sin ^2t+\sin t-1=0$

$\Rightarrow 2\sin ^2t+2\sin t-\sin t-1=0$

$\Rightarrow 2\sin t(\sin t+1)-1(\sin t+1)=0$

$\Rightarrow (\sin t+1)(2\sin t-1)=0$

$\Rightarrow \sin t+1=0 \;\text{and}\; 2\sin t-1=0$

$\Rightarrow \sin t =-1 \;\text{and}\; \sin t =\frac 12$

In the interval $0\leq t\leq \frac {\pi}2$ , the answer to this problem is $\frac {\pi}6$

Now find the second derivative of $f(t)$ w.r.t. $t$ ,

$f''(t)=-2\cos t-4\sin 2t$

$\Rightarrow \left[f''(t)\right]_{t=\frac {\pi}6}=-2\times \frac {\sqrt 3}2-4\times \frac{\sqrt 3}2=-3\sqrt 3$

Thus, $f(t)$ is maximum at $t=\frac {\pi}6$ and minimum at $t=0$

$\left[f(t)\right]_{t=\frac {\pi}6}=2\times \frac {\sqrt 3}2+\frac{\sqrt 3}2=\frac{3\sqrt 3}2\;\text{and}\;\left[f(t)\right]_{t=\frac{\pi}2}= 2\times 0+0=0$

Hence, the absolute maximum is $\frac{3\sqrt 3}2$ and the absolute minimum value is $0$ .

7 0

3 years ago