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Natali5045456 [20]
3 years ago
6

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

Mathematics
1 answer:
yuradex [85]3 years ago
8 0

Answer:

Step-by-step explanation:

Gggg

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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:

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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
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