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Oduvanchick [21]

3 years ago

10

The area of a rectangle whose base measures 4 cm is greater than 24 cm. Which graph represents all possible values for the heigh

t of the rectangle?
A number line going from 2 to 8. An open circle is at 6. Everything to the right of the circle is shaded.
A number line going from 2 to 8. An open circle is at 6. Everything to the left of the circle is shaded.
A number line going from 20 to 26. An open circle is at 24. Everything to the left of the circle is shaded.
A number line going from 20 to 26. An open circle is at 24. Everything to the right of the circle is shaded.

1 answer:

slamgirl [31]3 years ago

5 0

Answer:it’s A

Step-by-step explanation: I did it on edg

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Sveta_85 [38]

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3 years ago

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tensa zangetsu [6.8K]

M = 3
b = 8.25

you'll plug those numbers into this equation
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8 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

Please help i dont understand at all :(

andrey2020 [161]

180 - 46 - 24 = 110

d+ 80 = 110
d = 30

answer
d = 30

3 0

3 years ago