Ask question

Ask question

All categories

3 years ago

12

Match the set of figures with the postulate or theorem that can be used to prove the triangles congruent.

1 answer:

Scorpion4ik [409]3 years ago

6 0

It is ASA bc it shows the angle on left side in middle and angle on right :)

You might be interested in

F(x) = –22 + 10x - 6<br> Find f(-5)

Darina [25.2K]

Answer:

Step-by-step explanation:

-31

7 0

3 years ago

Read 2 more answers

!!PLEASE HELP!!

balandron [24]

Answer:

Step-by-step explanation:

second one

8 0

3 years ago

Which of the followiing units the Metric System Identify 5 units

tigry1 [53]

ANSWER:

MEASURE OF BASIC SI UNIT

Distance meter

volume(capacity) liter

mass(weight) gram

time. seconds

Temperature. Degree (Celsius)

5 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

FREE 11 POINTS<br>i will mark brainsit if i see crown

Murljashka [212]

Answer:

tank you

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers