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4 years ago

PLS HELP ME WITH THIS TY!

Mathematics

2 answers:

Inessa05 [86]4 years ago

8 0

Answer:

B.) 450

Step-by-step explanation:

I hope I could help

sammy [17]4 years ago

8 0

Answer:

450cm^2

Step-by-step explanation:

There are 4 triangles with 2 of each.

Triangle = $\frac{1}{2}bh$

Triangle 1 = $\frac{1}{2}(10)(15) = \frac{1}{2}(150) = \frac{150}{2} =75 *2=150$

Triangle 2 = $\frac{1}{2} (15)(20)=\frac{1}{2}(300)=\frac{300}{2} = 150*2=300$

300 + 150 = 450

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Step-by-step explanation:

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3 years ago

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3 years ago

-3,6,-9,12,-15 which of the following represents the general term for the sequence give?

Mademuasel [1]

The general term for the sequence $a_{n}=(-1)^{n} 3 n$

<h3><u>Solution:</u></h3>

Given sequence is -3, 6, -9, 12, -15

We have to find the general term of sequence

The terms in the sequence are found out by using the recursive definition:

$a_{n}=(-1)^{n} 3 n$

Let us use this definition to find out the terms and check if it matches our given sequence

$\begin{array}{l}{\text { For } n=1:-(-1)^{n} 3 n=(-1)^{1} \times 3 \times 1=-3} \\\\ {\text { For } n=2:-(-1)^{n} 3 n=(-1)^{2} \times 3 \times 2=1 \times 6=6} \\\\ {\text { For } n=3:-(-1)^{n} 3 n=(-1)^{3} \times 3 \times 3=-1 \times 9=-9} \\\\ {\text { For } n=4:-(-1)^{n} 3 n=(-1)^{4} \times 3 \times 4=1 \times 12=12} \\\\ {\text { For } n=5:-(-1)^{n} 3 n=(-1)^{5} \times 3 \times 5=-1 \times 15=-15}\end{array}$

Thus the general term is given by $a_{n}=(-1)^{n} 3 n$

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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 $0$ .

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3 years ago