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3 years ago

10

(8.9x108)

1 answer:

KatRina [158]3 years ago

8 0

Answer:

the correct answer is 3x104

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What is the following equal to 7 times 10^3

Masja [62]

Answer:

7000

Step-by-step explanation:

10^3=1000

1000*7=7000

3 0

4 years ago

Read 2 more answers

Tasya [4]

Answer:

That is an obtuse scalene triangle

Step-by-step explanation:

Scalene- has no congruent sides

Obtuse - has an angle that is greater than 90

we can use Pythagorean theorem to confirm this

The Pythagorean theorem states that in an obtuse triangle the hypotenuse squared is greater than the two legs squared added together

so basically $7^2>4^2+5^2\\7^2=49\\4^2=25\\5^2=25\\25+16=41\\49>41$

therefore it is an obtuse scalene

3 0

3 years ago

Read 2 more answers

82.50 +46.63 i need help

RSB [31]

129.13, I believe.

Hope this helps.

6 0

3 years ago

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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

Can u help me of this problem

icang [17]

I've been answering lots of questions and this one's hard

8 0

4 years ago