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4 years ago

15

Luis earns a base salary of \$150.00$150.00 every week with an additional 14\%14% commission on everything he sells. Luis sold \

$6050.00$6050.00 worth of items last week. What was Luis's total pay last week?

1 answer:

tester [92]4 years ago

4 0

Answer: Luis earned $997.00 for last week.

$6,050.00 - 14%= $5,203.00

Just figure the difference ($847)

Then add his base salary.

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Solve for p.<br><br> 12p − 9p − –7p − 8 = –18

Zielflug [23.3K]

Answer:

p=-1

Step-by-step explanation:

12p - 9p --7p -8 =-18

19p-9p-8=-18

10p - 8 = -18

10p = -10

p=-1

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3 years ago

70°<br> x<br> 82°<br> r<br> r<br> 6 -<br> - 4)º<br> an<br> 160°<br> (2y - 40<br> HELP ASAP

natulia [17]

Answer:

Step-by-step explanation:

x + 70 = 180 {Linear pair}

x = 180 - 70

x = 110

y + 82 = 180 {linear pair}

y = 180 -82

y = 98

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2 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

Which of the following statements about a plane is not true?

shusha [124]

Answer:

The last one is the answer.

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4 years ago

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pogonyaev

C - because root 80 squared is 80
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3 years ago