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2 years ago

Simon Simms takes out $20,000 of a twenty-payment life when he is twenty-five years old. a. annual premium b. quarterly premium

Mathematics

1 answer:

OLga [1]2 years ago

6 0

B. Quarterly premium

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Help!<br><br> Please look at the pic and help me!

inna [77]

Answer:

Step-by-step explanation:

equilateral triangle has all sides So,

perimeter = a+b+c= 12+12+12=36

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2 years ago

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I need help...I don't understand

Andrej [43]

Keep change change [keep the first integer change the opperation and change the integer on your number or fraction ... Rember the answer takes the sign of the larger number .

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3 years ago

Find 15 percent of 15 percent of 15 percent of 500 how does this compare to finding 45 percent of 500?

4vir4ik [10]

Completely different...

.15*.15*.15*500

500(.15^3)=1.6875

500(.45)=225

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2 years ago

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In the triangle pictured, let A, B, C be the angles at the three vertices, and let a,b,c be the sides opposite those angles. Acc

Troyanec [42]

Answer:

Step-by-step explanation:

(a)

Consider the following:

$A=\frac{\pi}{4}=45°\\\\B=\frac{\pi}{3}=60°$

Use sine rule,

$\frac{b}{a}=\frac{\sinB}{\sin A}
\\\\=\frac{\sin{\frac{\pi}{3}}
}{\sin{\frac{\pi}{4}}}\\\\=\frac{[\frac{\sqrt{3}}{2}]}{\frac{1}{\sqrt{2}}}\\\\=\frac{\sqrt{2}}{2}\times \frac{\sqrt{2}}{1}=\sqrt{\frac{3}{2}}$

Again consider,

$\frac{b}{a}=\frac{\sin{B}}{\sin{A}}
\\\\\sin{B}=\frac{b}{a}\times \sin{A}\\\\\sin{B}=\sqrt{\frac{3}{2}}\sin {A}\\\\B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{A}]$

Thus, the angle B is function of A is, $B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{A}]$

Now find $\frac{dB}{dA}$

Differentiate implicitly the function $\sin{B}=\sqrt{\frac{3}{2}}\sin{A}$ with respect to A to get,

$\cos {B}.\frac{dB}{dA}=\sqrt{\frac{3}{2}}\cos A\\\\\frac{dB}{dA}=\sqrt{\frac{3}{2}}.\frac{\cos A}{\cos B}$

When $A=\frac{\pi}{4},B=\frac{\pi}{3}$ , the value of $\frac{dB}{dA}$ is,

$\frac{dB}{dA}=\sqrt{\frac{3}{2}}.\frac{\cos {\frac{\pi}{4}}}{\cos {\frac{\pi}{3}}}\\\\=\sqrt{\frac{3}{2}}.\frac{\frac{1}{\sqrt{2}}}{\frac{1}{2}}\\\\=\sqrt{3}$

In general, the linear approximation at x= a is,

$f(x)=f'(x).(x-a)+f(a)$

Here the function $f(A)=B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{A}]$

At $A=\frac{\pi}{4}$

$f(\frac{\pi}{4})=B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{\frac{\pi}{4}}]\\\\=\sin^{-1}[\sqrt{\frac{3}{2}}.\frac{1}{\sqrt{2}}]\\\\\=\sin^{-1}(\frac{\sqrt{2}}{2})\\\\=\frac{\pi}{3}$

And,

$f'(A)=\frac{dB}{dA}=\sqrt{3}$ from part b

Therefore, the linear approximation at $A=\frac{\pi}{4}$ is,

$f(x)=f'(A).(x-A)+f(A)\\\\=f'(\frac{\pi}{4}).(x-\frac{\pi}{4})+f(\frac{\pi}{4})\\\\=\sqrt{3}.[x-\frac{\pi}{4}]+\frac{\pi}{3}$

Use part (c), when $A=46°$ , B is approximately,

$B=f(46°)=\sqrt{3}[46°-\frac{\pi}{4}]+\frac{\pi}{3}\\\\=\sqrt{3}(1°)+\frac{\pi}{3}\\\\=61.732°$

8 0

3 years ago

Lisa's test grades are 79, 89, and 90. there will be one more test this year. if lisa wants her test average to be at least 88,

erik [133]

For this case, what we are going to do first is to assume that all the exams are worth the same percentage of the final grade.
We have then that Lisa's average grade point equation is:
$88 = \frac{79 + 89 + 90 + x}{4}$
Where,
x: minimum note that lisa must obtain in the last exam.
Clearing x we have:
$x = (88 * 4) - (79 + 89 + 90)

x = 94$
Answer:
the lowest grade she can get on her last test is:
x = 94

5 0

3 years ago