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3 years ago

6

Which of the following plans include a savings plan as a part of the insurance? Whole Life Universal Life Variable Universal Lif

e All of the above

1 answer:

worty [1.4K]3 years ago

6 0

Answer:

Which of the following plans include a savings plan as a part of the insurance?

Whole Life

Universal Life

<u>Variable Universal Life</u>

All of the above

Step-by-step explanation:

You might be interested in

which of the following is equivalent to 3 sqrt 32x^3y^6 / 3 sqrt 2x^9y^2 where x is greater than or equal to 0 and y is greater

Nutka1998 [239]

Answer:

$\frac{\sqrt[3]{16y^4}}{x^2}$

Step-by-step explanation:

The options are missing; However, I'll simplify the given expression.

Given

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$

Required

Write Equivalent Expression

To solve this expression, we'll make use of laws of indices throughout.

From laws of indices $\sqrt[n]{a} = a^{\frac{1}{n}}$

So,

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$ gives

$\frac{(32x^3y^6)^{\frac{1}{3}}}{(2x^9y^2)^\frac{1}{3}}$

Also from laws of indices

$(ab)^n = a^nb^n$

So, the above expression can be further simplified to

$\frac{(32^\frac{1}{3}x^{3*\frac{1}{3}}y^{6*\frac{1}{3}})}{(2^\frac{1}{3}x^{9*\frac{1}{3}}y^{2*\frac{1}{3}})}$

Multiply the exponents gives

$\frac{(32^\frac{1}{3}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

Substitute $2^5$ for 32

$\frac{(2^{5*\frac{1}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

$\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

From laws of indices

$\frac{a^m}{a^n} = a^{m-n}$

This law can be applied to the expression above;

$\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$ becomes

$2^{\frac{5}{3}-\frac{1}{3}}x^{1-3}*y^{2-\frac{2}{3}}$

Solve exponents

$2^{\frac{5-1}{3}}*x^{-2}*y^{\frac{6-2}{3}}$

$2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}$

From laws of indices,

$a^{-n} = \frac{1}{a^n}$ ; So,

$2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}$ gives

$\frac{2^{\frac{4}{3}}*y^{\frac{4}{3}}}{x^2}$

The expression at the numerator can be combined to give

$\frac{(2y)^{\frac{4}{3}}}{x^2}$

Lastly, From laws of indices,

$a^{\frac{m}{n} = \sqrt[n]{a^m}$ ; So,

$\frac{(2y)^{\frac{4}{3}}}{x^2}$ becomes

$\frac{\sqrt[3]{(2y)}^{4}}{x^2}$

$\frac{\sqrt[3]{16y^4}}{x^2}$

Hence,

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$ is equivalent to $\frac{\sqrt[3]{16y^4}}{x^2}$

8 0

3 years ago

Determine the vertical asymptotes for the reciprocal of this function f(x)=x^2 + 3x -28

Gennadij [26K]

Answer:

x = -7 and x = 4.

Step-by-step explanation:

f(x) = x^2 + 3x - 28

= (x + 7)(x - 4).

The reciprocal of f(x) can be written as

1 / (x + 7)(x - 4).

When the denominator is zero we have vertical asymptotes

so they are x = -7 and x = 4.

7 0

1 year ago

If you place a 17 foot ladder against the top of a 8 foot building how many feet will the bottom of the ladder be from the botto

KatRina [158]

9 because if you subtract 17 from 8 you will get 9 I hope this works :)

7 0

3 years ago

Read 2 more answers

Can somebody help me please

inna [77]

Answer:

104 degrees because 7 and 8 are congruent meaning 8 is also 104

Step-by-step explanation:

6 0

3 years ago

Find distance (4,7) (-4,-3)

AURORKA [14]

∆x = (4--4)=8

∆y = (7--3)= 10

Step-by-step explanation:

distance is given by;

d² = (∆y)²+ (∆x)²

d = √((∆y)²+ (∆x)²)

d = √((4+4)² + (7+3)²)

d = √((8²) + (10²))

d = √(64+100)

d = √164

<em><u>d =12.806 units</u></em>

7 0

2 years ago