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vesna_86 [32]
4 years ago
12

When shopping for life insurance, you should look for: a. A company with a low premium rate and a good rating. b. A company that

gives you benefits, like rewards points. c. A company with an agent that you like and trust. d. A company that has many policies to choose from.
Mathematics
2 answers:
NemiM [27]4 years ago
8 0

The correct option is : A company with a low premium rate and a good rating.

Explanation:

When shopping for life insurance, you should look for: A company with a low premium rate and a good rating.

Life insurance is an insurance which covers a person in case of death. It pays the amount to the family of the deceased. This policy also pays after a certain period of time when the policy matures. So, while choosing such insurances, one looks for a company with good client ratings and low premium rates.

ra1l [238]4 years ago
3 0

Answer:

its actually a

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Which of the following is the measure of
jonny [76]

Answer:

The answer to your question is 31.4°

Step-by-step explanation:

Data

SR = 54 cm

SQ = 99 cm

RQ = 141 cm

∠QRS = ?

Process

1.- Use the law of cosines to find ∠QRS

          cos QRS = (SQ² - RS² - RQ²) / -2(SR)(RQ)

-Substitution

          cos QRS = (99² - 54² - 141²) / -2(54)(141)

-Simplification

          cos QRS = (9801 - 2916 - 19881) / -15228

          cos QRS = -12996 / -15228

          cos QRS = 0.85343

2.- Find QRS

            ∠QRS = 31.4°

4 0
3 years ago
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The quotient of 25 and 5 increased by3
Pavlova-9 [17]

Answer:

8, the answer is 8

5 0
3 years ago
I'm in Algebra 1 and I need help with this, (show all working as well please)
mafiozo [28]

Answer:

Step-by-step explanation:

All you do is  draw a line straight across 5x and then graph 1 which is right next to 5.

6 0
3 years ago
Consider the system of differential equations dx/dt=−2y dy/dt=−2x. . Convert this system to a second order differential equation
Musya8 [376]

Answer:

Step-by-step explanation:

we have the following differential equations

\frac{dx}{dt}=-2y\\\frac{dy}{dt}=-2x\\

by differentiating the second equation we have

\frac{d}{dt}(\frac{dy}{dt})=-2\frac{dx}{dt}\\\frac{d^{2}y}{dt^{2}}=-2\frac{dx}{dt}\\\frac{dx}{dt}=\frac{-1}{2}\frac{d^{2}y}{dt^{2}}

and we replace dx/dt in the first equation

\frac{-1}{2}\frac{d^{2}y}{dt^{2}}=-2y\\\frac{d^{2}y}{dt^{2}}-4y=0

and by using the characteristic polynomial

m^{2}+4=0\\m=\±2i

the solution is

y(t)=Acos(2t)+Bsin(2t)

and to compute x(t) we have

\frac{dx}{dt}=-2Acos(2t)-2Bsin(2t)\\\\\int dx = \int[-2Acos(2t)-2Bsin(2t)]dt\\\\x(t)=-Asin(2t)+Bcos(2t)

and if we use x(0)=4 and y(0)=3, we can calculate the constants A and B

x(0)=B=4\\y(0)=A=3

I hope this is useful for you

regards

4 0
3 years ago
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Show work step by step
beks73 [17]

Answer:

dont know srry

Step-by-step explanation:

5 0
3 years ago
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