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2 years ago

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An insurance company uses the formula T=3.5n + 5 to work out the cost, T pounds, for insuring a customer for n days. Work out th

e cost for travel insurance for 4 days?

1 answer:

artcher [175]2 years ago

8 0

Answer:

Cost of travel insurance for 4 days = 19 pounds

Step-by-step explanation:

The insurance company uses the formula T = 3.5n + 5 to work out the cost , T pounds for insuring the customer for n number of days . Where n is the number of days.

The algebraic equation given has two variables which are cost T of insuring and n which is the number of days for insuring. Mathematically,

T = cost of insuring in pounds

n = number of days

Therefore, finding the cost T for 4 days

T = 3.5n + 5

T = 3.5 × 4 + 5

T = 14 + 5

T = 19 pounds

Cost of travel insurance for 4 days = 19 pounds

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Dea writes the expression 0.25 + 0.76 + 0.75 + 0.24. She rewrites the expression using the commutative property. Which expressio

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Answer:

B ) 0.25 + 0.75 + 0.76 + 0.24

Step-by-step explanation:

<u>commutative law of addition:-</u>

According to the commutative property of addition,changing the order of the numbers.we are adding doesn't change the sum.

let a,b,c+d are two numbers

The commutative property condition in addition a+b+c+d = a+c+b+d

Given expression 0.25+0.76+0.75+0.24 = 0.25 + 0.75 + 0.76 + 0.24

conclusion:- Answer B.

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2 years ago

The general form of the equation of a circle is x2 + y2 + 42x + 38y − 47 = 0. The equation of this circle in standard form is (x

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Step 1

we know that

The equation of a circle in standard form is equal to

$(x-h)^{2} +(y-k)^{2}=r^{2}$

where

(h,k) is the center of the circle

r is the radius of the circle

In this problem we have

$x^{2} +y^{2} +42x+38y-47=0$

Convert to standard form

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$(x^{2}+42x)+(y^{2}+38y)=47$

Complete the square twice. Remember to balance the equation by adding the same constants to each side.

$(x^{2}+42x+21^{2})+(y^{2}+38y+19^{2})=47+21^{2}+19^{2}$

$(x^{2}+42x+21^{2})+(y^{2}+38y+19^{2})=849$

Rewrite as perfect squares

$(x+21)^{2}+(y+19)^{2}=849$

The center of the circle is the point $(-21,-19)$

The radius of the circle is $\sqrt{849}\ units$

<u>The answer Part a) is</u>

The equation of the circle in standard form is equal to

$(x+21)^{2}+(y+19)^{2}=849$

<u>The answer Part b) is</u>

The center of the circle is the point $(-21,-19)$

<u>The answer Part c) is</u>

The radius of the circle is $\sqrt{849}\ units$

Let's verify each case to determine the solution of the second part of the problem

Step 2

we have

$x^{2} +y^{2} +60x+14y+98=0$

Convert to standard form

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$(x^{2}+60x)+(y^{2}+14y)=-98$

Complete the square twice. Remember to balance the equation by adding the same constants to each side.

$(x^{2}+60x+30^{2})+(y^{2}+14y+7^{2})=-98+30^{2}+7^{2}$

$(x^{2}+60x+30^{2})+(y^{2}+14y+7^{2})=851$

Rewrite as perfect squares

$(x+30)^{2}+(y+7)^{2}=851$

The radius of the circle is $\sqrt{851}\ units$

therefore

This circle does not have the same radius of the circle above

Step 3

we have

$x^{2} +y^{2} +44x-44y+117=0$

Convert to standard form

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$(x^{2}+44x)+(y^{2}-44y)=-117$

Complete the square twice. Remember to balance the equation by adding the same constants to each side.

$(x^{2}+44x+22^{2})+(y^{2}-44y+22^{2})=-117+22^{2}+22^{2}$

$(x^{2}+44x+22^{2})+(y^{2}-44y+22^{2})=851$

Rewrite as perfect squares

$(x+22)^{2}+(y-22)^{2}=851$

The radius of the circle is $\sqrt{851}\ units$

therefore

This circle does not have the same radius of the circle above

Step 4

we have

$x^{2} +y^{2} -38x+42y+74=0$

Convert to standard form

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$(x^{2}-38x)+(y^{2}+42y)=-74$

Complete the square twice. Remember to balance the equation by adding the same constants to each side.

$(x^{2}-38x+19^{2})+(y^{2}+42y+21^{2})=-74+19^{2}+21^{2}$

$(x^{2}-38x+19^{2})+(y^{2}+42y+21^{2})=728$

Rewrite as perfect squares

$(x-19)^{2}+(y+21)^{2}=728$

The radius of the circle is $\sqrt{728}\ units$

therefore

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Step 5

we have

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Convert to standard form

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$(x^{2}-50x)+(y^{2}-30y)=-1$

Complete the square twice. Remember to balance the equation by adding the same constants to each side.

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Rewrite as perfect squares

$(x-25)^{2}+(y-15)^{2}=849$

The radius of the circle is $\sqrt{849}\ units$

therefore

This circle has the same radius of the circle above

therefore

<u>The answer is</u>

$x^{2} +y^{2} -50x-30y+1=0$ -----> has the same radio that the circle above

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