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zepelin [54]
3 years ago
9

Help meeee with this questionnnnnnnnnn!!!!!!!!!!!!

Mathematics
1 answer:
Dafna1 [17]3 years ago
3 0

Given:

The scale factor between two circles is \dfrac{2x}{5y}.

To find:

The ratio of their areas.

Solution:

We know that, all circles are similar.

The ratio of the areas of similar figures is equal to the ratio of squares of their corresponding sides or equal to the square of ratio of their corresponding sides.

The scale factor is the ratio of the corresponding sides.

Ratio of areas of circles = Square of scale factor between two circles

\text{Ratio of areas of circles}=\left(\dfrac{2x}{5y}\right)^2

\text{Ratio of areas of circles}=\dfrac{(2x)^2}{(5y)^2}

\text{Ratio of areas of circles}=\dfrac{4x^2}{25y^2}

Therefore, the correct option is D.

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Answer:

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ABC has side lenths 20, 24, and 31. Do the side lenghts form a Pythagorean triple?
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No they don't.
Phythagorean triple is when the sum of the squares of two numbers is equal to the square of another number.

20^2 + 24^2 = 400 + 576 = 976
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20^2 + 24^2 should have been equal to 31^2 but that's not the case. So, the side's length don't form a triple.

Hope it helped!
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Calvin's rug covers 1/8 of the floor space in his bedroom how much floor space would be covered if Calvin had four rugs of that
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Write a polynomial of degree 5 with zero x=0,i square root 7, -2i
professor190 [17]

Answer:

P(x)=x^5+11x^3+28x

Step-by-step explanation:

<u>Roots of a polynomial</u>

If we know the roots of a polynomial, say x1,x2,x3,...,xn, we can construct the polynomial using the formula

P(x)=a(x-x_1)(x-x_2)(x-x_3)...(x-x_n)

Where a is an arbitrary constant.

We know three of the roots of the degree-5 polynomial are:

x_1=0;\ x_2=\sqrt{7}\boldsymbol{i}:\ x_3=-2\boldsymbol{i}

We can complete the two remaining roots by knowing the complex roots in a polynomial with real coefficients, always come paired with their conjugates. This means that the fourth and fifth roots are:

x_4=-\sqrt{7}\boldsymbol{i}:\ x_3=+2\boldsymbol{i}

Let's build up the polynomial, assuming a=1:

P(x)=(x-0)(x-\sqrt{7}\boldsymbol{i})(x+\sqrt{7}\boldsymbol{i})(x-2\boldsymbol{i})(x+2\boldsymbol{i})

Since:

(a+b\boldsymbol{i})\cdot (a-b\boldsymbol{i})=a^2+b^2

P(x)=(x)(x^2+7)(x^2+4)

Operating the last two factors:

P(x)=(x)(x^4+11x^2+28)

Operating, we have the required polynomial:

\boxed{P(x)=x^5+11x^3+28x}

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3 years ago
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