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3 years ago

EF is tangent to circle P at G (1,-3). If the slope of EF is 7/9, what is the slope of GP?

1 answer:

7 0

The center of the circle is at P.
The tangent EF is tangent to the circle at point G. So GP will be the radius of the circle.

The slope of tangent EF = 7/9

The tangent of a circle is always perpendicular to the radius at the point where it touches the circle. So we can say that EF is perpendicular to GP.

The product of slopes of two perpendicular line is always equal to -1.

Slope of EF x Slope of GP = -1
So,

Slope of GP = - 1 / (Slope of EF)

Thus slope of GP = - 9/7

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3 years ago

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3 years ago

The length of each side of a regular pentagon is increased by 8 inches, so the perimeter is now 65 inches. What is the original

evablogger [386]

Here as the Pentagon is regular so it's all sides will be of equal length . And if we assume It's each side be s , then it's perimeter is going to be (s+s+s+s+s) = 5s.And as here , each side is increased by 8 inches and then it's perimeter is 65 inches , so we got that it's side after increament is (s+8) inches and original length is s inches . And if it's each side is (s+8) inches , so it's perimeter will be 5(s+8) and as it's equal to 65 inches . So , 5(s+8) = 65

${:\implies \quad \sf 5(s+8)=65}$

${:\implies \quad \sf 5s+5\times 8=65}$

${:\implies \quad \sf 5s+40=65}$

${:\implies \quad \sf 5s=65-40}$

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2 years ago

Verify that 2,-1 and 1⁄2 are the zeroes of the cubic polynomial

Mariulka [41]

Answer:

i) Since P(2), P(-1) and P(½) gives 0, then it's true that 2,-1 and 1⁄2 are the zeroes of the cubic polynomial.

ii) - the sum of the zeros and the corresponding coefficients are the same

-the Sum of the products of roots where 2 are taken at the same time is same as the corresponding coefficient.

-the product of the zeros of the polynomial is same as the corresponding coefficient

Step-by-step explanation:

We are given the cubic polynomial;

p(x) = 2x³ - 3x² - 3x + 2

For us to verify that 2,-1 and 1⁄2 are the zeroes of the cubic polynomial, we will plug them into the equation and they must give a value of zero.

Thus;

P(2) = 2(2)³ - 3(2)² - 3(2) + 2 = 16 - 12 - 6 + 2 = 0

P(-1) = 2(-1)³ - 3(-1)² - 3(-1) + 2 = -2 - 3 + 3 + 2 = 0

P(½) = 2(½)³ - 3(½)² - 3(½) + 2 = ¼ - ¾ - 3/2 + 2 = -½ + ½ = 0

Since, P(2), P(-1) and P(½) gives 0,then it's true that 2,-1 and 1⁄2 are the zeroes of the cubic polynomial.

Now, let's verify the relationship between the zeros and the coefficients.

Let the zeros be as follows;

α = 2

β = -1

γ = ½

The coefficients are;

a = 2

b = -3

c = -3

d = 2

So, the relationships are;

α + β + γ = -b/a

αβ + βγ + γα = c/a

αβγ = -d/a

Thus,

First relationship α + β + γ = -b/a gives;

2 - 1 + ½ = -(-3/2)

1½ = 3/2

3/2 = 3/2

LHS = RHS; So, the sum of the zeros and the coefficients are the same

For the second relationship, αβ + βγ + γα = c/a it gives;

2(-1) + (-1)(½) + (½)(2) = -3/2

-2 - 1½ + 1 = -3/2

-1½ - 1½ = -3/2

-3/2 = - 3/2

LHS = RHS, so the Sum of the products of roots where 2 are taken at the same time is same as the coefficient

For the third relationship, αβγ = -d/a gives;

2 * -1 * ½ = -2/2

-1 = - 1

LHS = RHS, so the product of the zeros(roots) is same as the corresponding coefficient

7 0

3 years ago