# 16) AB = 3.2cm

BC = 8.4cm

The area of the triangle ABC is 10cm².

Calculate the perimeter and give the answer correct to 3 significant figures.

19) Bill said that the line y=6 cuts the curve x² + y² =25 at two points.

a) By eliminating y, show that Bill is incorrect.

b) By eliminating y, find the solutions to the simultaneous equations

x² + y² = 25

y = 2x - 2

26)

c) Factorise (p+q)² + 5(p+q)

e) Simplify 2t² * 3r³t4

30) It is always possible to draw a circle which passes through all four vertices of a rectangle. Explain why.

(26) (c) factor out the (p+q) term and you get (p+q)(p+q+5)

(d) factor out the t term. I assume the last 4 is supposed to be an exponent.

(30) Draw the intersecting diagonals of the rectangle. Use geometry axioms and theorems to prove that the distance from the intersection of the diagonals to the corners are equal for all 4 corners. Therefore a circle centered on the intersection passes through all corners.

16) AB = 3.2cm

BC = 8.4cm

The area of the triangle ABC is 10cm².

Calculate the perimeter and give the answer correct to 3 significant figures

1--Triangle ABC has a base of BC = 8.4, a left side of AB = 3.2 and an unknown left side of AC.

2--Draw altitude AD perpendicular to BC

3--SInce the area A = 10, A = 8.4h/2 = 4.2h making h == 2.38095.

4--With AB = 3.2 and h == 2.38095, BD = sqrt(3.2^2 - 2.38095^2) = 2.138

5--CD therefore equals 8.4 - 2.138 = 6.262.

6--Thus, AC = sqrt(2.38095^2 + 6.262^2) = 6.699.

7--The perimeter is therefore 8.4 + 3.2 + 6.699 = 18.299.

16 years ago

9 months ago

## 19)

a) To eliminate y, we can substitute the value of y from equation (2) into equation (1).

Substituting y = 2x - 2 into x² + y² = 25:

x² + (2x - 2)² = 25

Simplifying the equation:

x² + 4x² - 8x + 4 = 25

Combining like terms:

5x² - 8x - 21 = 0

b) To solve the simultaneous equations, we can substitute the value of y from equation (2) into equation (1) and solve for x.

Substituting y = 2x - 2 into x² + y² = 25:

x² + (2x - 2)² = 25

Expanding and simplifying the equation:

x² + 4x² - 8x + 4 = 25

Combining like terms:

5x² - 8x - 21 = 0

We can now solve this quadratic equation using factoring, completing the square, or the quadratic formula.

9 months ago

## 19) Bill said that the line y=6 cuts the curve x² + y² =25 at two points.

a) By eliminating y, show that Bill is incorrect.

To eliminate y, we substitute the value of y from the line equation into the curve equation:

x² + (6)² = 25

x² + 36 = 25

x² = 25 - 36

x² = -11

As you can see, the equation x² = -11 does not have any real solutions, which means that the line y=6 does not intersect the curve x² + y² =25 at any point.

b) By eliminating y, find the solutions to the simultaneous equations

To eliminate y, we substitute the value of y from the line equation into the curve equation:

x² + (2x - 2)² = 25

x² + (4x² - 8x + 4) = 25

5x² - 8x + 4 = 25

5x² - 8x - 21 = 0

Now we can solve this quadratic equation using factoring, completing the square, or the quadratic formula. In this case, let's use factoring:

(5x + 3)(x - 7) = 0

Setting each factor equal to zero:

5x + 3 = 0 or x - 7 = 0

5x = -3 or x = 7

x = -3/5 or x = 7

Now, we substitute the values of x back into the line equation to find the corresponding y-values:

For x = -3/5:

y = 2(-3/5) - 2

y = -6/5 - 2

y = -16/5

For x = 7:

y = 2(7) - 2

y = 14 - 2

y = 12

Therefore, the solutions to the simultaneous equations x² + y² = 25 and y = 2x - 2 are (-3/5, -16/5) and (7, 12).

26)

c) Factorise (p+q)² + 5(p+q)

To factorize (p+q)² + 5(p+q), we can notice that (p+q) appears as a common factor. Therefore, we can factorize by factoring out (p+q):

(p+q)((p+q) + 5)

(p+q)(p+q + 5)

e) Simplify 2t² * 3r³t⁴

To simplify 2t² * 3r³t⁴, we can combine the like terms by multiplying the coefficients and adding the exponents of the variables:

2 * 3 = 6 (coefficients)

t² * t⁴ = t^(2+4) = t⁶ (exponents)

r³ (no change as there are no other r terms to combine)

Therefore, 2t² * 3r³t⁴ simplifies to 6r³t⁶.

30) It is always possible to draw a circle which passes through all four vertices of a rectangle.

To understand why it is always possible to draw such a circle, we can consider the properties of a rectangle and the definition of a circle.

A rectangle is defined as a quadrilateral with four right angles. It has two pairs of parallel sides and opposite sides that are equal in length. The diagonals of a rectangle are congruent and bisect each other.

A circle is defined as a shape with all points equidistant from a center point. The distance between the center point and any point on the circle's circumference is called the radius.

Now, let's consider a rectangle. When we draw the intersecting diagonals, they intersect at a point called the center of the rectangle. This point is equidistant from all four vertices of the rectangle.

Since a circle is defined as a shape with all points equidistant from a center point, we can draw a circle centered at the intersection of the diagonals. This circle will have a radius equal to the distance from the center point to any vertex of the rectangle.

By the properties of a rectangle, the distance from the intersection of the diagonals to each corner is equal. Hence, the circle centered on the intersection point will pass through all four vertices of the rectangle.

Therefore, it is always possible to draw a circle passing through all four vertices of a rectangle.