I can't exactly see if the 10 at the end is negative, but I'm going to assume that it is.
10x - 6y = 8
5x - 10y = -10
= (2, 2)
} I hope this helped! {
4 1/2(2)=9. Multiply by 2 to get total
9- 3 1/2=z. Subtract 3 1/2 from total to get Z
5 1/2= Z
To find<span> the </span>area<span> of a rectangle multiply its height by its width. For a square you only need to </span>find<span> the length of one of the sides (as each side is the same length) and then multiply this by itself to </span>find<span> the </span>area<span>.</span>
Answer
a. 28˚
b. 76˚
c. 104˚
d. 56˚
Step-by-step explanation
Given,
∠BCE=28° ∠ACD=31° & line AB=AC .
According To the Question,
- a. the angle between a chord and a tangent through one of the end points of the chord is equal to the angle in the alternate segment.(Alternate Segment Theorem) Thus, ∠BAC=28°
- b. We Know The Sum Of All Angles in a triangle is 180˚, 180°-∠CAB(28°)=152° and ΔABC is an isosceles triangle, So 152°/2=76˚
thus , ∠ABC=76° .
- c. We know the Sum of all angles in a triangle is 180° and opposite angles in a cyclic quadrilateral(ABCD) add up to 180˚,
Thus, ∠ACD + ∠ACB = 31° + 76° ⇔ 107°
Now, ∠DCB + ∠DAB = 180°(Cyclic Quadrilateral opposite angle)
∠DAB = 180° - 107° ⇔ 73°
& We Know, ∠DAC+∠CAB=∠DAB ⇔ ∠DAC = 73° - 28° ⇔ 45°
Now, In Triangle ADC Sum of angles in a triangle is 180°
∠ADC = 180° - (31° + 45°) ⇔ 104˚
- d. ∠COB = 28°×2 ⇔ 56˚ , because With the Same Arc(CB) The Angle at circumference are half of the angle at the centre
For Diagram, Please Find in Attachment