The equation in Slope-Intercept Form is y = -2/5 x + 4.
We have given that,
y−2=−2/5(x−5)
We have to rewrite the equation in Slope-Intercept Form.
<h3>What is the slope-intercept form of the line?</h3>
slope-intercept form is y = mx + b
for equation y - 2 = -2/5 (x - 5) we need to simplify/rearrange
so y - 2 = -2/5 x + -2/5 (-5)
= -2/5 x + 2
=> y = -2/5 x + 4
This line has a gradient of -2/5 (goes down 2 units and across 5 units) and a y-intercept of 4 so passes through the point (0,4).
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Answer:
x = - 1, x = 7
Step-by-step explanation:
Given
x² - 6x = 7
To complete the square
add ( half the coefficient of the x- term )² to both sides
x² + 2(- 3) x + 9 = 7 + 9
(x - 3)² = 16 ( take the square root of both sides )
x - 3 = ±
= ± 4
Add 3 to both sides
x = 3 ± 4, that is
x = 3 + 4 = 7 or x = 3 - 4 = - 1
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