Answer: y= x-2
Step-by-step explanation:
Perpendicularity entails the slope of line a is -1 divided by the slope of line b
m1= (-1/m2)
From y=Mx +c
We compare with x-4y=7
y=(1/4)x-7/4
Meaning m2= 1/4
m1= -1/(1/4)= -1*4= -4
((y-y1)/(x-x1))= ((y2-y1)/(x2-x1))=m1
x1= 2, y1= -4
((y-(-4))/(x-2))= -4
((y+4)/(x-2))= -4
Cross multiply
y+4= -4(x-2)
y= x-2
Answer:
x=18
Step-by-step explanation:
We need to find the areas of the rectangles
1st rectangle
A= l*w
A = (2x+4)*9
= 18x+36
2nd rectangle
A = l*w
= 15*(3x-30)
= 45x-450
Since the area of the rectangles are equal, we set these two equations equal
18x+36 = 45x-450
Subtract 18x from each side
18x-18x+36 = 45x-18x-450
36 = 27x-450
Add 450 to each side
36+450 =27x
486 = 27x
Divide by 27 on each side
486/27 = 27x/27
18=x
Answer:
you will get speed on a slope and you will get speed with speed (:
Step-by-step explanation:
Answer:
m = 3
n = 14
p = 2
Step-by-step explanation:
Given:

We need to find the values of m, n, p
Solution:
We will first solve the L.H.S
L.H.S = 
Now we will first make the denominator common by taking L.C.M

Now the denominator is common hence we will subtract the numerators.
L.H.S 
Now Comparing the the value of L.H.S with R.H.S we can say that;

m = 3
n = 14
p = 2
Answer:
Hence proved triangle ADE ≅ triangle BCE by Side Angle Side congruent property.
Step-by-step explanation:
Given:
AD ⊥ AB
CD
BC ⊥ AB
CD
AD = BC
∴ ∠ A = ∠ B = ∠ C = ∠ D =90°
∠ EDC = ∠ ECD
Solution
∠ C = ∠ BCE + ∠ ECD⇒ equation 1
∠ D = ∠ ADE + ∠ EDC⇒ equation 2
∠ C = ∠ D (given)
Substituting equation 1 and 2 in above equation we get
∠ BCE + ∠ ECD = ∠ ADE + ∠ EDC
But ∠ EDC = ∠ ECD (given)
∴ ∠ ADE = ∠ BCE
ED = EC (∵ base angles are same triangle is isosceles triangle)
Now, In Δ ADE and Δ BCE
AD =BC
∠ ADE = ∠ BDE
ED = EC
∴ By Side Angle Side congruent property
Δ ADE ≅ Δ BCE