Pqrs and abrs are parallelogram and x is any point on br show that 1) area(pqrs)=area(abrs) 2) area(axs)=1/2 area(pqrs)

Step-by-step explanation:

Given PQRS and ABRS are parallelogram and X is any point on BR. we have to prove that  

1)

2)

In ∆ASP and ΔBRQ  

∠SPA = ∠RQB [Corresponding angles]

∠PAS = ∠QBR [Corresponding angles]

PS = QR [Opposite sides of the parallelogram PQRS]

By AAS rule, ∆ASP≅ΔBRQ  

∴ ar(ASP) = ar(BRQ)      (Congruent triangles have equal area)

Now, ar (PQRS) = ar (PSA) + ar (ASRQ]  

                           = ar(QRB) + ar(ASRQ]  = ar(ABRS)  

So, ar(PQRS)=ar(ABRS)

(ii) Now, ∆AXS and ||gm ABRS are on the same base AS and between same parallels AS and BR

⇒   (∵ar(PQRS)=ar(ABRS))  

Hence Proved.

(- 1, 0 )

step-by-step explanation:

an x- intercept is where the function crosses the x- axis.

to find the x- intercept/s look for the value of x when f(x) = 0

from the given table there are 3 values of x where f(x) = 0

when x = - 1, x = 2, x = 3

the one that is listed is (- 1, 0 )

5=812a+9+3

step 1: simplify both sides of the equation.

5=812a+9+3

5=(812a)+(9+3)(combine like terms)

5=812a+12

5=812a+12

step 2: flip the equation.

812a+12=5

step 3: subtract 12 from both sides.

812a+12−12=5−12

812a=−7

step 4: divide both sides by 812.

812a /812 = -7/812

a= -1/116

so the answer is

a=   -1/116