Question

construct a shmbhu ABCD in which AB=7.4cm,LB=72°​

Answer 1

Step-by-step explanation:

it will help you to understand

Answer 2

Answer:

$\displaystyle \sf \longrightarrow \frac{dy}{dt} = 2.(t-1)$

Step-by-step explanation:

Given :

$\displaystyle \sf t=\sqrt{y}+1$

Rewrite as :

$\displaystyle \sf \longrightarrow t-1=\sqrt{y}$

$\displaystyle \sf \longrightarrow y= (t-1)^2$

Now diff. w.r.t. t we get :

$\displaystyle \sf \longrightarrow \frac{dy}{dt} =\frac{d}{dt} (t-1)^2$

$\displaystyle \sf \longrightarrow \frac{dy}{dt} = 2.(t-1)^{2-1}.(t-1)'$

$\displaystyle \sf \longrightarrow \frac{dy}{dt} = 2.(t-1).(t'-1')$

We know derivative of constant is zero!

$\displaystyle \sf \longrightarrow \frac{dy}{dt} = 2.(t-1).(1-0)$

$\displaystyle \sf \longrightarrow \frac{dy}{dt} = 2.(t-1)$

Hence we get required answer!

Answer 3

Answer:

$\displaystyle \sf \longrightarrow \frac{dy}{dt} = 2.(t-1)$

Step-by-step explanation:

Given :

$\displaystyle \sf t=\sqrt{y}+1$

Rewrite as :

$\displaystyle \sf \longrightarrow t-1=\sqrt{y}$

$\displaystyle \sf \longrightarrow y= (t-1)^2$

Now diff. w.r.t. t we get :

$\displaystyle \sf \longrightarrow \frac{dy}{dt} =\frac{d}{dt} (t-1)^2$

$\displaystyle \sf \longrightarrow \frac{dy}{dt} = 2.(t-1)^{2-1}.(t-1)'$

$\displaystyle \sf \longrightarrow \frac{dy}{dt} = 2.(t-1).(t'-1')$

We know derivative of constant is zero!

$\displaystyle \sf \longrightarrow \frac{dy}{dt} = 2.(t-1).(1-0)$

$\displaystyle \sf \longrightarrow \frac{dy}{dt} = 2.(t-1)$

Hence we get required answer!

Answer 4

Answer:

$\displaystyle \sf \longrightarrow \frac{dy}{dt} = 2.(t-1)$

Step-by-step explanation:

Given :

$\displaystyle \sf t=\sqrt{y}+1$

Rewrite as :

$\displaystyle \sf \longrightarrow t-1=\sqrt{y}$

$\displaystyle \sf \longrightarrow y= (t-1)^2$

Now diff. w.r.t. t we get :

$\displaystyle \sf \longrightarrow \frac{dy}{dt} =\frac{d}{dt} (t-1)^2$

$\displaystyle \sf \longrightarrow \frac{dy}{dt} = 2.(t-1)^{2-1}.(t-1)'$

$\displaystyle \sf \longrightarrow \frac{dy}{dt} = 2.(t-1).(t'-1')$

We know derivative of constant is zero!

$\displaystyle \sf \longrightarrow \frac{dy}{dt} = 2.(t-1).(1-0)$

$\displaystyle \sf \longrightarrow \frac{dy}{dt} = 2.(t-1)$

Hence we get required answer!

Answer 5

Answer:

$\displaystyle \sf \longrightarrow \frac{dy}{dt} = 2.(t-1)$

Step-by-step explanation:

Given :

$\displaystyle \sf t=\sqrt{y}+1$

Rewrite as :

$\displaystyle \sf \longrightarrow t-1=\sqrt{y}$

$\displaystyle \sf \longrightarrow y= (t-1)^2$

Now diff. w.r.t. t we get :

$\displaystyle \sf \longrightarrow \frac{dy}{dt} =\frac{d}{dt} (t-1)^2$

$\displaystyle \sf \longrightarrow \frac{dy}{dt} = 2.(t-1)^{2-1}.(t-1)'$

$\displaystyle \sf \longrightarrow \frac{dy}{dt} = 2.(t-1).(t'-1')$

We know derivative of constant is zero!

$\displaystyle \sf \longrightarrow \frac{dy}{dt} = 2.(t-1).(1-0)$

$\displaystyle \sf \longrightarrow \frac{dy}{dt} = 2.(t-1)$

Hence we get required answer!

Answer 6

Answer:

$\displaystyle \sf \longrightarrow \frac{dy}{dt} = 2.(t-1)$

Step-by-step explanation:

Given :

$\displaystyle \sf t=\sqrt{y}+1$

Rewrite as :

$\displaystyle \sf \longrightarrow t-1=\sqrt{y}$

$\displaystyle \sf \longrightarrow y= (t-1)^2$

Now diff. w.r.t. t we get :

$\displaystyle \sf \longrightarrow \frac{dy}{dt} =\frac{d}{dt} (t-1)^2$

$\displaystyle \sf \longrightarrow \frac{dy}{dt} = 2.(t-1)^{2-1}.(t-1)'$

$\displaystyle \sf \longrightarrow \frac{dy}{dt} = 2.(t-1).(t'-1')$

We know derivative of constant is zero!

$\displaystyle \sf \longrightarrow \frac{dy}{dt} = 2.(t-1).(1-0)$

$\displaystyle \sf \longrightarrow \frac{dy}{dt} = 2.(t-1)$

Hence we get required answer!

Answer 7

Answer:

$\displaystyle \sf \longrightarrow \frac{dy}{dt} = 2.(t-1)$

Step-by-step explanation:

Given :

$\displaystyle \sf t=\sqrt{y}+1$

Rewrite as :

$\displaystyle \sf \longrightarrow t-1=\sqrt{y}$

$\displaystyle \sf \longrightarrow y= (t-1)^2$

Now diff. w.r.t. t we get :

$\displaystyle \sf \longrightarrow \frac{dy}{dt} =\frac{d}{dt} (t-1)^2$

$\displaystyle \sf \longrightarrow \frac{dy}{dt} = 2.(t-1)^{2-1}.(t-1)'$

$\displaystyle \sf \longrightarrow \frac{dy}{dt} = 2.(t-1).(t'-1')$

We know derivative of constant is zero!

$\displaystyle \sf \longrightarrow \frac{dy}{dt} = 2.(t-1).(1-0)$

$\displaystyle \sf \longrightarrow \frac{dy}{dt} = 2.(t-1)$

Hence we get required answer!