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Featured 1 week ago

New coordinates of

Change (reduction) in distance from

A (1, -1), B (-2,1) and rotated by

Point A moves from IV quadrant to I quadrant.

Distance

Distance

Change (reduction) in distance from

Featured 1 week ago

Area of a trapizoin

Now base a has been halved or the new base is a/2

Dividing Eqn(2) by (1),

Featured 6 days ago

**Case 1** : ED is the altitude (perpendicular to GH) **ft**

**Case 2** : ED is the median **ft**

**Case 3** :ED is the angular bisector of **ft**

Assuming

Since it has not been given, ED is perpendicular to GH or D is the mid point of GH or ED is the angular bisector of

Let's consider all these three cases.

**Case 1** : ED is the altitude (perpendicular to GH)

In right triangle DEG,

In right triangle DEH,

Comparing Eqns (1), (2),

**ft**

**Case 2** : ED is the median

i.e DH = DG #

**ft**

**Case 3** : ED is the angular bisector of #GhatEH

As per angular bisector theorem,

**ft**

Featured 1 week ago

The length of perpendicular from

Now as circle cuts off an intercept of length

Hence radius of circle is

and equation of circle is

or

graph{(x^2+y^2-6x+2y-28)(2x-5y+18)((x-3)^2+(y+1)^2-0.03)=0 [-9.75, 10.25, -4.04, 6.38]}

Featured yesterday

#"the length of the arc is calculated using"#

#â€¢ " length of arc "="circumference "xx"fraction of circle"#

#color(white)(xxxxxxxxxxxx)=2pirxx60/360#

#color(white)(xxxxxxxxxxxx)=6pixxcancel(60)^1/cancel(360)^6#

#color(white)(xxxxxxxxxxxx)=(cancel(6) pi)/cancel(6)=pi~~3.14#

Featured yesterday

See below.

Here we are formulating an equation to solve for

We know that the interior angles of any triangle adds up

We have three angles given:

This means that :

Now we collect like terms to simplify.

Now we solve like any linear equation by isolating the variable on one side of the equation with the constant on the other.

Here we must subtract **both** sides to isolate the

We want one

Here we divide by

We can check if we are right by putting our value of

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