Please help

abd=58 and bdc=32 what is the measure of adb​

<ADB = 61degrees

Step-by-step explanation:

From the diagram shown, triangle ADB is isosceles since they projected out from the chord AD. Hence;

<ABD + <ADB + <BAD = 180

Since <ADB = <BAD, then;

<ABD + <ADB + <ADB = 180

Given that:

<ABD = 58degrees

58 + 2<ADB = 180

2<ADB = 180 - 58

2<ADB = 122

<ADB = 122/2

<ADB = 61degrees

3x^2−14x+3y  (if you didn't know, "^2" means that its an exponent)

Step-by-step explanation:

Let's simplify step-by-step.

2x−4y+3x^2+7y−16x

=2x+−4y+3x^2+7y+−16x

Combine Like Terms:

=2x+−4y+3x^2+7y+−16x

=(3x^2)+(2x+−16x)+(−4y+7y)

=3x^2+−14x+3y