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1 year ago

Simplify the expression (4a^2 +2ab) - (2a^2 -ab + b^2) what is the coefficient of the term b^2

Mathematics

1 answer:

iragen [17]1 year ago

7 0

In the simplified expression, the coefficient of the term b² is -1.

<h3>Order to solve Math Operations</h3>

When you solve any Math Operations, it is necessary to be attentive to the order of the operations.

First, you should solve the operation insides of Parentheses. After that priority is given to Exponents. Thereon, you can solve multiplication or division operations from left to right. Finally, addition or subtraction from left to right

See the order below

Parentheses
Exponents
Multiplication or Division from left to right
Addition or Subtraction from left to right

The question gives the expression (4a² +2ab) - (2a² -ab + b²). Therefore, for solving the expression you should follow the order of the math operations.

Remove the Parentheses

4a² +2ab -2a² +ab - b²

Add similar terms

2a²+3ab-b²

Like, there are no more similar terms. Therefore, you can represent the expression (4a² +2ab) - (2a² -ab + b²) by the expression 2a²+3ab-b².

Learn more about the algebraic expression here:

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A store sells grape jelly in 16-ounces jars for $ 4.68. How much is the jelly per ounce?

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3 years ago

Read 2 more answers

PLEASE PLEASE HELP! Find the measure of APB^.

Whitepunk [10]

Answer:

65°

Step-by-step explanation:

Radii CA and CB are perpendicular to tangent lines AT and BT, so

$\angle CAT=\angle CBT=90^{\circ}$

Since angle BAT is equal to 65°, angle CAB has measure

$\angle CAB=90^{\circ}-65^{\circ}=25^{\circ}.$

Consider triangle ACB. This triangle is isosceles, because CA=CB as radii of the circle. Two angles adjacent to the base are congruent, thus

$\angle CBA=\angle CAB=25^{\circ}$

The sum of the measures of all interior angles in triangle is always 180°, so

$\angle CAB+\angle CBA+\angle ACB=180^{\circ}\\ \\25^{\circ}+25^{\circ}+\angle ACB=180^{\circ}\\ \\\angle ACB=130^{\circ}$

Angle ACB is central angle subtended on the minor arc AB, angle APB is inscribed angle subtended on the same minor arc AB. The measure of inscribed angle is half the measure of central angle subtended on the same arc, so

$x=\angle APB=\dfrac{1}{2}\cdot 130^{\circ}=65^{\circ}$