Answer:
<h2>
∠PQT = 72°</h2>
Step-by-step explanation:
According to the diagram shown, ∠OPQ = ∠OQP = 18°. If PQT is a tangent to the circle, it can be inferred that line OQ is perpendicular to line QT. Ths shows that ∠OQT = 90°.
Also from the diagram, ∠OQP + ∠PQT = ∠OQT;
∠PQT = ∠OQT - ∠OQP
Given ∠OQP = 18° and ∠OQT = 90°
∠PQT = 90°-18°
∠PQT = 72°
Answer: 28
Step-by-step explanation:
Answer:
x = -7/4
General Formulas and Concepts:
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
Equality Properties
- Multiplication Property of Equality
- Division Property of Equality
- Addition Property of Equality
- Subtract Property of Equality
<u>Algebra I</u>
Step-by-step explanation:
<u>Step 1: Define</u>
5x + 3 - 2x = 12 + 7x - 2
<u>Step 2: Solve for </u><em><u>x</u></em>
- Combine like terms: 3x + 3 = 7x + 10
- [SPE] Subtract 3x on both sides: 3 = 4x + 10
- [SPE] Subtract 10 on both sides: -7 = 4x
- [DPE] Divide 4 on both sides: -7/4 = x
- Rewrite: x = -7/4
<u>Step 3: Check</u>
<em>Plug in x into the original equation to verify it's a solution.</em>
- Substitute in <em>x</em>: 5(-7/4) + 3 - 2(-7/4) = 12 + 7(-7/4) - 2
- Multiply: -35/4 + 3 + 7/2 = 12 - 49/4 - 2
- Add: -23/4 + 7/2 = 12 - 49/4 - 2
- Add: -9/4 = 12 - 49/4 - 2
- Subtract: -9/4 = -1/4 - 2
- Subtract: -9/4 = -9/4
Here we see that -9/4 does indeed equal -9/4.
∴ x = -7/4 is the solution to the equation.
Answer:
a - b2
Step-by-step explanation:
STEP 1
:
Trying to factor as a Difference of Squares:
1.1 Factoring: a-b2
Theory : A difference of two perfect squares, A2 - B2 can be factored into (A+B) • (A-B)
Proof : (A+B) • (A-B) =
A2 - AB + BA - B2 =
A2 - AB + AB - B2 =
A2 - B2
Note : AB = BA is the commutative property of multiplication.
Note : - AB + AB equals zero and is therefore eliminated from the expression.
Check : a1 is not a square !!
Ruling : Binomial can not be factored as the difference of two perfect squares
Final result :
a - b2
<u><em>HOPE THIS HELPS!</em></u>
<u><em>PLEASE MARK BRAINLIEST! :)</em></u>
