The area of a triangle with vertices at (0, −2) ,(8, −2) and (9, 1) is 12 square units
Solution:
Given, vertices of the triangle are A(0, -2), B(8, -2) and C(9, 1).
We have to find the area of the given triangle.
The area of triangle when vertices are given is:
Here in our problem,
Now, substitute the above values in the formula:
Hence, the area of the triangle is 12 square units.
Answer:
Express as exponents) 1) 81 / 2) 4/9 3) -64
Challange: Use parenthesis) 1) 3b 2) 8x^3y^3 3)15x^2y^2
Evaluate) 1) 125 2) 81/256
Challenge) 9a^2b^2
True or flase) 1) flase 2) false 3) true 4) false
Step-by-step explanation:
Answer:
a = 9, b = 3
Step-by-step explanation:
The exponential function is
y = a
Find a and b by using the points it passes through
Using (0, 9 ) then
9 = a ( = 1 )
a = 9
y = 9
Using (3, 243 ) then
243 = 9b³ ( divide both sides by 9 )
27 = b³ ( take the cube root of both sides )
b = = 3
y = 9 ← exponential equation
Answer:
Step-by-step explanation:
Note the equal sign, what you do to one side, you do to the other side. Do the opposite of PEMDAS. Isolate the variable, x.
PEMDAS is the order of operation, and =
Parenthesis
Exponents (& Roots)
Multiplication
Division
Addition
Subtraction
~
First, subtract 7y from both sides:
3x + 7y (-7y) = 27 (-7y)
3x = -7y + 27
Next, divide 3 from all terms on both sides:
(3x)/3 = (-7y + 27)/3
x = (-7y/3) + 9
The square of 0.9 will result to 0.81, and the squareroot of 0.81 is equal to 0.9.
Squares are the numbers gotten after multiplying a value by itself. While the squareroot of a number is a number whose square is the given number. Hence, both are vice versa methods.
We shall solve for the square root of the number 0.81 as follows;
0.81 = (0.9 × 0.9)
by taking the squareroot of both sides we have;
√0.81 = √(0.9 × 0.9)
√0.81 = √(0.9)²
√0.81 = 0.9
Therefore, the square of 0.9 will result to 0.81, and the squareroot of 0.81 is equal to 0.9
Know more about square and squareroot here:brainly.com/question/428672
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