Answer:
The area of the putting green is 1133.5 square feet.
Step-by-step explanation:
Area of a circle:
The area of a circle is given by:
In which r is the radius, which is half the diameter.
The circular putting green has a diameter of 38 ft.
This means that
What is the area of the putting green?
The area of the putting green is 1133.5 square feet.
Answer:
1133.5 is the answer.
Step-by-step explanation:
Hope its right
It worked for me
and
B. are between 37.25 and 101.75
When y = -6 and z = 5, the value of the expression y + z + 2 is 1.
To calculate the expression y + z + 2 when y = -6 and z = 5, you simply substitute these values into the expression and perform the arithmetic operations:
y + z + 2 = (-6) + (5) + 2
Now, perform the addition:
(-6) + (5) + 2 = -1 + 2 = 1
So, when y = -6 and z = 5, the value of the expression y + z + 2 is 1.
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Answer:
1
Step-by-step explanation:
y + z + 2
first you have to replace the variables with the numbers
This is your equation now.
-6 + 5 + 2
Now you just have to solve it.
-6 + 5 = -1+ 2 = 1
or
7 - 6 = 1
The quadratic value function is y = -x² - 4x + 16
Given data ,
Let the quadratic equation be represented as A
Now , the value of A is
Let the function passes through the given points. (-1,-11) (2,-26) (-3,-31)
And , using the point (-1,-11) , we get
-11 = a(-1)² + b(-1) + c
-11 = a - b + c
Using the point (2,-26) , we get
-26 = a(2)² + b(2) + c
-26 = 4a + 2b + c
Using the point (-3,-31) , we get
-31 = a(-3)² + b(-3) + c
-31 = 9a - 3b + c
And , the three set of equations are
-11 = a - b + c
-26 = 4a + 2b + c
-31 = 9a - 3b + c
From the first equation, we can solve for c:
c = 11 - a + b
Substituting this expression for c into the other two equations, we get:
-26 = 4a + 2b + 11 - a + b
-31 = 9a - 3b + 11 - a + b
Simplifying these equations, we get:
-15 = 3a + 3b
-21 = 8a - 2b
Solving the first equation for b in terms of a, we get:
b = -5 - a
Substituting this expression for b into the second equation, we get:
-21 = 8a - 2(-5 - a)
Simplifying and solving for a, we get:
a = -1
Substituting this value of a into the equation for b that we found earlier, we get:
b = -5 - (-1) = -4
Finally, substituting these values of a and b into the equation for c that we found earlier, we get:
c = 11 - (-1) - (-4) = 16
Hence , the quadratic function that passes through the given points is given by y = -x² - 4x + 16
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Answer:
88
Step-by-step explanation:
72 + 95 + 100 + 85 + 88=440
440/5=88
3. AABC is made of (chords in, tangents to) OX.
4. ZDEF is an (intercepted arc, inscribed angle) of OX.
Answer:
1 . EF is a secant of OX
2. DF is a chord of OX.
3. AABC is made of tangents to OX
4. ∠DEF is an inscribed angle of OX.
Step-by-step explanation:
Secant is line which intersects a circle from two distant points. Locus is a set of points which is same distance from the center of a circle. Chord is a straight line whose end points lie on circular arc of a circle. Tangent is a line which touches the circle at one point only.
B) quadrilateral
C) right triangle
D) scalene triangle