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VikaD [51]
2 years ago
15

Solve for x,y,z on the triangle

Mathematics
2 answers:
nordsb [41]2 years ago
4 0
X = 64.5
Y= 25.5
Z= 129
Hatshy [7]2 years ago
4 0

Answer:

x=64.5

y=25.5

z=129

Step-by-step explanation:

should be correct if you'd like to check each isosceles triangle equals 180 degree

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This graph represents a linear relationship between x and y. Which equation best represents the relationship shown in this graph
kirza4 [7]

Answer:

y= 5/3x

Step-by-step explanation:

4 0
3 years ago
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Show all of your work, even though the question may not explicitly remind you to do so. Clearly label any functions, graphs, tab
Leona [35]

Answer:

a. 5 b. y = -\frac{3}{4}x + \frac{1}{2} c. 148.5 d. 1/7

Step-by-step explanation:

Here is the complete question

Show all of your work, even though the question may not explicitly remind you to do so. Clearly label any functions, graphs, tables, or other objects that you use. Justifications require that you give mathematical reasons, and that you verify the needed conditions under which relevant theorems, properties, definitions, or tests are applied. Your work will be scored on the correctness and completeness of your methods as well as your answers. Answers without supporting work will usually not receive credit. Unless otherwise specified, answers (numeric or algebraic) need not be simplified. If your answer is given as a decimal approximation, it should be correct to three places after the decimal point. Unless otherwise specified, the domain of a function f is assumed to be the set of all real numbers for which f() is a real number Let f be an increasing function with f(0) = 2. The derivative of f is given by f'(x) = sin(πx) + x² +3. (a) Find f" (-2) (b) Write an equation for the line tangent to the graph of y = 1/f(x) at x = 0. (c) Let I be the function defined by g(x) = f (√(3x² + 4). Find g(2). (d) Let h be the inverse function of f. Find h' (2). Please respond on separate paper, following directions from your teacher.

Solution

a. f"(2)

f"(x) = df'(x)/dx = d(sin(πx) + x² +3)/dx = cos(πx) + 2x

f"(2) = cos(π × 2) + 2 × 2

f"(2) = cos(2π) + 4

f"(2) = 1 + 4

f"(2) = 5

b. Equation for the line tangent to the graph of y = 1/f(x) at x = 0

We first find f(x) by integrating f'(x)

f(x) = ∫f'(x)dx = ∫(sin(πx) + x² +3)dx = -cos(πx)/π + x³/3 +3x + C

f(0) = 2 so,

2 = -cos(π × 0)/π + 0³/3 +3 × 0 + C

2 = -cos(0)/π + 0 + 0 + C

2 = -1/π + C

C = 2 + 1/π

f(x) = -cos(πx)/π + x³/3 +3x + 2 + 1/π

f(x) = [1-cos(πx)]/π + x³/3 +3x + 2

y = 1/f(x) = 1/([1-cos(πx)]/π + x³/3 +3x + 2)

The tangent to y is thus dy/dx

dy/dx = d1/([1-cos(πx)]/π + x³/3 +3x + 2)/dx

dy/dx = -([1-cos(πx)]/π + x³/3 +3x + 2)⁻²(sin(πx) + x² +3)

at x = 0,

dy/dx = -([1-cos(π × 0)]/π + 0³/3 +3 × 0 + 2)⁻²(sin(π × 0) + 0² +3)

dy/dx = -([1-cos(0)]/π + 0 + 0 + 2)⁻²(sin(0) + 0 +3)

dy/dx = -([1 - 1]/π + 0 + 0 + 2)⁻²(0 + 0 +3)

dy/dx = -(0/π + 2)⁻²(3)

dy/dx = -(0 + 2)⁻²(3)

dy/dx = -(2)⁻²(3)

dy/dx = -3/4

At x = 0,

y = 1/([1-cos(π × 0)]/π + 0³/3 +3 × 0 + 2)

y = 1/([1-cos(0)]/π + 0 + 0 + 2)

y = 1/([1 - 1]/π + 2)

y = 1/(0/π + 2)

y = 1/(0 + 2)

y = 1/2

So, the equation of the tangent at (0, 1/2) is

\frac{y - \frac{1}{2} }{x - 0} = -\frac{3}{4}  \\y - \frac{1}{2} = -\frac{3}{4}x\\y = -\frac{3}{4}x + \frac{1}{2}

c. If g(x) = f (√(3x² + 4). Find g'(2)

g(x) = f (√(3x² + 4) = [1-cos(π√(3x² + 4)]/π + √(3x² + 4)³/3 +3√(3x² + 4) + 2

g'(x) = [3xsinπ√(3x² + 4) + 18x(3x² + 4) + 9x]/√(3x² + 4)

g'(2) = [3(2)sinπ√(3(2)² + 4) + 18(2)(3(2)² + 4) + 9(2)]/√(3(2)² + 4)

g'(2) = [6sinπ√(12 + 4) + 36(12 + 4) + 18]/√12 + 4)

g'(2) = [6sinπ√(16) + 36(16) + 18]/√16)

g'(2) = [6sin4π + 576 + 18]/4)

g'(2) = [6 × 0 + 576 + 18]/4)

g'(2) = [0 + 576 + 18]/4)

g'(2) = 594/4

g'(2) = 148.5

d. If h be the inverse function of f. Find h' (2)

If h(x) = f⁻¹(x)

then h'(x) = 1/f'(x)

h'(x) = 1/(sin(πx) + x² +3)

h'(2) = 1/(sin(π2) + 2² +3)

h'(2) = 1/(sin(2π) + 4 +3)

h'(2) = 1/(0 + 4 +3)

h'(2) = 1/7

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3 years ago
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The circumference is:

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Answer:

y = 8 degrees

Step-by-step explanation:

128 + 7y - 4 = 180

7y - 4 = 52

7y + 56

y = 8

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ANSWER:

Segment TS is a Chord

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