All categories

3 years ago

Find mRT. Assume that segments that appear to be tangent are tangent.

Mathematics

1 answer:

IceJOKER [234]3 years ago

7 0

Answer:

m(arc RT) = 148°

Step-by-step explanation:

From the picture attached,

Segment RT is a chord and segment RS is a tangent of a circle O meeting at R.

By the property of tangent chord angle,

"Angle formed by an intersecting tangent and chord measures half of the intercepted minor arc"

m(∠SRT) = $\frac{1}{2}(\text{minor arc RT})$

$74^0=\frac{1}{2}m(RT)$

$m(\text{arc RT)}=2(74^0)$

$=148^0$

Therefore, measure of minor arc RT is 148°.

You might be interested in

Please help me! I don't understand this! DUE TODAY

pishuonlain [190]

Answer: See the answers below.

The first equation that needs to be solve is: 10 = -16t^2 + 18
If you use the quadratic equation, you will get 0.707 seconds.

For the second equation, you need to solve 0 = -16t^2 + 18.
If you use the quadratic equation, you will get 1.061 seconds.

No, the rate of change is not constant because this is a quadratic equation.

5 0

3 years ago

3х – 6+1 = - 2x – 5+5х

marysya [2.9K]

Answer:0=12

Step-by-step explanation:

Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation:

3x-7=3x-5

Combine like terms:

3x-3x=7+5

0=12

4 0

3 years ago

Read 2 more answers

What is 20 divided by 10 but to the power or 2

Klio2033 [76]

4 is the answer for your question

5 0

2 years ago

Read 2 more answers

Jeremy has a total of $250. If he spends 30% of his money, how much will he have left?

vagabundo [1.1K]

If you look at the picture, you set it up as part over whole is equal to percent over 100. then you just multiply across.

5 0

3 years ago

The profit P (in thousands of dollars) for a company spending an amount s (in thousands of dollars on advertising is

sattari [20]

Answer:

The company should spend $40 to yield a maximum profit.

The point of diminishing returns is (40, 3600).

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Equality Properties

Multiplication Property of Equality
Division Property of Equality
Addition Property of Equality
Subtraction Property of Equality

Algebra I

Coordinate Planes

Coordinates (x, y) → (s, P)

Functions

Function Notation

Terms/Coefficients

Factoring/Expanding

Quadratics

Algebra II

Coordinate Planes

Maximums/Minimums

Calculus

Derivatives

Derivative Notation

Derivative Property [Multiplied Constant]: $\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)$

Derivative Property [Addition/Subtraction]: $\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)]$

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

1st Derivative Test - tells us where on the function f(x) does it have a relative maximum or minimum

Critical Numbers

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle P = \frac{-1}{10}s^3 + 6s^2 + 400$

Step 2: Differentiate

[Function] Derivative Property [Addition/Subtraction]: $\displaystyle P' = \frac{dP}{ds} \bigg[ \frac{-1}{10}s^3 \bigg] + \frac{dP}{ds} [ 6s^2 ] + \frac{dP}{ds} [ 400 ]$
[Derivative] Rewrite [Derivative Property - Multiplied Constant]: $\displaystyle P' = \frac{-1}{10} \frac{dP}{ds} \bigg[ s^3 \bigg] + 6 \frac{dP}{ds} [ s^2 ] + \frac{dP}{ds} [ 400 ]$
[Derivative] Basic Power Rule: $\displaystyle P' = \frac{-1}{10}(3s^2) + 6(2s)$
[Derivative] Simplify: $\displaystyle P' = -\frac{3s^2}{10} + 12s$

Step 3: 1st Derivative Test

[Derivative] Set up: $\displaystyle 0 = -\frac{3s^2}{10} + 12s$
[Derivative] Factor: $\displaystyle 0 = \frac{-3s(s - 40)}{10}$
[Multiplication Property of Equality] Isolate s terms: $\displaystyle 0 = -3s(s - 40)$
[Solve] Find quadratic roots: $\displaystyle s = 0, 40$

∴ s = 0, 40 are our critical numbers.

Step 4: Find Profit

[Function] Substitute in s = 0: $\displaystyle P(0) = \frac{-1}{10}(0)^3 + 6(0)^2 + 400$
[Order of Operations] Evaluate: $\displaystyle P(0) = 400$
[Function] Substitute in s = 40: $\displaystyle P(40) = \frac{-1}{10}(40)^3 + 6(40)^2 + 400$
[Order of Operations] Evaluate: $\displaystyle P(40) = 3600$

We see that we will have a bigger profit when we spend s = $40.

∴ The maximum profit is $3600.

∴ The point of diminishing returns is ($40, $3600).

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Differentiation (Applications)

5 0

2 years ago