In this diagram, circle A had a radius = 11.3 and BC = 13.3. BC is a tangent line, calculate the distance of CD.

How do you find the midpoint formula

koban [17]

The formula is x1+x2 divided by 2 and y1+y2 divided by 2.I hope this is what you where looking for

6 0

3 years ago

NEED HELP HURRY

Serga [27]

Answer:

6.3+1.1

Step-by-step explanation:

The communitive property of addition states that no matter what order of the numerical values the answer will be the same. hope this helps

8 0

3 years ago

2. Which algebraic expression represents the following phrase?

oksian1 [2.3K]

The algebraic expression which represents the square of the difference of s and 6 is (s - 6)²

Step-by-step explanation:

Let us represent some words by mathematics expressions

Square a number ⇒ x²
Square the sum of two numbers ⇒ (x + y)²
Sum of the squares of two numbers ⇒ x² + y²
Square of difference of two numbers (x - y)²
The difference of square two numbers x² - y²

∵ The expression is the square of the difference of s and 6

- That means find the difference between s and 6 at first,

then square the difference

∵ The difference of s and 6 = s - 6

- Square this difference means but the difference in a bracket

and then square the bracket

∴ The square of the difference = (s - 6)²

The algebraic expression which represents the square of the difference of s and 6 is (s - 6)²

Learn more:

You can learn more about the algebraic expressions in brainly.com/question/10771256

#LearnwithBrainly

3 0

3 years ago

Which of the following verifies that AABC is similar to ADEF?

nasty-shy [4]

Answer:

B. SAS Theorem

Step-by-step explanation:

SAS means that there are 2 corresponding sides with an angle in between. This condition is satisfied in this case because AC corresponds to DF (18/2 = 9), angle c and angle f are congruent, and BC corresponds to EF

6 0

3 years ago

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In a previous exercise we formulated a model for learning in the form of the differential equation dp dt = k(m − p) where p(t) m

DaniilM [7]

I assume you mean

$\dfrac{dP}{dt} = k(M-P)$

ANSWER
An expression for P(t) is

$P = M - Me^{-kt}$

EXPLANATION
This is a separable differential equation. Treat M and k as constants. Then we can divide both sides by M - P to get the P term with the differential dP and multiply both sides by dt to separate dt from the P terms

$\begin{aligned} \dfrac{dP}{dt} &= k(M-P) \\ \dfrac{dP}{M-P} &= k\, dt
\end{aligned}$

Integrate both sides of the equation.

$\begin{aligned}
\int \dfrac{dP}{M-P} &= \int k\, dt \\
-\ln|M-P| &= kt + C \\
\ln|M-P| &= -kt - C\end{aligned}$

Note that for the left-hand side, u-substitution gives us

$u = M - P \implies du = -1dP \implies dP = -du$

hence why $\int \frac{dP}{M-P} \ne \ln|M - P|$

Now we use the definition of the logarithm to convert into exponential form.

The definition is

$\ln(a) = b \iff \log_e(a) = b \iff e^b = a$

so applying it here, we get

$\begin{aligned} \ln|M-P| &= -kt - C \\ |M - P| &= e^{-kt - C} \\ 
M - P &= \pm e^{-kt - C} 
 \end{aligned}$

Exponent properties can be used to address the constant C. We use $x^{a} \cdot x^{b} = x^{a+b}$ here:

$\begin{aligned}
 M - P &= \pm e^{-kt - C} \\
M - P &= \pm e^{- C - kt} \\ 
M - P &= \pm e^{- C + (- kt)} \\ 
M - P &= \pm e^{- C} \cdot e^{- kt} \\ 
M - P &= Ke^{- kt} && (\text{\footnotesize Let $K = \pm e^{-kt}$ }) \\ 
M &= Ke^{- kt} + P\\
P &= M - Ke^{- kt}
\end{aligned}$

If we assume that P(0) = 0, then set t = 0 and P = 0

$\begin{aligned} 
0 &= M - Ke^{- k\cdot 0} \\
0 &= M - K \cdot 1 \\
M &= K
 \end{aligned}
$

Substituting into our original equation, we get our final answer of

$P = M - Me^{-kt}$

6 0

3 years ago

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