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Vera_Pavlovna [14]
3 years ago
10

Which statements about the function f are true? Check all that apply.

Mathematics
2 answers:
goblinko [34]3 years ago
8 0

Answer:

The domain (or input values) of a continuous line is all real numbers

The range (or output values) of a continuous line is all real numbers

f(2) =4

Step-by-step explanation:

The domain (or input values) of a continuous line is all real numbers

x can be positive and negative

The range (or output values) of a continuous line is all real numbers

y can be positive and negative

f(2) is the value of y at x =2

f(2) =4

g(2) = 1/2(2) +3

Delicious77 [7]3 years ago
6 0

Answer:

1, 2 ,5

Step-by-step explanation:

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An elementary school is offering 3 language classes: one in Spanish, one in French,and one in German. The classes are open to an
Alona [7]

Answer:

A. 0.5

B. 0.32

C. 0.75

Step-by-step explanation:

There are

  • 28 students in the Spanish class,
  • 26 in the French class,
  • 16 in the German class,
  • 12 students that are in both Spanish and French,
  • 4 that are in both Spanish and German,
  • 6 that are in both French and German,
  • 2 students taking all 3 classes.

So,

  • 2 students taking all 3 classes,
  • 6 - 2 = 4 students are in French and German, bu are not in Spanish,
  • 4 - 2 = 2 students are in Spanish and German, but are not in French,
  • 12 - 2 = 10 students are in Spanish and French but are not in German,
  • 16 - 2 - 4 - 2 = 8 students are only in German,
  • 26 - 2 - 4 - 10 = 10 students are only in French,
  • 28 - 2 - 2 - 10 = 14 students are only in Spanish.

In total, there are

2 + 4 + 2 + 10 + 8 + 10 +14 = 50 students.

The classes are open to any of the 100 students in the school, so

100 - 50 = 50 students are not in any of the languages classes.

A. If a student is chosen randomly, the probability that he or she is not in any of the language classes is

\dfrac{50}{100} =0.5

B. If a student is chosen randomly,  the probability that he or she is taking exactly one language class is

\dfrac{8+10+14}{100}=0.32

C. If 2 students are chosen randomly,  the probability that both are not taking any language classes is

0.5\cdot 0.5=0.25

So,  the probability that at least 1 is taking a language class is

1-0.25=0.75

3 0
4 years ago
Tell whether the triangle with the given side lengths is a right triangle.
Lerok [7]

Answer:

No

Step-by-step explanation:

6 0
3 years ago
What is the constant in the expression 12q + 9e + 15
Talja [164]

Answer:

Step-by-step explanation:

the constant in this expression is 15

4 0
3 years ago
Read 2 more answers
(5×24) ÷ (14×16) what's the answer
taurus [48]
Times the ones in () which becomes 120 * 224 = 0.535
6 0
3 years ago
Read 2 more answers
A large corporation starts at time t = 0 to invest part of its receipts continuously at a rate of P dollars per year in a fund f
Andrews [41]

Answer:

A = \frac{P}{r}\left( e^{rt} -1 \right)

Step-by-step explanation:

This is <em>a separable differential equation</em>. Rearranging terms in the equation gives

                                                \frac{dA}{rA+P} = dt

Integration on both sides gives

                                            \int \frac{dA}{rA+P} = \int  dt

where c is a constant of integration.

The steps for solving the integral on the right hand side are presented below.

                               \int \frac{dA}{rA+P} = \begin{vmatrix} rA+P = m \implies rdA = dm\end{vmatrix} \\\\\phantom{\int \frac{dA}{rA+P} } = \int \frac{1}{m} \frac{1}{r} \, dm \\\\\phantom{\int \frac{dA}{rA+P} } = \frac{1}{r} \int \frac{1}{m} \, dm\\\\\phantom{\int \frac{dA}{rA+P} } = \frac{1}{r} \ln |m| + c \\\\&\phantom{\int \frac{dA}{rA+P} } = \frac{1}{r} \ln |rA+P| +c

Therefore,

                                        \frac{1}{r} \ln |rA+P| = t+c

Multiply both sides by r.

                               \ln |rA+P| = rt+c_1, \quad c_1 := rc

By taking exponents, we obtain

      e^{\ln |rA+P|} = e^{rt+c_1} \implies  |rA+P| = e^{rt} \cdot e^{c_1} rA+P = Ce^{rt}, \quad C:= \pm e^{c_1}

Isolate A.

                 rA+P = Ce^{rt} \implies rA = Ce^{rt} - P \implies A = \frac{C}{r}e^{rt} - \frac{P}{r}

Since A = 0  when t=0, we obtain an initial condition A(0) = 0.

We can use it to find the numeric value of the constant c.

Substituting 0 for A and t in the equation gives

                         0 = \frac{C}{r}e^{0} - \frac{P}{r} \implies \frac{P}{r} = \frac{C}{r} \implies C=P

Therefore, the solution of the given differential equation is

                                   A = \frac{P}{r}e^{rt} - \frac{P}{r} = \frac{P}{r}\left( e^{rt} -1 \right)

4 0
3 years ago
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