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Tanya [424]
3 years ago
5

Help help help help pls

Mathematics
1 answer:
Yanka [14]3 years ago
3 0

Answer:

x - 3 > 10

Step-by-step explanation:

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Can I please have the answers.
Brums [2.3K]

Question 22.). Write the equation in Standard Form:

10y = 12x + 0.7

Identify: A, B, C

Standard Form for Linear Equation: Ax + By = C

10y = 12x + 0.7

The Standard Form of the solution is

Ax + By = C, Where, A ≥ 0

A and B, are Non- Zero numbers.

Therefore, Your answer, Letter Choice, (D), 120x - 100y = - 7, Where,

A = 120, B = - 100, and C = - 7

Question 24.). Described as, Inconsistent, Consistent, and Independent, Consistent, and Dependent, or None of These.

9x - 8y = 15, and 27x - 24y = 3

Solve:

9x - 8y = 15; 27x - 24y = 3

Steps: I will try to solve the system of equation:

9x - 8y = 15; 27x - 24y = 3

Step, Solve: 9x - 8y = 15 for x

9x - 8y + 8y = 15 + 8y

Add 8y to both sides:

9x = 8y + 15

9x/9 = 8y + 15/9 ( Divide both sides by 9):

x = 8/9y + 5/3

Step Substitute:

8/9y + 5/3 for X in 27x - 24y = 3

27x - 24y = 3

27 ( 8/9y + 5/3) - 24y = 3

45 = 3

Simplify Both sides of equation:

45 + - 45 = 3 + - 45 ( Add - 45, to both sides.):

0 = - 42

Therefore, No real Solutions, to both Equation.

Hence; 9x - 8y = 15; and 27x - 24 = 3, have No Real Solutions, and Therefore, They are, Answer Letter Choice, (B), Inconsistent.

Hope that helps!!! Sorry, I couldn't figure out Number (23.). : )

8 0
3 years ago
For positive test result, the number of those who did not lie and lie are 11 and 41, respectively. Those of the negative test re
vagabundo [1.1K]

P(subject lied | negative results) = 4/19

P(negative results | subject lied) = 8/49

 

I am hoping that these answers have satisfied your queries and it will be able to help you in your endeavors, and if you would like, feel free to ask another question.

3 0
3 years ago
Question 2 of 10<br> Which describes the graph of y = (x-4)2 - 1?
WITCHER [35]

y=(x-4)^2-1:   y=-\frac{4}{5}

Sure hope this helps you

y = -4/5

4 0
2 years ago
4. The reflecting dish of a parabolic microphone has a cross-section in the shape of a parabola. The microphone itself is placed
elixir [45]
Assume the parabola is placed on a graph where the x-axis is the top of the dish.
The vertex is then at (0,-30)  The x-intercepts or zeros are at (-30,0) and (30,0)

The equation of such parabola would be:
y = a(x+30)(x-30)
Plug in vertex to find value of 'a'
-30 = a(0+30)(0-30) \\  \\ a = \frac{-30}{(-30)(30)} = \frac{1}{30}

Now find the focus given that p = \frac{1}{4a}
p = \frac{1}{4(1/30)} = \frac{30}{4} = 7.5

Answer: the microphone should be placed 7.5 inches from vertex.
8 0
3 years ago
The graph of an exponential function is given. Which of the following is the correct equation of the function?
katen-ka-za [31]

Answer:

If one of the data points has the form  

(

0

,

a

)

, then a is the initial value. Using a, substitute the second point into the equation  

f

(

x

)

=

a

(

b

)

x

, and solve for b.

If neither of the data points have the form  

(

0

,

a

)

, substitute both points into two equations with the form  

f

(

x

)

=

a

(

b

)

x

. Solve the resulting system of two equations in two unknowns to find a and b.

Using the a and b found in the steps above, write the exponential function in the form  

f

(

x

)

=

a

(

b

)

x

.

EXAMPLE 3: WRITING AN EXPONENTIAL MODEL WHEN THE INITIAL VALUE IS KNOWN

In 2006, 80 deer were introduced into a wildlife refuge. By 2012, the population had grown to 180 deer. The population was growing exponentially. Write an algebraic function N(t) representing the population N of deer over time t.

SOLUTION

We let our independent variable t be the number of years after 2006. Thus, the information given in the problem can be written as input-output pairs: (0, 80) and (6, 180). Notice that by choosing our input variable to be measured as years after 2006, we have given ourselves the initial value for the function, a = 80. We can now substitute the second point into the equation  

N

(

t

)

=

80

b

t

to find b:

⎧

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎨

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎩

N

(

t

)

=

80

b

t

180

=

80

b

6

Substitute using point  

(

6

,

180

)

.

9

4

=

b

6

Divide and write in lowest terms

.

b

=

(

9

4

)

1

6

Isolate  

b

using properties of exponents

.

b

≈

1.1447

Round to 4 decimal places

.

NOTE: Unless otherwise stated, do not round any intermediate calculations. Then round the final answer to four places for the remainder of this section.

The exponential model for the population of deer is  

N

(

t

)

=

80

(

1.1447

)

t

. (Note that this exponential function models short-term growth. As the inputs gets large, the output will get increasingly larger, so much so that the model may not be useful in the long term.)

We can graph our model to observe the population growth of deer in the refuge over time. Notice that the graph below passes through the initial points given in the problem,  

(

0

,

8

0

)

and  

(

6

,

18

0

)

. We can also see that the domain for the function is  

[

0

,

∞

)

, and the range for the function is  

[

80

,

∞

)

.

Graph of the exponential function, N(t) = 80(1.1447)^t, with labeled points at (0, 80) and (6, 180).If one of the data points has the form  

(

0

,

a

)

, then a is the initial value. Using a, substitute the second point into the equation  

f

(

x

)

=

a

(

b

)

x

, and solve for b.

If neither of the data points have the form  

(

0

,

a

)

, substitute both points into two equations with the form  

f

(

x

)

=

a

(

b

)

x

. Solve the resulting system of two equations in two unknowns to find a and b.

Using the a and b found in the steps above, write the exponential function in the form  

f

(

x

)

=

a

(

b

)

x

.

EXAMPLE 3: WRITING AN EXPONENTIAL MODEL WHEN THE INITIAL VALUE IS KNOWN

In 2006, 80 deer were introduced into a wildlife refuge. By 2012, the population had grown to 180 deer. The population was growing exponentially. Write an algebraic function N(t) representing the population N of deer over time t.

SOLUTION

We let our independent variable t be the number of years after 2006. Thus, the information given in the problem can be written as input-output pairs: (0, 80) and (6, 180). Notice that by choosing our input variable to be measured as years after 2006, we have given ourselves the initial value for the function, a = 80. We can now substitute the second point into the equation  

N

(

t

)

=

80

b

t

to find b:

⎧

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎨

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎩

N

(

t

)

=

80

b

t

180

=

80

b

6

Substitute using point  

(

6

,

180

)

.

9

4

=

b

6

Divide and write in lowest terms

.

b

=

(

9

4

)

1

6

Isolate  

b

using properties of exponents

.

b

≈

1.1447

Round to 4 decimal places

.

NOTE: Unless otherwise stated, do not round any intermediate calculations. Then round the final answer to four places for the remainder of this section.

The exponential model for the population of deer is  

N

(

t

)

=

80

(

1.1447

)

t

. (Note that this exponential function models short-term growth. As the inputs gets large, the output will get increasingly larger, so much so that the model may not be useful in the long term.)

We can graph our model to observe the population growth of deer in the refuge over time. Notice that the graph below passes through the initial points given in the problem,  

(

0

,

8

0

)

and  

(

6

,

18

0

)

. We can also see that the domain for the function is  

[

0

,

∞

)

, and the range for the function is  

[

80

,

∞

)

.

Graph of the exponential function, N(t) = 80(1.1447)^t, with labeled points at (0, 80) and (6, 180).

Step-by-step explanation:

4 0
3 years ago
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