4. Evaluate e^-2 to one decimal place
A) 0.7
B) -1.4
C) 0.1
D) 1.4

Considering the definition of exponential function, the value of the car is 11,287.25 dollars.

An exponential function is one in which the independent variable x appears in the exponent and has a constant a as its base. Its expression is:

f(x)= aˣ

Being a a positive real, a > 0, and different from 1, a ≠ 1.

When 0 < a < 1, then the exponential function is a decreasing function and when a > 1, it is an increasing function.

In this case, the value of that car depreciates based on the function f(t)=12,000×[tex]0.96^{t}[/tex] where t is measured in years after purchase.

After 3/2 years, the value of the car is calculated as:

f(3/2)=12,000×[tex]0.96^{3/2}[/tex]

Solving:

f(3/2)= 11,287.25

Finally, the value of the car is 11,287.25 dollars.

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Answer:

1) 5
2) 0.8
3)  -1.3

4) -[tex]\frac{4}{9}[/tex]

5) ±10

6) ±[tex]\frac{7}{12}[/tex]

Step-by-step explanation:

Answer:

51 pounds, 48 pounds, 44 pounds and 40 pounds in that chronological order.

Answer:

i believe the answer is A

Step-by-step explanation:

The values of a, b and c from the given exponential function are 7, 9 and 4 respectively

According to the exponential law of indices

[tex]\sqrt[c]{a^b}[/tex]

This can be written as;

[tex]\sqrt[c]{a^b}=a^{\frac{b}{c} }[/tex]

Given the exponential expression

[tex]7^\frac{9}{4}[/tex]

Compare with the original expression

a = 7, b = 9 and c = 4

Hence the values of a, b and c from the given exponential function are 7, 9 and 4 respectively

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Answer:

tiookvgvc. jbjvth kivtcth jjvf h. bkbgv

Answer:

The IQR of the given distribution is

Step-by-step explanation:

The given distribution has the five-number

28, 34, 43, 59, 62

Divide these numbers in two equal parts.

(28, 34), 43,( 59, 62)

Now divide each parenthesis in two equal parts.

(28), (34), 43,( 59), (62)

It means first quartile is the average of 28 and 34. Third quartile is the average of 59 and 62.

The interquartile range (IQR) of this distribution is

Therefore the IQR of the given distribution is 29.5.

Answer:

6, 8, 10

Step-by-step explanation:

Where:

a = 6

b = 8

c = 10

Pythagorean theorem: [tex]a^2 + b^2 = c^2[/tex]

[tex]6^2 + 8^2 = c^2[/tex]

[tex]36 + 64 = c^2[/tex]

[tex]100 = c^2[/tex]

c = [tex]\sqrt{100}[/tex]

c = 10

Answer:

  4 square units

Step-by-step explanation:

The vertices of the figure are on grid points, so it is appropriate to use Pick's theorem to find the area.

__

Pick's theorem tells you the area is ...

  A = i +b/2 -1

where i is the number of grid points interior to the figure (0), and b is the number of grid points on the boundary (10).

Using the counted values in the formula, we find the area to be ...

  A = 0 +10/2 -1 = 4

The area of the polygon is 4 square units.

_____

Additional comment

There are several other ways to find the area. Here are a couple:

decompose the figure

A horizontal line 1 unit up from the bottom will divide the figure into a trapezoid and a triangle. The trapezoid has bases 4 and 1, and height 1, so its area is ...

  A = 1/2(b1 +b2)h = 1/2(4 +1)(1) = 5/2

The triangle has base 1 and height 3, so its area is ...

  A = 1/2bh = 1/2(1)(3) = 3/2

Then the total area is 5/2 +3/2 = 8/2 = 4 square units.

subtract empty space

The figure occupies a 4×4 square with triangles removed from the left side and the top. Each of those triangles has a base of 4 and a height of 3. The remaining (shaded) area is ...

  A = s² -1/2bh -1/2bh

  A = 4² -1/2(4)(3) -1/2(4)(3) = 16 -12 = 4 square units

Answer:

C.) The variable x represents the independent variable

Step-by-step explanation:

The variable "x" is the independent variable because that is the value that we can manipulate. We cannot manipulate the dependent variable because it is determined by the independent variable. In that case, "y" would be the dependent variable because its value is dependent on what "x" is.

A.) is incorrect because g(x) symbolizes the input of the variable "x" into the g function.

B.) is incorrect because "x" is the input.

D.) is incorrect because "x" is the input.

(1) The missing term in the sequence, a₁₂  = 0.8.

(2) The missing term in the sequence, a₈ = 102.5.

(3) The missing term in the sequence, a₈ = 111.

(4) The missing term in the sequence, a₁₂ = -19.

(5) The missing term in the sequence, a₁₂ = 94.

(6)  The missing term in the sequence, a₆ = 40.

(7)  The missing term in the sequence, a₃₆ = -52.

(8)  The missing term in the sequence, a₂₁ = -58.

The missing term in the sequence is determined as follows;

Tₙ = a + (n - 1)d

T₄ = a + 3d

18.4 = a + 3d  ---(1)

T₅ = a + 4d

16.2 = a + 4d  ---(2)

subtract (1) from (2)

-2.2 = d

18.4 = a + 3(-2.2)

a = 25

a₁₂  = a + 11d

a₁₂  = 25 + 11(-2.2)

a₁₂  = 0.8

a₂ = a + d

57.5 = a + d -- (1)

a₅ = a + 4d

80 = a + 4d  --- (2)

solve (1) and (2)

d = 7.5

a = 50

a₈ =  a + 7d

a₈ = 50 + 7(7.5)

a₈ = 102.5

a₁₀ = a + 9d

141 = a + 9d --- (1)

a₁₃ = a + 12d

186 = a + 12d --- (2)

Subtract (1) from (2)

d = 15

a = 6

a₈ = a + 7d

a₈ = 6 + 7(15)

a₈ = 111

a₂₂ = a + 21d

-49 = a + 21d ---- (1)

a₂₅ = a + 24d

-58 = a + 24d --- (2)

subtract (1) from (2)

d = -3

a = 14

a₁₂ = a + 11d

a₁₂ = 14 + 11(-3)

a₁₂ = -19

a₄ = a + 3d

-2 = a + 3d --- (1)

a₈ = a + 7d

46 = a + 7d ---- (2)

Subtract (1) from (2)

d = 12

a = -38

a₁₂ = a + 11d

a₁₂ = -38 + 11(12)

a₁₂ = 94

a₉ = a + 8d

64 = a + 8d --- (1)

a₁₂ = a + 11d

88 = a + 11d --- (2)

Subtract (1) from (2)

d = 8

a = 0

a₆ = a + 5d

a₆ = 0 + 5(8)

a₆ = 40

a₂₀ = a + 19d

-4 = a + 19d ---- (1)

a₂₃ = a + 22d

-13 = a + 22d --- (2)

Subtract (1) from (2)

d = -3

a = 53

a₃₆ = a + 35d

a₃₆ = 53 + 35(-3)

a₃₆ = -52

a₂₈ = a + 27d

5 = a + 27d ---- (1)

a₃₃ = a + 32d

50 = a + 32d --- (2)

Subtract (1) from (2)

d = 9

a = -238

a₂₁ = a + 20d

a₂₁ = -238 + 20(9)

a₂₁ = -58

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The value of this equation is x = 0

The representation of an equation of the first degree - is given by:

[tex] \boxed{ \large \sf ax + b = 0}[/tex]

This representation is also defined in a first degree function.

— To solve this expression, let's: add or subtract the real terms. Then we will do the division.

⚘ Resolution

[tex] \large \sf{-3+8x-5=-8}[/tex]

[tex] \large \sf{-8+8x=-8}[/tex]

[tex] \large \sf{8x=0}[/tex]

[tex] \large \sf{x=0 \div 8}[/tex]

[tex] \green{ \boxed{ \boxed{ \blue{ \large \sf{x=0}}}}} \\ [/tex]

Therefore, the value of this equation will be x = 0

Answer:

D) 9

Step-by-step explanation:

x = ½ × 13 = 6.5

y = ½ × 5 = 2.5

6.5 + 2.5 = 9

So the length of PC is 9

HOPE THIS HELPS AND HAVE A NICE DAY <3

Answer:

By proportional method

[tex] \frac{4}{6} = \frac{5}{x} [/tex]

4x= 30

x= 7.5

Answer:

175 ft^2

Step-by-step explanation:

split the pentagon into 5 triangles with base length 10ft and height 7ft. each triangle then has an area of 10 * 7 * 1/2 = 35 ft^2

then the pentagon has an area 35*5 = 175 ft^2

Answer: 0.5 or - 0.834

Step-by-step explanation: Here is the explanation!

Answer:

r = - 18/5

I hope this helps! ^^

( This problem has 13 steps, so it would take a lot of time to type them in ^^)

Answer:

[tex]x=\frac{3}{4},\:x=-1[/tex]

Keys:

For this problem, you need the quadratic formula(listed below).

When you see ± in a quadratic equation, you must know there is going to be at least 2 solutions.

Step-by-step explanation:

solving for x₁ and x₂

[tex]4x^2+x=3\\4x^2+x-3=3-3\\4x^2+x-3=0\\x_{1,\:2}=\frac{-1\pm \sqrt{1^2-4\cdot 4\left(-3\right)}}{2\cdot 4}\\[/tex]

[tex]1^2=1\\=\sqrt{1-4\cdot \:4\left(-3\right)}\\=\sqrt{1+4\cdot \:4\cdot \:3}\\=\sqrt{1+48}\\=\sqrt{49}\\=\sqrt{7^2}\\\sqrt{7^2}=7\\=7[/tex]

[tex]x_{1,\:2}=\frac{-1\pm \:7}{2\cdot \:4}\\x_1=\frac{-1+7}{2\cdot \:4},\:x_2=\frac{-1-7}{2\cdot \:4}\\[/tex]

solve for x₁

[tex]\frac{-1+7}{2\cdot \:4}[/tex]

[tex]=\frac{6}{2\cdot \:4}[/tex]

[tex]=\frac{6}{8}[/tex]

[tex]= \frac{6\div2}{8\div2}[/tex]

[tex]=\frac{3}{4}[/tex]

solve for x₂

[tex]\frac{-1-7}{2\cdot \:4}[/tex]

[tex]=\frac{-8}{2\cdot \:4}[/tex]

[tex]=\frac{-8}{8}[/tex]

[tex]=-\frac{8}{8}[/tex]

[tex]=-1[/tex]

Hope this helps!

A function assigns the values. The function will move down by 3 units.

A function assigns the value of each element of one set to the other specific element of another set.

The graph of the function is given below. Now if 3 is replaced by a 0, then the y-intercept of the function will become zero while keeping the slope the same. Therefore, the function will move down by 3 units.

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Answer: Yes it is possible

Example: Yeast is a single-celled fungus.

There are probably other types of fungus that are single-celled. However, some other fungi are multi-celled. You will likely need more information about the organism under the microscrope before you can classify it properly.

Answer: x=15, y=3

Step-by-step explanation:

Subtracting the two equations, we get -3y=-9, meaning y=3.

Substituting this into the first equation, we get that x-3=12, and thus, x=15.

Answer:

w = 5

Step-by-step explanation:

24w = (14)10

2w + 4 = 10 + 4

2w = 10

w = 5

Answer:

area = 62ft

Step-by-step explanation

so you section the polygon into a big rectangle and 2 small rectangles

the area of the small rectangle will be 2ft x 3ft which will be 6 ft, and as there are 2 of them the area of it will be 12ft

the area of the big rectangle will be 10ft x 5ft, as 8ft - 3ft is 5ft so 5x10=50ft

now add the areas you got, 50+12 which is 62ft

hope this helps:)

Answer:

Hi,

Step-by-step explanation:

sin(a-b)=sin(a) cos(b)+ cos(a) sin(b)

a=45° and b=30°

[tex]sin(45^o-30^o)=sin(45^o)*cos(30^o)+cos(45^o)*sin(30^o)\\\\=\dfrac{\sqrt{2}}{2}* \dfrac{\sqrt{3}}{2}+\dfrac{\sqrt{2}}{2}*\dfrac{1}{2}\\\\\\\boxed{sin(15^o)=\dfrac{\sqrt{2}}{4}*(1+\sqrt{3} )}\\[/tex]

If the definitions of type I and type II regions is the same as in the link provided, then as a type I region the integration domain is the set

[tex]R_{\rm I} = \left\{(x,y) \mid 0 \le x \le 3 \text{ and } \sqrt{x+1} \le y \le 2\right\}[/tex]

and as a type II region,

[tex]R_{\rm II} = \left\{(x,y) \mid 0 \le x \le y^2-1 \text{ and } 1 \le y \le 2\right\}[/tex]

where we solve y = √(x + 1) for x to get x as a function of y.

A. The area of the type I region is

[tex]\displaystyle \iint_{R_{\rm I}} dA = \int_0^3 \int_{\sqrt{x+1}}^2 dy \, dx = \int_0^3 (2 - \sqrt{x+1}) \, dx = \boxed{\frac43}[/tex]

B. The area of the type II region is of course also

[tex]\displaystyle \iint_{R_{\rm II}} dA = \int_1^2 \int_0^{y^2-1} dx \, dy = \int_1^2 (y^2-1) \, dy = \boxed{\frac43}[/tex]

I've attached a plot of the type II region to give an idea of how it was determined. The black arrows indicate the domain of x as it varies from the line x = 0 (y-axis) to the curve y = √(x + 1).

Answer:

67m²

Step-by-step explanation:

12m + 6m + 4m + 36 m 8m = 67m²

Two numbers multiply the value on top (-84) and add to the value on the bottom (17) is 21 and -4.

Multiplication is one of the four basic arithmetic operations, alongside addition, subtraction, and division. Multiplication essentially means the repeated addition.

The two given numbers are -84 and 17.

We need to write two numbers that multiply to the value on top (-84) and add to the value on the bottom (17).

If we multiply 21 and -4, we get the product as -84.

That is, 21×(-4)=-84

If we add 21 and -4, we get the sum as -17.

That is, 21+(-4)=21-4=17

Therefore, two numbers multiply the value on top (-84) and add to the value on the bottom (17) is 21 and -4.

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Answer:

The angle Sita here is called Amplitude