how to Solve for x. -1/2 (x + 2) + 1 1/2 x = 3 ​

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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the exact value of sin A  to the nearest ten- thousandth is 0. 6625

a² + b² = c²

The opposite side is unknown, so use the pythagorean theorem to find it

c = hypotenuse = 4

a= opposite site = ?

b= adjacent side = 3

Substitute into the formula

4² = a² + 3²

16 = a² + 9

a² = 16 -9 = 7

Find the square root

a =√7 = 2. 65

To find Sin A, use

Sin A = opposite side ÷ hypotenuse

Sin A = 2. 65 ÷ 4 = 0. 6625

Thus,  the value of sin A is 0. 6625

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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!

(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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Answer: 0.5 or - 0.834

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

Answer:

67m²

Step-by-step explanation:

12m + 6m + 4m + 36 m 8m = 67m²

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.

[tex]35 - 12 + \frac{6 - 36}{4 } \\ 35 - 12 + \frac{216}{4} [/tex]

[tex] 35 - 12 + \frac{216}{4} \\ 35 - 12 + 54[/tex]

[tex]35 - 12 + 54 \\ 77[/tex]

Answer:

hope you can understand

Answer:

3m²n⁴

Step-by-step explanation:

Given :

⇒ [tex]\sqrt[4]{81m^{8}n^{16} }[/tex]

=============================================================

Solving :

⇒ [tex]\sqrt[4]{81}[/tex] × [tex]\sqrt[4]{m^{8} }[/tex] × [tex]\sqrt[4]{n^{16} }[/tex]

⇒ [tex]\sqrt[4]{3^{4} }[/tex] × [tex]\sqrt[4]{(m^{2})^{4} }[/tex] × [tex]\sqrt[4]{(n^{4})^{4} }[/tex]

⇒ 3 × m² × n⁴

⇒ 3m²n⁴

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:

[tex] - \infty < x < \infty [/tex]

Step-by-step explanation:

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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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).

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:

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:

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

The ratio of the lengths of and is 1: 1

PQ parallel to RS

PR and PS; perpendicular to PQ and RS.

Perpendicular lines are lines that intersect at a 90 degrees angle. Parallel lines are lines that are always the same distance apart.

FOR parallel lines, the perpendicular line joining then is always the same; here the 4 angles formed are all 90°

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Answer:

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

For the first image, I would recommend using Desmos; it is an online graphing calculator. You can find it on online and it is free; I use it all the time for my school work

I would input each function and sees which one matches the image on the test.

As for the second image, I believe the answer is C; x is greater than or equal to -1.

For the third and last image, I believe the answer is A; x is less than -1.

I 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]

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:

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.

Answer:

$130

Step-by-step explanation:

120 x (1 + 1/12) = 120 x 13/12 = 130

Answer:

a negative linear association (C)

Re-Written Problem [Formatted]

Teams with varying numbers of players are playing a group card game. the time taken by each team to complete one game is given in the table. What is the relationship between the two variables?

Numbers of players on team are going to be 1, 2 , 3, 4, 5.

time taken to complete game (in minutes) 32, 26, 20, 14, 8

(Select the correct answer)

A. Positive linear association

B. Exponential relationship

C. Negative linear association

D. No relationship

----------------------------------------------------------------------------------

Step-by-step explanation:

Lower x values (numbers of players on team) are correlated with higher y values (time taken to complete game)

[note: I know which values are x and y because of a cause-and-effect relationship. x causes y. number of players causes a change in minutes {it wouldn't make sense for the number of time it took to cause the number of players} ]

So, if we were to graph this relationship, we would see a downward-slope / negative slope of a line, because greater x-values are associated with lower y-values.

So, you would see a negative line, showing a negative linear association.

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:

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.