what is the solution to the equation?

Answer:

4y

They would have the same value if a number was substituted for y

Step-by-step explanation:

y+y+2y =

Combine like terms

4y

These are all like terms

They would have the same value if a number was substituted for y

Let y = 5

5+5+2(5) = 5+5+10 = 20

4(5) =20

Answer:

582-600

1,234-1,200

640-600

770-800

1,104-1,100

Answer:

Step-by-step explanation:

Assuming the system is solvable in the first place, create an augmented matrix of coefficients from the equations. Then put the matrix into reduced row echelon form.

Example is attached.

"Demonstrate by solving the following system."

You need to provide the system of equations.

Answer:

b

Step-by-step explanation:

b is  correct.

Answer:

[tex]A(x) = 12500(1.054)^x[/tex]

Step-by-step explanation:

Exponential equation for population growth:

Considering a constant growth rate, the population, in x years after 2005, is given by:

[tex]A(x) = A(0)(1 + r)^x[/tex]

In which A(0) is the population in 2005 and r is the growth rate, as a decimal.

The initial population of the town was estimated to be 12,500 in 2005.

This means that [tex]A(0) = 12500[/tex]

The population has increased by about 5.4% per year since 2005.

This means that [tex]r = 0.054[/tex]

So

[tex]A(x) = A(0)(1 + r)^x[/tex]

[tex]A(x) = 12500(1 + 0.054)^x[/tex]

[tex]A(x) = 12500(1.054)^x[/tex]

Answer:

1.083

Step-by-step explanation:

Exact form: 13/12

Decimal form: 1.083 (put a line above the 3)

Mixed number form: 1 1/12

Answer:

We are given the function:

[tex]f(x)=2x^2-x-10[/tex]

[tex]Here,\\a=2, b=-1,c=-10[/tex]

1. X-intercepts are the points at which the graph of a function intersects or cuts the x-axis. Since the x-intercept always lies on the x-axis, its ordinate or y-coordinate will always be 0. Since the function is quadratic, it will have at most 2 x-intercepts.

In order to find the x intercept, we basically solve for x at y=0:

[tex]f(x)=2x^2-x-10\\As\ y=0,\\0=2x^2-x-10\\2x^2-x-10=0\\ 2x^2-5x+4x-10=0\\x(2x-5)+2(2x-5)=0\\(x+2)(2x-5)=0\\Hence,\\Individually:\\x=-2,\ x=\frac{5}{2}[/tex]

Hence, the x-intercepts of the parabola of f(x) is (-2,0),(2.5,0)

2. The vertex of parabola is determined as maximum or minimum, solely on how it opens. This depends on the nature of the co-efficient of the x^2 term or 'a'. If a is positive the parabola opens upwards (minimum point) and downwards (maximum point) if negative. Hence, here as a=2, the parabola opens upwards and its vertex is minimum.

[tex]Vertex=(\frac{-b}{2a},\frac{-D}{4a})\\Hence,\\D=b^2-4ac\\Substituting\ a=2,b=-1,c=-10:\\D=(-1)^2-4*2*-10=1+80=81\\Hence,\\Vertex\ of\ f(x)=(\frac{-(-1)}{2*2},\frac{-81}{4*2})=(\frac{1}{4},\frac{-81}{8})[/tex]

3. [Please refer to the attachment]

From the graph, we observe that the parabola cuts the x-axis at (-2,0),(2.5,0). Also, its clear that the axis of symmetry passes through [tex](\frac{1}{4},\frac{-81}{8})[/tex], which is its minimum point.

Answer:

A chord of a circle is 9cm long if it's distance from the centre of the circle is 5cm calculate the radius of the circle

Answer:

0.5

Step-by-step explanation:

E to E'

(0, 3) to (0, 1.5)  each term of E' is ½ of the corresponding term of E

N to N'

(-1, 1) to (-0.5, 0.5)  each term of N' is ½ of the corresponding term of N

U to U'

(2, -2) to (1, -1)  each term of U' is ½ of the corresponding term of U

V to V'

(1, -3) to (0.5, -1.5)  each term of V' is ½ of the corresponding term of V

Answer:

3.5 inches

Step-by-step explanation:

Sample mean basically means that we need to find the average of the samples.

So the formula for finding average is

Number of observations/ Number of Occurrences

So when we add the values together we get

42.

So there are 12 numbers

So, 42/12 =

3.5 inches

The sample mean of the heights of the plants in Susan's garden is

3.5 inches.

Here,

Susan randomly selected a sample of plants to determine the average height of the total 35 plants in her garden.

She measured the heights (in inches) of 12 randomly selected plants and recorded the data:

1.0, 1.4, 1.8, 2.0, 2.5, 3.5, 4.2, 4.5, 4.8, 5.0, 5.3, 6.0

We have to find the sample mean of the heights of the plants in Susan's garden.

What is Average?

Average value in a set of given numbers is the middle value, calculate as dividing the total of all values by the number of values.

Now,

The recorded data is;

1.0, 1.4, 1.8, 2.0, 2.5, 3.5, 4.2, 4.5, 4.8, 5.0, 5.3, 6.0

To find the sample mean of the heights of the plants in Susan's garden we have to find the average of the recorded data.

Formula for average = [tex]\frac{sum of number of observation}{ number of occurrence}[/tex]

Hence, Average = [tex]\frac{1.0+ 1.4+1.8+2.0+ 2.5+3.5+4.2+4.5+ 4.8+ 5.0+ 5.3+ 6.0}{12} = \frac{42}{12} = 3.5[/tex]

Therefore, The sample mean of the heights of the plants in Susan's garden is 3.5 inches.

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Answer:

(-2,0)

Step-by-step explanation:

The x intercept is the value when it crosses the x axis ( the y value is zero)

x = -2 and y =0

(-2,0)

50 - (16 + 19)

= 50 - 35

= 15m

9514 1404 393

Answer:

Step-by-step explanation:

For the purpose of selecting the appropriate tile, it is only necessary to figure the real part of the sum or product.

We notice that the second product (-xy) is -1/2 times the first product (2xy). This can let you find the answers on that basis alone. The only tiles with a (-1) : (2) relationship are (-29 -53i) : (58 +106i).

__

The sum -5x +y has a real part of -5(3) +7 = -8.

The sum 2x -3y has a real part of 2(3) -3(7) = 6 -21 = -15.

Hence the sequence of answers needed on the right side is as shown above.

_____

Additional comment

You know that arithmetic operations with complex numbers (multiplication and addition) are identical to those operations performed on any polynomials. That is, "i" can be treated as a variable. The simplification comes at the end, where any instances of i² can be replaced by -1.

  xy = (3 +8i)(7 -i) = 3·7 -3·i +8·7·i -8·i·i = 21 +53i -8i²

  = (21 +8) +53i . . . . replaced i² with -1, so -8i² = +8

  = 29 +53i

Answer:

11 meters

Step-by-step explanation:

First, we can say that the square has a side length of x. The perimeter of the square is 4x, and that is how much wire goes into the square. To maximize the area, we should use all the wire possible, so the remaining wire goes into the triangle, or (11-4x).

The area of the square is x², and the area of an equilateral triangle with side length a is (√3/4)a². Next, 11-4x is equal to the perimeter of the triangle, and since it is equilateral, each side has (11-4x)/3 length. Plugging that in for a, we get the area of the equilateral triangle is

(√3/4)((11-4x)/3)²

= (√3/4)(11/3 - 4x/3)²

= (√3/4)(121/9  - 88x/9 + 16x²/9)

= (16√3/36)x² - (88√3/36)x + (121√3/36)

The total area is then

(16√3/36)x² - (88√3/36)x + (121√3/36) + x²

= (16√3/36 + 1)x² -  (88√3/36)x + (121√3/36)

Because the coefficient for x² is positive, the parabola would open up and the derivative of the parabola would be the local minimum. Therefore, to find the maximum area, we need to go to the absolute minimum/maximum points of x (x=0 or x=2.75)

When x=0, each side of the triangle is 11/3 meters long and its area is

(√3/4)a² ≈ 5.82

When x=2.75, each side of the square is 2.75 meters long and its area is

2.75² = 7.5625

Therefore, a maximum is reached when x=2.75, or the wire used for the square is 2.75 * 4 = 11 meters

The length of the square must be 4 m in order to maximize the total area.

When we put the differentiation of the given function as zero and find the value of the variable we get maxima and minima.

We have,

Length of the wire = 11 m

Let the length of the wire bent into a square = x.

The length of the wire bent into an equilateral triangle = (11 - x)

Now,

The perimeter of a square = 4 side

4 side = x

side = x/4

The perimeter of an equilateral triangle = 3 side

11 - x = 3 side

side = (11 - x)/3

Area of square = side²

Area of equilateral triangle = (√3/4) side²

Total area:

T = (x/4)² + √3/4 {(11 -x)/3}² _____(1)

Now,

To find the maximum we will differentiate (1)

dT/dx = 0

2x/4 + (√3/4) x 2(11 - x)/3 x -1 = 0

2x / 4 - (√3/4) x 2(11 - x)/3 = 0

2x/4 - (√3/6)(11 - x) = 0

2x / 4 = (√3/6)(11 - x)

√3x = 11 - x

√3x + x = 11

x (√3 + 1) = 11

x = 11 / (1.732 + 1)

x = 11/2.732

x = 4

Thus,

The length of the square must be 4 m in order to maximize the total area.

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Answer:

4 units

Step-by-step explanation:

A'B' = 2 × AB

8 = 2×AB

AB = 8/2 = 4

Answer:

16 units

Step-by-step explanation:

The triangle will get bigger because it has a scale factor that is greater than 1.

k=scale factor

k<1=reduction

k>1=enlargement

Answer:

Perimeter

[tex]202 = x + x + 26 + 4x + 26 + 4x\\202-26-26=10x\\150=10x\\x=15[/tex]

Therefore, the dimensions are

Answer:

628 mi^3

Step-by-step explanation:

the volume of a cylinder is given by:

V = base area x height

thus,

V = (3.14)(5)^2(8)

V = (3.14)(25)(8)

V = 628 mi^3

the volume of the cylinder is 628 cubic miles

The volume of the cylinder is 628 cubic miles.

We have a cylinder of radius 5 mi and height 8 mi.

We have to find the volume of the figure and round it to nearest tenth.

The volume of cylinder is given by the formula -

Volume [tex]=\pi r^{2} h[/tex]

We can use the above formula to calculate the volume of cylinder. In our case -

r = 5 mi

h = 8 mi

Substituting the values in the formula -

Volume [tex]=\pi\times5^{2}\times 8\\\\[/tex] = 628 cubic miles.

Hence, the volume of the cylinder is 628 cubic miles.

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

Explanation:

Add up the values to get

1.96+1.81+1.97+1.83+1.87+1.84+1.85+1.94+1.96+1.81+1.86+1.95+1.89= 24.54

Then divide over 13 (the number of values) to get 24.54/13 = 1.8876923 which is approximate.

So the mean is approximately 1.8876923

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

Now make a spreadsheet as shown below

We have the first column as the x values, which are the original numbers your teacher provided. The second column is of the form (x-M)^2, where M is the mean we computed earlier. We subtract off the mean and square the result.

After we compute that column of (x-M)^2 values, we add them up to get what is shown in the highlighted yellow cell at the bottom of the column.

That sum is approximately 0.04403076924

Next, we divide that over n-1 = 13-1 = 12

0.04403076924 /12 = 0.00366923077

That is the sample variance. Apply the square root to this to get the sample standard deviation. This is the point estimate of the population standard deviation. As the name implies, it works for samples that estimate population parameters.

sqrt(0.00366923077) = 0.06057417576822

This rounds to 0.061 which is the final answer.

Answer:

C. 2x+11

Step-by-step explanation:

2x+11≈21 -55

2x+11≈21 -55

2x+11≈21 -55

2x+11≈21 -55

2x+11≈21 -55

Answer:

it simple as look

Step-by-step explanation:

He y I a m r a v I t ha n k S

Wowowowowowowowowk

The system of inequalities that is graphed is:

y ≤ - (2/3)*x

y < x - 3

So the correct option is B.

First, we can see that for both of the inequalities the shaded part is below the lines.

You also can see that the solid line (correspondent to the symbol ≤) is the one with a negative slope, and the dashed line (correspondent with the line <) is the one with a positive slope.

Only with that, we conclude that the correct option is B.

y ≤ - (2/3)*x

y < x - 3

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Solution :

Given information :

A sample of n = 10 adults

The mean failure was 24 and the standard deviation was 3.2

a). The formula to calculate the 95% confidence interval is given by :

[tex]$\overline x \pm t_{\alpha/2,-1} \times \frac{s}{\sqrt n}$[/tex]

Here, [tex]$t_{\alpha/2,n-1} = t_{0.05/2,10-1}$[/tex]

                       = 2.145

Substitute the values

[tex]$24 \pm 2.145 \times \frac{3.2}{\sqrt {10}}$[/tex]

(26.17, 21.83)

When the [tex]\text{sampling of the same size}[/tex] is repeated from the [tex]\text{population}[/tex] [tex]n[/tex] infinite number of [tex]\text{times}[/tex], and the [tex]\text{confidence intervals}[/tex] are constructed, then [tex]95\%[/tex] of them contains the [tex]\text{true value of the population mean}[/tex], μ in between [tex](26.17, 21.83)[/tex]

b). The formula to calculate 95% prediction interval is given by :

    [tex]$\overline x \pm t_{\alpha/2,-1} \times s \sqrt{1+\frac{1}{n}}$[/tex]

   [tex]$24 \pm 2.145 \times 3.2 \sqrt{1+\frac{1}{10}}$[/tex]

   (31.13, 16.87)

Answer:

soory i dont know just report me if you angry

Answer:

The volume of the resulting solid is π/2 cubic units.

Step-by-step explanation:

Please refer to the diagram below.

The shell method is given by:

[tex]\displaystyle V = 2\pi \int _a ^b r(x) h(x)\, dx[/tex]

Where the representative rectangle is parallel to the axis of revolution, r(x) is the distance from the axis of revolution to the center of the rectangle, and h(x) is the height of the rectangle.

From the diagram, we can see that r(x) = (2 - x) and that h(x) is simply y. The limits of integration are from a = 0 to b = 1. Therefore:

[tex]\displaystyle V = 2\pi \int_0^1\underbrace{\left(2-x\right)}_{r(x)}\underbrace{\left(x - x^2\right)}_{h(x)}\, dx[/tex]

Evaluate:

[tex]\displaystyle \begin{aligned} V&= 2\pi \int_0 ^1 \left(2x-2x^2-x^2+x^3\right) \, dx\\ \\ &= 2\pi\int _0^1 x^3 -3x^2 + 2x \, dx \\ \\ &= 2\pi\left(\frac{x^4}{4} - x^3 + x^2 \Bigg|_0^1\right) \\ \\ &= 2\pi \left(\frac{1}{4} - 1 + 1 \right) \\ \\ &= \frac{\pi}{2}\end{aligned}[/tex]

The volume of the resulting solid is π/2 cubic units.

Answer:

pi/2

Step-by-step explanation:

I always like to draw an illustration for these problems.

For shells method think volume of cylinder=2pi×r×h

Integrate(2pi(2-x)(x-x^2) ,x=0...1)

Multiply

Integrate(2pi(2x-2x^2-x^2+x^3 ,x=0...1)

Combine like terms

Integrate(2pi(2x-3x^2+x^3) ,x=0...1)

Begin to evaluate

2pi(2x^2/2-3x^3/3+x^4/4) ,x=0...1

2pi(x^2-x^3+x^4/4), x=0...1

2pi(1-1+1/4)

2pi/4

pi/2

Answer:

[tex]f(x) = \sqrt{x} + 2 \\ \\ g(x) = {x}^{2} + 1 \\ \\ f{g(x)} = \sqrt{ {x}^{2} + 1 } + 2 \\ \\ g{f(x)} = {( \sqrt{x} + 2 )}^{2} + 1[/tex]

Answer:

Step-by-step explanation:

I assume the sequence is 0, 2, 4, 6

nth term = 2(n-1)

Answer:

Step-by-step explanation:

Answer:

Step-by-step explanation:

Parabola: y=2x²

Let say (a,b) the focus and y=k the directrix.

[tex]formula\ to\ use:\ \boxed{y=\dfrac{(x-a)^2}{2(b-k)} +\dfrac{b+k}{2} }\\\\\left \{\begin {array}{ccc}a&=&0\\2&=&\dfrac{1}{2(b-k)} \\\dfrac{b+k}{2} &=&0\\\end {array} \right.\\\\\\\left \{\begin {array}{ccc}a&=&0\\b-k&=&\dfrac{1}{4} \\b+k &=&0\\\end {array} \right.\\\\\\\left \{\begin {array}{ccc}a&=&0\\b&=&\dfrac{1}{8} \\k &=&\dfrac{-1}{8} \\\end {array} \right.\\\\\\focus=(0,\dfrac{1}{8} )\\\\directrix:\ y=\dfrac{-1}{8}[/tex]