In 2015, Tamya spent 100 hours reading. In 2016, she has only read 9.5 hours. How many more hours does she have to read in order to also read 100 hours in 2016?​

The given Quadrilateral D is enlarged by a scale factor of 3.5 to produce its image D'.

The given quadrilateral D and its image D' have corresponding side lengths of 6, 8, 6, 8 and 21, 28, 21, 28 respectively.  the type of dilation shown and the scale factor used, we can compare the corresponding side lengths.

We can see that all the corresponding side lengths of D and D' are multiplied by the same factor of 3.5. Since the scale factor is greater than 1, the dilation is an enlargement.

Therefore, the type of dilation shown is an enlargement and the scale factor used is 3.5.  the given quadrilateral D is enlarged by a scale factor of 3.5 to produce its image D'.

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Sally's average annual percentage gain is approximately 5.26%.

To calculate Sally's average annual percentage gain, we can use the formula:

Average Annual Percentage Gain = (Profit / Initial Investment) * (1 / Time) * 100

Profit = $2500

Initial Investment = $19000

Time = 2.5 years

Substituting the values into the formula:

Average Annual Percentage Gain = (2500 / 19000) * (1 / 2.5) * 100

= (0.1316) * (0.4) * 100

= 5.26

Therefore, Sally's average annual percentage gain is approximately 5.26%.

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The vertex of point A' in the dilated triangle A'B'C' is (6, 4.5).

1. Start by calculating the distance between the vertices of the original triangle ABC:

  - Distance between A(4, 3) and B(3, -2):

    Δx = 3 - 4 = -1

    Δy = -2 - 3 = -5

    Distance = √((-\(1)^2\) + (-\(5)^2\)) = √26

  - Distance between B(3, -2) and C(-3, 1):

    Δx = -3 - 3 = -6

    Δy = 1 - (-2) = 3

    Distance = √((-6)² + 3²) = √45 = 3√5

  - Distance between C(-3, 1) and A(4, 3):

    Δx = 4 - (-3) = 7

    Δy = 3 - 1 = 2

    Distance = √(7² + 2²) = √53

2. Apply the scale factor of 1.5 to the distances calculated above:

  - Distance between A' and B' = 1.5 * √26

  - Distance between B' and C' = 1.5 * 3√5

  - Distance between C' and A' = 1.5 * √53

3. Determine the coordinates of A' by using the distance formula and the given coordinates of A(4, 3):

  - A' is located Δx units horizontally and Δy units vertically from A.

  - Δx = 1.5 * (-1) = -1.5

  - Δy = 1.5 * (-5) = -7.5

  - Coordinates of A':

    x-coordinate: 4 + (-1.5) = 2.5

    y-coordinate: 3 + (-7.5) = -4.5

4. Thus, the vertex of point A' in the dilated triangle A'B'C' is (2.5, -4.5).

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Answer:

30

Step-by-step explanation:

It took Emily 33.4 minutes to make dinner.

A mixed number is a kind of fraction that also has a proper fraction and a whole number. The number of whole units is represented by the whole number, and the fraction of a unit is represented by the proper fraction.

To solve the problem, we have to convert the mixed number  \(9 \frac{1}{10}\)   to a decimal number:

⇒ \(9 \frac{1}{10} = 9 +\frac{1}{10} = \frac{(9*10)+1}{10}\)

⇒ \(\frac{91}{10}\)  =  9.1

This means that Emily finished making dinner 9.1 minutes sooner than the cake.

To find out how long it took Emily to make dinner, we can subtract 9.1 from the cake baking time:

⇒ 42.5 - 9.1 = 33.4 minutes

Therefore, it took Emily 33.4 minutes to make dinner.

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The amounts of koalas are given as follows:

a. Initially: 47.

b. 13 weeks: 21.

c. 23 weeks: 19.

d. The virus will not kill all the koalas, as when t -> ∞, N(t) -> ∞.

To calculate the numeric value of a function or of an expression, we substitute each instance of any variable or unknown on the function by the value at which we want to find the numeric value of the function or of the expression presented in the context of a problem.

The function for this problem is defined as follows:

N(t) = 15 + 96/(t + 3).

For the initial number of koalas, we have that t = 0, hence:

N(0) = 15 + 96/3 = 47.

After 13 weeks, the number is given as follows:

N(13) = 15 + 96/16 = 21.

After 23 weeks, the number is given as follows:

N(23) = 15 + 96/26 = 19.

As t -> ∞, we have that:

N(∞) = 15 + 96/∞ = 15 + 0 = 15.

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Answer:

110 feet

Step-by-step explanation:

Take x as the width of the field.

4x = 440 feet

x = 440 feet ÷ 4

x = 110 feet

So, the field is 110 feet wide #

Answer:

Step-by-step explanation:

\(5^5 - 2^3\\\\\mathrm{Follow\:the\:PEMDAS\:order\:of\:operations}\\\\\mathrm{Calculate\:exponents}\:5^5\::\quad 3125\\\\=3125-2^3\\\\\mathrm{Calculate\:exponents}\:2^3\::\quad 8\\\\=3125-8\\\\\mathrm{Add\:and\:subtract\:\left(left\:to\:right\right)}\:3125-8\:\\\\:\quad 3117\)

Answer:

7 nights

Step-by-step explanation:

If he does 1/3 every night then you must turn 2 1/3 into a mixed fraction.

7/3

Then divide.

7/3 by 1/3

your answer is 7/1 or just 7 nights

Just divide

No of nights

He need 7 nights

Answer:

c. -20x and 20x

Step-by-step explanation:

x² is the square of x.

100 is the square of 10 and of -10.

The middle term of (a + b)² is 2ab.

For b = 10:

2ab = 20x

for b = -10:

2ab = -20x

Answer: c. -20x and 20x

Answer:

Step-by-step explanation:

The vertex is at (-3, -1)  and the leading coefficient is negative ( because  the curve is inverted)

y = -a(x + 3)^2 - 1

One point on the curve is (-2, -3) so

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

-3 = -a - 1

-a = -2

So we have

y = -2(x + 3)^2 -  1

y = -2(x^2 + 6x + 9) - 1

y = -2x^2 - 12x - 19.

Answer:

.

Step-by-step explanation:

Answer:

Step-by-step explanation:

If the diameter is 4 meters, then the radius has to be 2 because the radius is half of the diameter.

so the idea being, we have a system of equations of two variables and 4 equations, each one rendering a line, for this case these aren't equations per se, they're INEquations, so pretty much the function will be the same for an equation but we'll use > or < instead of =, but fairly the function is basically the same, the behaviour differs a bit.

we have a line passing through (-6,0) and (0,8), side one

we have a line passing through the x-axis and -6, namely (-6,0) and the y-axis and -4, namely (0,-4), side two

we have a line passing through (0,-4) and (6,4), side three

now, side four is simply the line connecting one and three.

the intersection of all four lines looks like the one in the picture below, so what are those lines with their shading producing that quadrilateral?

well, we have two points for all four, and that's all we need to get the equation of a line, once we get the equation, with its shading like that in the picture, we'll make it an inequality.

\((\stackrel{x_1}{-6}~,~\stackrel{y_1}{0})\qquad (\stackrel{x_2}{0}~,~\stackrel{y_2}{8}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{\textit{\large rise}} {\stackrel{y_2}{8}-\stackrel{y1}{0}}}{\underset{\textit{\large run}} {\underset{x_2}{0}-\underset{x_1}{(-6)}}} \implies \cfrac{8 -0}{0 +6} \implies \cfrac{ 8 }{ 6 } \implies \cfrac{4}{3}\)

\(\begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{0}=\stackrel{m}{ \cfrac{4}{3}}(x-\stackrel{x_1}{(-6)}) \implies y -0 = \cfrac{4}{3} ( x +6) \\\\\\ y=\cfrac{4}{3}x+8\hspace{5em}\stackrel{\textit{side one} }{\boxed{y < \cfrac{4}{3}x+8}}\)

\(\rule{34em}{0.25pt}\\\\ (\stackrel{x_1}{-6}~,~\stackrel{y_1}{0})\qquad (\stackrel{x_2}{0}~,~\stackrel{y_2}{-4}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{\textit{\large rise}} {\stackrel{y_2}{-4}-\stackrel{y1}{0}}}{\underset{\textit{\large run}} {\underset{x_2}{0}-\underset{x_1}{(-6)}}} \implies \cfrac{-4 -0}{0 +6} \implies \cfrac{ -4 }{ 6 } \implies - \cfrac{2}{3}\)

\(\begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{0}=\stackrel{m}{- \cfrac{2}{3}}(x-\stackrel{x_1}{(-6)}) \implies y -0 = - \cfrac{2}{3} ( x +6) \\\\\\ y=-\cfrac{2}{3}x-4\hspace{5em}\stackrel{\textit{side two} }{\boxed{y > -\cfrac{2}{3}x-4}} \\\\[-0.35em] \rule{34em}{0.25pt}\)

\(\stackrel{slope}{m}\implies \cfrac{\stackrel{\textit{\large rise}} {\stackrel{y_2}{4}-\stackrel{y1}{(-4)}}}{\underset{\textit{\large run}} {\underset{x_2}{6}-\underset{x_1}{0}}} \implies \cfrac{4 +4}{6 -0} \implies \cfrac{ 8 }{ 6 } \implies \cfrac{4}{3}\)

\(\begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{(-4)}=\stackrel{m}{ \cfrac{4}{3}}(x-\stackrel{x_1}{0}) \implies y +4 = \cfrac{4}{3} ( x -0) \\\\\\ y=\cfrac{4}{3}x-4\hspace{5em}\stackrel{ \textit{side three} }{\boxed{y > \cfrac{4}{3}x-4}} \\\\[-0.35em] \rule{34em}{0.25pt}\)

\((\stackrel{x_1}{6}~,~\stackrel{y_1}{4})\qquad (\stackrel{x_2}{0}~,~\stackrel{y_2}{8}) ~\hfill~ \stackrel{slope}{m}\implies \cfrac{\stackrel{\textit{\large rise}} {\stackrel{y_2}{8}-\stackrel{y1}{4}}}{\underset{\textit{\large run}} {\underset{x_2}{0}-\underset{x_1}{6}}} \implies \cfrac{ 4 }{ -6 } \implies - \cfrac{2}{3}\)

\(\begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{4}=\stackrel{m}{- \cfrac{2}{3}}(x-\stackrel{x_1}{6}) \\\\\\ y=-\cfrac{2}{3}x+8\hspace{5em}\stackrel{ \textit{side four} }{\boxed{y < -\cfrac{2}{3}x+8}}\)

now, we can make that quadrilateral a trapezoid by simply moving one point for "side four", say we change the point (0 , 8) and in essence slide it down over the line to  (-3 , 4).  Notice, all we did was slide it down the line of side one, that means the equation for side one never changed and thus its inequality is the same function.

now, with the new points for side for of (-3,4) and (6,4), let's rewrite its inequality

\((\stackrel{x_1}{-3}~,~\stackrel{y_1}{4})\qquad (\stackrel{x_2}{6}~,~\stackrel{y_2}{4}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{\textit{\large rise}} {\stackrel{y_2}{4}-\stackrel{y1}{4}}}{\underset{\textit{\large run}} {\underset{x_2}{6}-\underset{x_1}{(-3)}}} \implies \cfrac{4 -4}{6 +3} \implies \cfrac{ 0 }{ 9 } \implies 0\)

\(\begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{4}=\stackrel{m}{ 0}(x-\stackrel{x_1}{(-3)}) \implies y -4 = 0 ( x +3) \\\\\\ y=4\hspace{5em}\stackrel{ \textit{side four changed} }{\boxed{y < 4}}\)

1a. The percentage total return is -19.56%

1b. The dividend yield is 2.42%.

1c. The capital gains yield is -21.98%.

2a. The arithmetic average annual return on large-company stocks in nominal terms is 14.5%.

2b. The arithmetic average annual return on large-company stocks in real terms is 9.67%.

3a. The real return on long-term government bonds is 3.195%

3b. The real return on long-term corporate bonds is 3.291%

In Financial accounting, the percentage total return (P) can be calculated by using this formula;

P = [(Ending price - Initial price) + Dividend] ÷ Initial price

P = [(71 - 91) + 2.20] ÷ 91

P = -0.1956 or -19.56%

1b. For the dividend yield, we have:

Dividend yield = Dividend ÷ Initial price

Dividend yield = 2.20 ÷ 91

Dividend yield = 0.0242 or 2.42%

1c. For capital gains yield, we have:

Capital gains yield = (Ending price - Initial price) ÷ Initial price

Capital gains yield = (71 - 91) ÷ 91

Capital gains yield = -0.2198 or -21.98%.

Part 2.

a. The arithmetic average of annual return on large-company stocks in nominal terms is equal to 14.5%.

b. For arithmetic average annual return in real terms, we have:

(1 + 0.145) = (1 + r)(1 + 0.044)

r = (1.145/1.044) - 1

r = 9.67%

Part 3.

a. For real return on the long-term government bonds, we would apply Fisher equation:

(1 + i) = (1 + r)(1 + h)

Where:

(1 + 0.066) = (1 + r)(1 + 0.033)

1 + r = 1.066/1.033

r = 3.195%

b. For real return on long-term corporate bonds:

(1 + 0.067) = (1 + r)(1 + 0.033)

1 + r = 1.067/1.033

r = 3.291%

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Answer:

31.4 cm

Step-by-step explanation:

\(C=2\pi r\)

\(=2*\pi *5\)

\(\neq 31.41593\)

≈ 31.4

Answer:

hummmm

Step-by-step explanation:

Most of the time, an inequality has more than one or even infinity solutions. For example the inequality: x>3 . The solutions of this inequality are "all numbers strictly greater than 3". ... The inequality has an infinite amount of solutions

in inequality u can not cross multiply

The value of 4 Superscript negative 4 is negative StartFraction 1 Over 4 Superscript negative 4 EndFraction.

In the given table, Tasha observed a pattern in the powers of 4. When the exponent decreases by 1, the previous value is divided by 4. Using this pattern, she determined the values for 4 squared, 4 Superscript 1, 4 Superscript 0, 4 Superscript negative 1, and 4 Superscript negative 2.

To find the value of 4 Superscript negative 3, she divided the previous value (StartFraction 1 Over 16 EndFraction) by 4, resulting in StartFraction 1 Over 64 EndFraction.

Similarly, for 4 Superscript negative 4, she divided the previous value (StartFraction 1 Over 64 EndFraction) by 4, yielding StartFraction 1 Over 256 EndFraction.

Finally, to rewrite the value for 4 Superscript negative 4, she expressed it as negative StartFraction 1 Over 4 Superscript negative 4 EndFraction.

Therefore, the value of 4 Superscript negative 4 is negative StartFraction 1 Over 4 Superscript negative 4 EndFraction, which simplifies to StartFraction 1 Over 256 EndFraction

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Using linear approximation, the expression √125.2 can be approximated as approximately 10.02.

To find the equation of the tangent line to the function f(x) = √x at x = 125, determine the slope (m) and the y-intercept (b) of the tangent line.

First, find the slope (m) using the derivative of the function f(x) = √x:

f'(x) = 1 / (2√x)

Evaluate the derivative at x = 125:

f'(125) = 1 / (2√125) = 1 / (2 × 5) = 1/10

Now, find the y-coordinate of the point on the function f(x) at x = 125:

f(125) = √125 = 5

So, the tangent line at x = 125 passes through the point (125, 5) and has a slope of 1/10.

Using the point-slope form of a linear equation, the equation of the tangent line can be written as:

y - 5 = (1/10)(x - 125)

To simplify, multiply through by 10:

10y - 50 = x - 125

Rearranging the equation, express it in the form y = mx + b:

y = (1/10)x - 75/10 + 5

Simplifying further:

y = (1/10)x - 75/10 + 50/10

Combining the terms:

y = (1/10)x - 25/10

Therefore, the equation of the tangent line to f(x) = √x at x = 125 is y = (1/10)x - 25/10.

Now, use this tangent line to approximate the value of √125.2:

Substitute x = 125.2 into the equation:

y = (1/10)(125.2) - 25/10

y = 12.52 - 2.5

y ≈ 10.02

So, using linear approximation, approximate √125.2 as approximately 10.02.

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can you see my messages?

Answer:

x=-1, y=4. (-1, 4).

Step-by-step explanation:

-6x-y=2

-7x-4y=-9

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

y=-6x-2

-7x-4(-6x-2)=-9

-7x+24x+8=-9

17x=-9-8

17x=-17

x=-17/17

x=-1

-6(-1)-y=2

6-y=2

y=6-2=4

Answer:

Step-by-step explanation:

Based on this information, you could conclude that the annual rainfall in Falls City is generally more than that of the annual rainfall in Rockaway Beach. This is based on the information provided since the numbers given are the mean annual rainfall. Meaning that this is the average amount of rain that falls in any given year in that specific location. Since the amount provided for Falls City is a larger number then it means it gets more rainfall than Rockaway Beach on average and would therefore be the safest conclusion that can be made.

Answer:

I hope this helps

Step-by-step explanation:

y=-3x+9

explanation

m=-3 (X1,Y1) = (4,-3)

(y-y1)= m×(x-x1)

(y-(-3))=-3(x-4)

y+3= -3x+12

y=-3x+12-3

y=-3x+9

Answer: I can't really tell you the answer because you didn't show the graphs...

Step-by-step explanation:

A boat's value over time is given as the exponential function f(x) = 400(0.6)ˣ

An exponential function is in the form:

y = abˣ

Where a is the initial value and b is the multiplication factor.

Given that the boat's value decreasing at a rate of 40% per year, hence:

b = 100% - 40% = 60% = 0.6

A boat's value over time is given as the exponential function f(x) = 400(0.6)ˣ

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Answer:

24

Step-by-step explanation:

the two times the cut into one