Keith has $500 in a savings account at the beginning of the summer. He wants to have at least $200 in the account by the end of the summer. He withdraws $35 each week for food, clothes and movie tickets. How many weeks can Keith withdraw money from his account?

Answer:

The answer options appear garbled.  If A is 4/9, then this is equal to 2/3.  The others I can't see/understand.

Step-by-step explanation:

1. The time taken for the water to fall that much is 0.2 s

2. The initial speed of the water is 4 m/s

h = ½gt²

0.2 = ½ × 9.8 × t²

0.2 = 4.9 × t²

Divide both side by 4.9

t² = 0.2 / 4.9

Take the square root of both side

t = √(0.2 / 4.9)

t = 0.2 s

s = ut

Divide both sides by t

u = s / t

u = 0.8 / 0.2

u = 4 m/s

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Question 4

[tex]\sin 30^{\circ}=\frac{y}{14\sqrt{3}}\\\\\frac{1}{2}=\frac{y}{14\sqrt{3}}\\\\\boxed{y=7\sqrt{3}}\\\\\\\\\cos 30^{\circ}=\frac{x}{14\sqrt{3}}\\\\\frac{\sqrt{3}}{2}=\frac{x}{14\sqrt{3}}\\\\\boxed{x=21}[/tex]

Question 5

[tex]\cos 45^{\circ}=\frac{x}{28}\\\\\frac{1}{\sqrt{2}}=\frac{x}{28}\\\\x=\frac{28}{\sqrt{2}}\\\\\boxed{x=14\sqrt{2}}\\\\\\\\\sin 45^{\circ}=\frac{y}{28}\\\\\frac{1}{\sqrt{2}}=\frac{y}{28}\\\\y=\frac{28}{\sqrt{2}}\\\\\boxed{y=14\sqrt{2}}[/tex]

The survey shows that adding up the daily media use figures to obtain a report of weekly media use leads to the staggering number of more than 60 hours a week of media use by 11- to 14-year-olds.

It should be noted that a survey simply means a list of questions that are used to get information about a group.

In this case, the daily media use figures to obtain a report of weekly media use leads to the staggering number of more than 60 hours a week of media use by 11- to 14-year-olds.

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a.

b.

A quadratic function is a function of the form y = ax² + bx + c

To write y = 5(x - 2)² + 7, we expand the bracket and simplify.

So, y = 5(x - 2)² + 7

y = 5(x² - 4x + 4) + 7

y = 5x² - 20x + 20 + 7

y = 5x² - 20x + 27

So,  the quadratic function y = 5(x - 2)² + 7 in standard form is y = 5x² - 20x + 27

The y intercept is gotten when x = 0.

So, y = 5x² - 20x + 27

y(0) = 5(0)² - 20(0) + 27

y(0) = 0 - 0 + 27

y(0) = 27

The y intercept is (0, 27)

Find the graph in the attachment

To write y = 3(x + 1)² + 12, we expand the bracket and simplify.

So, y = 3(x + 1)² + 12

y = 3(x² + 2x + 1) + 12

y = 3x² + 6x + 3 + 12

y = 3x² + 6x + 15

So, the quadratic function y = 3(x + 1)² + 12 in standard form is y = 3x² + 6x + 15

The y intercept is gotten when x = 0.

So, y = 3x² + 6x + 15

y(0) = 3(0)² - 6(0) + 15

y(0) = 0 - 0 + 15

y(0) = 15

The y-intercept is (0, 15)

Find the graph in the attachment

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The Polynomial Roots are given as follows:
 

x₁=−1.2085

x₂=0.2709+1.2576i

x₃=0.2709−1.2576i

It is to be noted that the Polynomial 3x³+ 2x2 + 3x + 6 has no real rational roots that can be found using the rational root text. Hence, this was solved using the Qubic Formulas.

The roots (also known as zeroes or solutions) of a polynomial P (x) P(x) P(x) are the x values for which P (x) P(x) P(x) equals zero.

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​

Answer: 12

Step-by-step explanation:

I got it right

Answer:

Step-by-step explanation:

Answer:

a.1

b. 17

Step-by-step explanation:

So the | | around an equation makes it an absolute value which can be defined as the "distance from zero" or in other words, if a value is positive, it stays positive, and if a value is negative, it becomes positive.

|-8 + 9| = |1| = 1

|-8| + 9 = 8 + 9 = 17

The answers to all the parts are given below.

The height equation is:

h(t) = - 16t^2 + 100t + 10

A) To find the maximum height, we must find the time where the velocity is 0, so we must derive it with respect to the time.

v(t) = - 2 * 16 * t + 100 = 0

t = 100/(2*16) = 3.125s

Now we put this time in our height equation:

h(3.125s) = -16*3.125^2 + 100*3.125 + 10 = 166.25 ft.

B) The lowest height of the potato will be h = 0ft when the potato hits the ground.

C) the lowest time that works for that function is t = 0s when the potato is fired by the gun.

D) the maximum time can be found when the potato hits the ground, after that point the equation does not work anymore, let's find it,

h(t) = 0 = -16t^2 +100t+10

we can solve it using Bhaskara's equation:

[tex]t=\frac{-100+-\sqrt{100^{2} +4*16*10} }{-2*16} =\frac{-100+-103.2}{-32}[/tex]

So the two solutions are:

t = 6.3s

t = -0.1s

We need to choose the positive time, as we already discussed that the minimum time that works for the equation is t = 0s.

so here the answer is t = 6.3s

E) the domain is  0s ≤ t ≤ 6.3s

The range is 0ft ≤ h ≤ 166.25 ft.

Therefore, the answers to all the parts are shown.

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The high school graduation is the dependent variable.

In this question,

Jordan is studying whether or not race affects high school graduation in utah.

A dependent variable is a variable in an expression that depends on the value of another variable. It is a variable that represents a quantity that changes based on other quantities being manipulated in an experiment. It is the variable being tested, and therefore, it is called the dependent variable.

In the above statement, the high school graduation gets affected by race. So, the the high school graduation is the dependent variable.

Hence we can conclude that the high school graduation is the dependent variable.

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The expanded form of the expression (x+3)(x-3) is x² - 9

Given the expression;

(x+3)(x-3)

We multiply each term in the second bracket by each term in the first bracket

x(x-3) + 3(x-3)

x² - 3x + 3x - 9

x² - 9

Therefore, the expanded form of the expression (x+3)(x-3) is x² - 9.

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the population of province is growing at the rate of 2.91%

the formula of the rate of growth of a population =

[(future population- initial population/initial population)*100]/time

population in 2016 = 4.2 milllion = 4200000

Population predicted in 20 years = 6.65 million = 6650000

so, initial population = 4200000

future population = 6650000

rate of growth = {(6650000- 4200000/4200000)*100]/20

                       = 58.33/20

                     = 2.91%

the population is growing at the rate of 2.91%

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Distributive property is the term for breaking a larger number apart into smaller numbers that can be multiplied together to get a specific result.

Firstly, to distribute means to divide something or give a share or part of something.

Distributive property explains that multiplying the sum of two or more addends by a number will give the same result as multiplying each addend individually by the number and then adding the products together.

Therefore, distributive property is the term for breaking a larger number apart into smaller numbers that can be multiplied together to get a specific result.

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Quadratic expressions are mathematical expressions that have the power of one of its alphabets being equal to and not more than two. The given expression in the question can be simplified as 81[tex]x^{4}[/tex] + 162[tex]x^{3}[/tex]  - 45[tex]x^{2}[/tex] - 54x - 112

Algebraic expressions are mathematical expressions that consist of given number(s) and alphabet(s). When the power (exponential) of one of the alphabets is given as two, then the expression can be classified as a quadratic expression.

Thus the given expansion of the quadratic expression can be simplified as follows:

([tex]9x^{2}[/tex] + 9x - 4)([tex]9x^{2}[/tex] + 9x - 10) - 72

But,

([tex]9x^{2}[/tex] + 9x - 4)([tex]9x^{2}[/tex] + 9x - 10) = [tex]9x^{2}[/tex]([tex]9x^{2}[/tex] + 9x - 10) + 9x([tex]9x^{2}[/tex] + 9x - 10) - 4([tex]9x^{2}[/tex] + 9x - 10)

               = 81[tex]x^{4}[/tex] + 81[tex]x^{3}[/tex] - 90[tex]x^{2}[/tex] + 81[tex]x^{3}[/tex] + 81 [tex]x^{2}[/tex] - 90x - 36 [tex]x^{2}[/tex] + 36x - 40

               = 81[tex]x^{4}[/tex] + 162[tex]x^{3}[/tex]  - 45[tex]x^{2}[/tex] - 54x - 40

So that we have;

(81[tex]x^{4}[/tex] + 162[tex]x^{3}[/tex]  - 45[tex]x^{2}[/tex] - 54x - 40) - 72  = 81[tex]x^{4}[/tex] + 162[tex]x^{3}[/tex]  - 45[tex]x^{2}[/tex] - 54x - 40 -72

                  = 81[tex]x^{4}[/tex] + 162[tex]x^{3}[/tex]  - 45[tex]x^{2}[/tex] - 54x - 112

Thus the simplification of the given expression is 81[tex]x^{4}[/tex] + 162[tex]x^{3}[/tex]  - 45[tex]x^{2}[/tex] - 54x - 112

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The area of the convex polygon is 21.5 square units

The vertices are given as:

(0,5), (-1,2), (4,4), (-3,-4) and (2,0)

The area is then calculated as:

A = 0.5 * |x1y2 - x2y1 +x2y3 - x3y2 + ....... |

So, we have:

A = 0.5 * |-1 * 5 - 2 * 0 + 0 * 4 - 5 * 4 + 4 * 0 - 4 * 2 +2 * -4 - 0 * -3 - 3 * 2 + 4 * 1|

Evaluate

A = 0.5 * |-43|

Remove the absolute bracket

A = 0.5 * 43

This gives

A = 21.5

Hence, the area of the convex polygon is 21.5 square units

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Answer:

[tex]h(x)=-0.11(3)^{-x}[/tex]

Step-by-step explanation:

Given function:

[tex]f(x)=0.11(3)^x[/tex]

Reflection in the x-axis:

When reflecting a function in the x-axis, make the whole function negative:

[tex]\implies g(x)=-f(x)[/tex]

[tex]\implies g(x)=-0.11(3)^x[/tex]

Reflection in the y-axis:

When reflecting a function in the y-axis, make the x variable negative:

[tex]\implies h(x)=g(-x)[/tex]

[tex]\implies h(x)=-0.11(3)^{-x}[/tex]

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Answer:

the answer to this question is d

Step-by-step explanation:

h(x)=-0.11(3)^-x

Jhon was between 6:12 p.m and 6:52 p.m. So, Jhon left home for exact 40 minutes. Using the angle made by the hands in the clock, the required value is calculated.

The general formula for finding the angle between the minute hand and the hour hand is

M = [tex]\frac{2}{11}[/tex] (H × 30 ± θ)

It is given that,

Jhon left home after 6 p.m with an angle of 110° formed by the minute hand and hour hand.

So,

M = [tex]\frac{2}{11}[/tex] (H × 30 - θ)

   = [tex]\frac{2}{11}[/tex] (6 × 30 - 110)

   = [tex]\frac{2}{11}[/tex] (180 - 110)

   = [tex]\frac{2}{11}[/tex] × 70

   = 140/11

So, he left after 6:12 p.m

Jhon returned home before 7 p.m with angle of 110° formed by the minute hand and hour hand.

M = [tex]\frac{2}{11}[/tex] (H × 30 + θ)

   = [tex]\frac{2}{11}[/tex] (6 × 30 + 110)

   = [tex]\frac{2}{11}[/tex] (180 + 110)

   = [tex]\frac{2}{11}[/tex] × 290

   = 580/11

So, he returned before 6:52 p.m

Then,

The time taken for returning home = 580/11 - 140/11 = 440/11 = 40 minutes.

Therefore, he took 40 minutes to return home.

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This follows from the identity

[tex]\cot(x) = \dfrac1{\tan(x)} = \tan(90^\circ - x)[/tex]

The overall product is 1 because

[tex]\tan(10^\circ) \tan(80^\circ) = \cot(80^\circ) \tan(80^\circ) = 1[/tex]

[tex]\tan(20^\circ) \tan(70^\circ) = \cot(70^\circ) \tan(70^\circ) = 1[/tex]

[tex]\tan(30^\circ) \tan(60^\circ) = \cot(60^\circ) \tan(60^\circ) = 1[/tex]

[tex]\tan(40^\circ) \tan(50^\circ) = \cot(50^\circ) \tan(50^\circ) = 1[/tex]

Answer:

Step-by-step explanation:

From the right triangle with cot theta = 4/7  = (adjacent side / opposite side), Fatima can find the length of the hypotenuse using Pythagoras. This gives  a value of 8.0823. So cos theta = 4 / 8.0823 = 0,49614,

Then she could find the values of sin theta  using

sine squared theta + cos squared theta = 1.

Answer: its D or 1 + cotangent squared theta = cosecant squared theta

Step-by-step explanation:

The test that represents the best choice if you wanted to compare the average mean time to repair (mttr) of your office for the last month to the division’s mttr is one sample t-test

The purpose of the one sample t-test is to determine if a sample observations could follow a specific parameter like the mean.

This test is used to determine whether an unknown population mean is different from a specific value.

one sample t-test is used to compare the mean of a single group against that of a known mean.

We wanted to compare the average mean time to repair (mttr) of your office for the last month to the division’s mttr

So this test (one sample t-test ) would be the best choice to compare mttr.

Therefore, the test that represents the best choice if you wanted to compare the average mean time to repair (mttr) of your office for the last month to the division’s mttr is one sample t-test

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Answer:

1st option

Step-by-step explanation:

let y = h(x) and rearrange making x the subject , that is

y = 6x - 6 ( add 6 to both sides )

y + 6 = 6x ( divide both sides by 6 )

[tex]\frac{y+6}{6}[/tex] = x

change y back into terms of x with x = [tex]h^{-1}[/tex] (x)

[tex]h^{-1}[/tex] (x) = [tex]\frac{x+6}{6}[/tex]

The equation that is represented by the graph below is y = e^x - 4.

From the graph, it can be seen that the only exponential function n to hat intercept the function at equal to 3 is y = e^x - 4.

Let's equate y to 0.

y = e^x - 4.

e^x - 4 = 0

e^x = 4

In(e^x) = In(4)

x = 1.386

The x intercept is located at (1.386, 0).

In conclusion, the correct option is D.

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Si triplicamos la cantidad de cada artículo, el subtotal será de 2 940 soles, el impuesto al consumo será de 529,20 soles y el total será de 3 469,20 soles.

En esta pregunta tenemos un recibo con una lista de productos, cantidades, precios unitarios y precios de compra, además de un impuesto que se aplica al consumo por concepto de los productos comprados y el total, el cual refleja tanto el subtotal como el concepto por impuestos.

En cuanto a la compra de una determinada cantidad para un artículo, el coste total respectivo es igual a producto del número de artículos por el precio unitario.

El subtotal se determina mediante la suma de los costes totales asociados a cada artículo y luego se le recarga con el impuesto al consumo. Por último, el total de compra es la suma del subtotal y los pagos tributarios.

Si triplicamos el número de cada artículo comprado, tenemos los siguientes costes para cada artículo:

Chompas de dralom

12 / 4 × 280 = 840

Chompas de Orlon

6 / 2 × 100

300

Chompas de lana de alpaca

15 / 5 × 600

1 800

El subtotal es 2 940 soles y el impuesto al consumo es 18 / 100 × 2 940 = 529,20 soles. Por tanto, el total es 3 469,20 soles.

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Answer: There were 78 bags of chips.

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