a projectile is launched straight up from ground level with an initial velocity of 320 ft/sec when will it's height above ground be
1538 feet

The cost of one Sociology workbook is $140, and the cost of one Economy 101 textbook is $220, determined by solving two simultaneous equations based on the initial and additional orders placed by the bookstore.

To solve for the cost of one Sociology workbook and one Economy 101 textbook, let's label the cost of one Sociology workbook as S and the cost of one Economy 101 textbook as E. Initially, the bookstore ordered 15 Sociology workbooks and 10 Economy 101 textbooks, spending a total of $4300. This gives us the equation: 15S + 10E = 4300. Later, the store ordered an additional 1 Sociology workbook and 3 Economy 101 textbooks for $800, leading to the equation: 1S + 3E = 800.

To solve these equations simultaneously, we can multiply the second equation by 15 to get 15S + 45E = 12000 and subtract the first equation from it: (15S + 45E) - (15S + 10E) = 12000 - 4300, simplifying to 35E = 7700. Dividing both sides by 35 gives us E = 220. Substituting E = 220 back into the first equation 15S + 10(220) = 4300 gives us 15S + 2200 = 4300, thus 15S = 2100 and S = 140.

Therefore, the cost of one Sociology workbook is $140 and the cost of one Economy 101 textbook is $220.

Donna will finish reviewing in about 4.05 hours, while Phyllis will take approximately 6.08 hours.

Donna and Phyllis are reviewing for a math final and have 150 pages to cover. Let's calculate how long it takes each girl to finish reviewing:

Donna can review 37 pages an hour. To find the time she needs, we use the formula:

Time = Total Pages / Pages per Hour
Time = 150 / 37
Time ≈ 4.05 hours

Phyllis reviews at 2/3 of Donna's speed. So, her rate is:

Phyllis' rate = (2/3) × 37 ≈ 24.67 pages per hour

To find the time Phyllis needs, we use the same formula:

Time = Total Pages / Pages per Hour
Time = 150 / 24.67
Time ≈ 6.08 hours

In summary, Donna will finish reviewing in about 4.05 hours, while Phyllis will take approximately 6.08 hours.

Answer:

  (-1, 1)

Step-by-step explanation:

The midpoint (M) is found by averaging the coordinates of the end points:

  M = ((-3, -2) +(1, 4))/2 = ((-3+1)/2, (-2+4)/2) = (-2/2, 2/2)

  M = (-1, 1)

The midpoint of the line segment is (-1, 1).

The indicated function y1(x) is a solution of the given differential equation.The general solution is [tex]y = c_1 e^{2x}- c_2e^{-6x}/8[/tex]

An equation containing derivatives of a variable with respect to some other variable quantity is called differential equations.

The derivatives might be of any order, some terms might contain the product of derivatives and the variable itself, or with derivatives themselves. They can also be for multiple variables.

Given differential equation is

y''-4y'+4y=0

and

[tex]y_1(x) = e^{2x}[/tex]

[tex]y_2(x) = y_1(x) \int\limits^a_b {e^{\int pdx} \, / y_1 ^2(x)dx[/tex]

The general form of equation

y''+P(x)y'+Q(x)y=0

Comparing both the equation

So, P(x)= - 4

[tex]y_2(x) = y_1(x) \int\limits^a_b {e^{\int pdx} \, / y_1 ^2(x)dx\\\\\\[/tex]

[tex]y_2(x) = e^{2x}\int e^{-4x} \, / e^{4x}dx[/tex]

[tex]y_2(x) = e^{2x}\int e^{-8x}dx\\\\y_2(x) = -e^{-6x}/8[/tex]

The general solution is

[tex]y = c_1 e^{2x}- c_2e^{-6x}/8[/tex]

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The probability with the condition of the spinner landing on a number less than 3 is 0.2

A conditional probability is a probability of an event occuring with a condition that another event had previously occurred. The event in the question is spinning the spinner once while the condition is that the number landed on is less than 3.

The spinner has 10 equal sections numbered 1 through 10.

The conditional probability of landing on a number less than 3 is the same as the probability of landing on either 1 or 2.

There are two sections out of ten that corresponds to numbers less than 3. The probability of landing on a number less than 3 is therefore;

P(Landing on a number less than 3) = P(Landing on 1) + P(Landing on 2)

P((Landing on 1) = 1/10

P(Landing on 2) = 1/10

P(Landing on 1) + P(Landing on 2) = (1/10) + (1/10) = 2/10

(1/10) + (1/10) = 2/10 = 0.2

The probability of landing on a number less than 3 is 0.2

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Answer:

3.66% ( approx )

Step-by-step explanation:

Since, the formula of annual percentage yield is,

[tex]APY = (1+\frac{r}{n})^n-1[/tex]

Where,

r = stated annual interest rate,

n = number of compounding periods,

Here, r = 3.6% = 0.036,

n = 12 ( ∵ 1 year = 12 months )

Hence, the annual percentage yield is,

[tex]APY=(1+\frac{0.036}{12})^{12}-1=1.03659 - 1 = 0.036599\approx 0.0366 = 3.66\%[/tex]

Answer:

Because basically, you are multiplying by 1

Step-by-step explanation:

Let me explain this with an example. Rationalize the following expression:

[tex]\frac{5}{\sqrt{7} }[/tex]

In order to rationalize the denominator, the numerator and denominator of the fraction must be multiplied by the root of the denominator. So, what happen if you do that? Well first of all you aren't altering the expression because:

[tex]\frac{\sqrt{7} }{\sqrt{7} } =1[/tex]

Right? Because a certain quantity divided by itself is always equal to 1. So basically you are doing this because you want to rewrite the expression without altering its original value, it is the same when you do this:

[tex]9=3^2=3+3+3=\sqrt{81}[/tex]

Therefore, the only thing you do when you rationalize is remove radicals from the denominator of a fraction. Take a look:

[tex]\frac{5}{\sqrt{7} } *\frac{\sqrt{7} }{\sqrt{7} } =\frac{5*\sqrt{7} }{\sqrt{7}*\sqrt{7} } =\frac{5*\sqrt{7}}{(\sqrt{7} )^2} =\frac{5*\sqrt{7} }{7}[/tex]

You can check this new expression is equal to the original using a calculator:

[tex]\frac{5}{\sqrt{7} } \approx1.8898\\\\\frac{5*\sqrt{7} }{7} \approx1.8898[/tex]

The rate of petrol per liter is 11.2 liter

The unitary method is a technique for solving a problem by first finding the value of a single unit, and then finding the necessary value by multiplying the single unit value.

Given:

In 25l car travels = 280km

For finding rate of petrol per liter we have to divide as

Rate of petrol = 280/25

Rate of petrol = 11.2 liter

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The rate of petrol usage for the car is 11.2 kilometers per litre, calculated by dividing the distance travelled, 280km, by the amount of petrol used, 25L.

The question asks to find the rate of petrol usage in kilometers per litre for a car that uses 25L of petrol to travel 280km. To calculate the fuel efficiency, you divide the distance travelled by the amount of petrol used.

Step-by-step calculation:

Distance travelled = 280 km

Amount of petrol used = 25L

Fuel efficiency (km/L) = Distance travelled / Amount of petrol used

Fuel efficiency = 280 km / 25L = 11.2 km/L

Therefore, the car's fuel efficiency is 11.2 kilometers per litre.

Answer:

b

Step-by-step explanation:

The value of x=-5 is a discontinuity of the function x^2+7x+1 / x^2+2x-15 since it makes the denominator of this function equal to zero.

The subject of this question is the discontinuity of a rational function. In Mathematics, a function f(x) = (p(x))/(q(x)), where p(x) and q(x) are polynomials, is said to be discontinuous at a particular value of x if and only if q(x) = 0 at that value. From the equation in the question; x^2+7x+1/x^2+2x-15, we can determine its discontinuity by finding the values of x that would make the denominator equal to zero. This is done by solving the polynomial equation x^2+2x-15 = 0 for x. The solutions to this equation represent the values at which the function is discontinuous.

By applying the quadratic formula, (-b ± sqrt(b^2 -4ac))/(2a), where a = 1, b = 2, and c = -15, we get that x = -5, and 3. However, the values given in the question are x=-1, x=-2, x=-5, and x=-4. From these options, only x=-5 makes the denominator zero, thus, x = -5 is a point of discontinuity in the function x^2+7x+1 / x^2+2x-15.

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The formula for volume of cone is:

V = π r^2 h / 3

or

π r^2 h / 3 = 36 cm^3

Simplfying in terms of r:

r^2 = 108 / π h

To find for the smallest amount of paper that can create this cone, we call for the formula for the surface area of cone:

S = π r sqrt (h^2 + r^2)

S = π sqrt(108 / π h) * sqrt(h^2 + 108 / π h) 

S = π sqrt(108 / π h) * sqrt[(π h^3 + 108) / π h] 

Surface area = sqrt (108) * sqrt[(π h + 108 / h^2)] 

Getting the 1st derivative dS / dh then equating to 0 to get the maxima value:

dS/dh = sqrt (108) ((π – 216 / h^3) * [(π h + 108/h^2)^-1/2] 

Let dS/dh = 0 so,

 π – 216 / h^3 = 0 

h^3 = 216 / π

h = 4.10 cm

Calculating for r:

r^2 = 108 / π (4.10)

r = 2.90 cm

Answers:

 h = 4.10 cm

r = 2.90 cm

The height of the cone is [tex]\boxed{4.10}[/tex] and the radius of the cone is [tex]\boxed{2.90}.[/tex]

Further explanation:

The volume of the cone is [tex]\boxed{V = \dfrac{1}{3}\left( {\pi {r^2}h} \right)}.[/tex]

The surface area of the cone is [tex]\boxed{S=\pi \times r\times l}[/tex]

Here l is the slant height of the cone.

The value of the slant height can be obtained as,

[tex]\boxed{l = \sqrt {{h^2} + {r^2}} }[/tex].

Given:

The volume of the cone shaped paper drinking cup is [tex]36{\text{ c}}{{\text{m}}^3}[/tex].

Explanation:

The volume of the cone shaped paper drinking cup  [tex]36{\text{ c}}{{\text{m}}^3}[/tex].

[tex]\begin{aligned}V&=36\\\frac{1}{3}\left({\pi {r^2}h}\right) &= 36\\{r^2}&= \frac{{108}}{{\pi h}}\\\end{aligned}[/tex]

The surface area of the cone is,

[tex]\begin{aligned}S &= \pi\times\sqrt {\frac{{108}}{{\pi h}}}\times\sqrt {{h^2} + \frac{{108}}{{\pi h}}}\\&= \sqrt{108}\times\sqrt{\frac{{\pi {h^3} + 108}}{{\pi h}}}\\&=\sqrt {108}\times\sqrt {\pi h + \frac{{108}}{{{h^2}}}}\\\end{aligned}[/tex]

Differentiate above equation with respect to h.

[tex]\dfrac{{dS}}{{dh}}=\sqrt {108}\times \left( {\pi  - \dfrac{{216}}{{{h^3}}}}\right)\times {\left( {\pi h + \dfrac{{108}}{{{h^2}}}}\right)^{ - \dfrac{1}{2}}}[/tex]

Substitute 0 for [tex]\dfrac{{dS}}{{dh}}[/tex].

[tex]\begin{aligned}\pi- \dfrac{{216}}{{{h^3}}}&= 0\\\dfrac{{216}}{{{h^3}}}&= \pi\\\dfrac{{216}}{{3.14}} &= {h^3}\\h &= 4.10\\\end{aligned}[/tex]

The radius of the cone can be obtained as,

[tex]\begin{aligned}{r^2}&=\frac{{108}}{{\pi \left({4.10} \right)}}\\{r^2}&= \frac{{108}}{{3.14 \times 4.10}}\\{r^2}&= 8.40\\r&= \sqrt {8.40}\\r &= 2.90\\\end{aligned}[/tex]

Hence, the height of the cone is  [tex]\boxed{4.10}[/tex]and the radius of the cone is [tex]\boxed{2.90}[/tex].

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Answer details:

Grade: High School

Subject: Mathematics

Chapter: Mensuration

Keywords: cone shaped paper, drinking cup, volume, [tex]36{\text{ c}}{{\text{m}}^3}[/tex], height of cone, cup, smallest amount of paper, water.

The expected number of rolls, e[n], to get doubles on a pair of fair dice is 6. This is calculated by recognizing that getting doubles has a probability of 1/6, and the expectation for a geometric distribution is the inverse of the probability (1/p).

To solve the problem of finding the expected number of rolls, e[n], to get doubles on a pair of fair dice, we first need to calculate the probability of rolling doubles. Since there are 6 faces on each die, there are a total of 6 x 6 = 36 possible outcomes when rolling two dice.

Out of these 36 possible rolls, there are 6 outcomes that result in doubles: (1,1), (2,2), (3,3), (4,4), (5,5), and (6,6). This means the probability of getting doubles in one roll is 6/36, which simplifies to 1/6.

Because each roll is independent, we can model the scenario using a geometric distribution, where the expected value, or mean, is given by 1/p, where p is the success probability. Substituting p with 1/6, the expected number of rolls to get doubles would be 1/(1/6) = 6.

Therefore, the expected number of rolls needed to roll doubles is 6.

Using the normal distribution table, which shows the percentage of the areas to the left a normal distribution, the area to the left z = - 2.15 and area to the right of z = 1.62 are 0.0158 and 0.0526 respectively.

1.)

The area under the normal distribution curve to the left of z = 2.15 can be expressed thus :

P(Z ≤ -2.15)

Using a normal distribution table ; the area to the left is

P(Z ≤ -2.15) = 0.0158

2.)

The area under the normal distribution curve to the right of z = 1.62 can be expressed thus :

P(Z ≥ 1.62) = 1 - P(Z ≤ 1.62)

Using a normal distribution table ; the area to the left is P(Z ≤ 1.62) = 0.94738

P(Z ≥ 1.62) = 1 - 0.94738 = 0.0526

Therefore, the area to the left z = - 2.15 and area to the right of z = 1.62 are 0.0158 and 0.0526 respectively.

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