PLEASE I NEED HELP ON THIS T^T
An online media service offers video and music downloads. Each download of a song costs $1. Each download of a video costs 5 times as much as the song download. You have a credit of $70. How many songs can you purchase after buying 12 videos?

Answer:

The fraction of 6% is 6/10 and the model is 10 squares and only color 6 of them

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Answer:the fraction of 6% is 6/10

Step-by-step explanation:and the model is 10 squares and only color 6 of them

Answer:

x = \(\frac{5}{2}\) , x = \(\frac{1}{3}\)

Step-by-step explanation:

(2x - 5)(3x - 1) = 0

Equate each factor to zero and solve for x

2x - 5 = 0 ⇒ 2x = 5 ⇒ x = \(\frac{5}{2}\)

3x - 1 = 0 ⇒ 3x = 1 ⇒ x = \(\frac{1}{3}\)

Answer: 30

Step-by-step explanation:

It to be 30 but this look hard so I’m not sure

To answer this question we have to evaluate the given equation using each ordered pair and see if it's true:

(0, sqrt(13))

This point does not lie on the circle.

(0, -sqrt(13))

This point does not lie on the circle.

(4, 3)

This point does lie on the circle.

(-2, 1)

Answer:

600 m³

Step-by-step explanation:

Attached Below

If it helps, mark as brainliest. : )

Answer:

The correct answer is: 2. t=0.45a

Step-by-step explanation:

Let t be the price of apples

and a be the number of apples

The price and quantity are directly proportional

\(t \propto a\)

Removing the proportionality symbol introduces a proportionality constant in the equation

\(t = ka\)

So far we know that cost of 6 apples is $2.70

Putting in the equation

\(2.70 = k (6)\\k = \frac{2.70}{6}\\k = 0.45\)

Putting the values of k will give us:

\(t = 0.45a\)

Hence,

The correct answer is: 2. t=0.45a

Answer: x is greater than or equal to 18

Step-by-step explanation:

Answer:

x ≤ - 2

Step-by-step explanation:

- x/6 ≤ 3

- x/2 ≤ 1

- x ≤ 2

x ≤ - 2

The function which represents the cost to drive a car from this agency miles x a day is :

C(x) = 15, if 0 ≤ x ≤ 200

      = 15 + 0.05x, if x > 200

Given that,

A car rental agency charges $15 a day for driving a car 200 miles or less.

The function can be written as,

C(x) = 15 if 0 ≤ x ≤ 200

If a car is driven over 200 miles, the renter must pay $0.05 for each mile over 200 driven.

C(x) = 15 + 0.05x, if x > 200

Hence the correct option is D.

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The probability of getting a sample mean selling price between $113,000 and $117,000 is approximately 0.4332.

A normal distribution with a mean of $115,000 and a standard deviation of - is used to estimate the distribution of the sample mean using the central limit theorem.

= 25,000/√100

= 2,500.

a. To find the probability that the sample means selling price was more than $110,000, we can calculate the z-score:

z = (110,000 - 115,000) / 2,500

= -5,000/2,500

= -2

As per the standard normal distribution table, the probability of getting a z-score of -2 or lower is approximately 0.0228. Therefore, the probability of getting a sample mean selling price of more than $110,000 is approximately 0.0228.

b.

Calculating the z-scores for each value:

z1

= (113,000 - 115,000) / 2,500

= -0.8

z2

= (117,000 - 115,000) / 2,500

= 0.8

The likelihood of receiving a z-score between -0.8 and 0.8, as determined by a typical normal distribution table or calculator, is found to be roughly 0.4332.

Complete Question:

The mean selling price of new homes in a city over a year was $115,000. The population standard deviation was $25,000. A random samples of 100 new homes is taken. A. What is the probability that the sample mean selling price was more than $110,000? b. What is the probability that the sample mean selling price was between $113,000 and $117,000

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The volume of the solid obtained by rotating the region bounded by the graphs of:

a. \(y=x^2-9, y=0\) about the x-axis is 0.

b. \(y=16-x,y=3x+12,x=-1\) about the x-axis is (82π / 9) cubic units.

c. \(y=x^2+2, y=x^2+10.x > 0\) about the y-axis is(160π/3) cubic units.

To find the volume of the solid obtained by rotating the region bounded by the given graphs, we can use the method of cylindrical shells. The volume is calculated as the integral of the shell's volume over the specified interval.

a.) For the region bounded by \(y=x^2-9, y=0\) , rotating about the x-axis:

V = ∫[a, b] 2πx * (f(x) - g(x)) dx

where a and b are the x-values where the curves intersect.

To find the intersection points, we set the two functions equal to each other:

x² - 9 = 0

x² = 9

x = ±3

So, a = -3 and b = 3.

V = ∫[-3, 3] 2πx * (x² - 9) dx

V = 2π ∫[-3, 3] (x³ - 9x) dx

= 2π [ (1/4)x⁴ - (9/2)x² ] | [-3, 3]

= 2π [ ((1/4)(3⁴) - (9/2)(3²)) - ((1/4)(-3⁴) - (9/2)(-3²)) ]

= 2π [ (81/4 - 81/2) - (81/4 - 81/2) ]

= 2π (0)

= 0

Therefore, the volume of the solid obtained by rotating the region bounded by y = x² - 9 and y = 0 about the x-axis is 0.

b. For the region bounded by y = 16 - x, y = 3x + 12, and x = -1, rotating about the x-axis:

V = ∫[a, b] 2πx * (f(x) - g(x)) dx

In this case, we have two curves intersecting at x = -1. So, we can split the integral into two parts.

For the first part, we have:

V1 = ∫[-1, a] 2πx * (f(x) - g(x)) dx

where a is the x-value where y = 16 - x and y = 3x + 12 intersect.

The two equations equal to each other:

16 - x = 3x + 12

x = 1

So, a = 1.

The integral becomes:

V1 = ∫[-1, 1] 2πx * ((16 - x) - (3x + 12)) dx

V1 = 2π ∫[-1, 1] (16x - x² - 3x - 12) dx

= 2π [ (8x² - (1/3)x³ - (3/2)x² - 12x) ] | [-1, 1]

= 2π [ (8(1)² - (1/3)(1)³ - (3/2)(1)² - 12(1)) - (8(-1)² - (1/3)(-1)³ - (3/2)(-1)² - 12(-1)) ]

= 2π [ (-19/6) - (49/6) ]

= 2π [ -68/6 ]

= -68π/3

For the second part, we have:

V2 = ∫[a, b] 2πx * (f(x) - g(x)) dx

where b is the x-value where y = 3x + 12 intersects with the x-axis (y = 0).

3x + 12 = 0

3x = -12

x = -4

So, b = -4.

V2 = ∫[1, -4] 2πx * (0 - (3x + 12)) dx

V2 = 2π ∫[1, -4] (3x² + 12x) dx

= 2π [ (x³ + 6x²) ] | [1, -4]

= 2π [ ((-4)³ + 6(-4)²) - (1³ + 6(1)²) ]

= 2π [ 32 - 7 ]

= 50π

The total volume is given by:

V = V1 + V2 = -68π/3 + 50π = (50π - 68π/3) / 3

V = (150π - 68π) / 9 = 82π / 9

Therefore, the volume of the solid obtained by rotating the region bounded by y = 16 - x, y = 3x + 12, and x = -1 about the x-axis is (82π / 9) cubic units.

c. For the region bounded by y = x² + 2, y = -x² + 10, and x > 0, rotating about the y-axis:

To find the volume, we need to determine the limits of integration by finding the x-values where the two curves intersect.

x² + 2 = -x² + 10:

x² = 4

x = ±2

Since we are only interested in the region where x > 0, the limits of integration are from 0 to 2.

V = ∫[0, 2] 2πx * (f(x) - g(x)) dx

V = ∫[0, 2] 2πx * ((x² + 2) - (-x² + 10)) dx

V = 2π ∫[0, 2] (2x² - x² + 12) dx

= 2π [ (2/3)x³ - (1/3)x³ + 12x ] | [0, 2]

= 2π [ (2/3)(2)³ - (1/3)(2)³ + 12(2) - (2/3)(0)³ - (1/3)(0)³ + 12(0) ]

= 2π [ (16/3 - 8/3 + 24) - (0) ]

= 2π [ (24 + 8/3) ]

= 2π [ (72/3 + 8/3) ]

= 2π [ 80/3 ]

= 160π/3

Therefore, the volume of the solid obtained by rotating the region bounded by y = x² + 2, y = -x² + 10, and x > 0 about the y-axis is (160π/3) cubic units.

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\(\\ \sf\longmapsto D=b^2-4ac\)

#1

\(\\ \sf\longmapsto x^2+6x+9=0\)

Now

\(\\ \sf\longmapsto D=(6)^2-4(1)(9)=36-36=0\)

#2

\(\\ \sf\longmapsto x^2+9x+20=0\)

Now

\(\\ \sf\longmapsto D=(9)^2-4(1)(20)=81-80=1\)

#3

\(\\ \sf\longmapsto x^2+6x+3=0\)

Now

\(\\ \sf\longmapsto D=6^2-4(1)(3)=36-12=24\)

#4

\(\\ \sf\longmapsto 2x^2-10x+8=0\)

Now

\(\\ \sf\longmapsto D=(-10)^2-4(1)(8)=100-32=68\)

#5

Now

\(\\ \sf\longmapsto D=5^2-4(1)(10)=25-40=-15\)

#6

\(\\ \sf\longmapsto D=(-3)^2-4(1)(1)=9-4=5\)

A 11.544 acre-feet of this ice field will melt from the beginning of day 1 (t = 0) to the beginning of day 4 (t = 3). So, correct option is C.

To solve the problem, we need to integrate the given rate of melting with respect to time over the interval [0,3] to find the total amount of ice that melts during this time.

First, we can simplify the given rate of melting by using the identity: sin³(t) = (3sin(t) - sin(3t))/4

So, M(t) = 4 - (3sin(t) - sin(3t))/4 = 16/4 - 3sin(t)/4 + sin(3t)/4 = 4 - 0.75sin(t) + 0.25sin(3t)

Integrating this expression with respect to t over the interval [0,3], we get:

\(\int\limits^3_0\) M(t) dt = \(\int\limits^3_0\) (4 - 0.75sin(t) + 0.25sin(3t)) dt

= [4t + 0.75cos(t) - (1/3)cos(3t)]|[0,3]

= (12 + 0.75cos(3) - (1/3)cos(9)) - (0 + 0.75cos(0) - (1/3)cos(0))

= 11.544

Therefore, the answer is (C) 11.544.

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Answer:

Step-by-step explanation:

-4.6+( -1.4)=

-4.6-1.4=-6 maureen is wrong, because both have the minus , so she should add with the minus symbol.

Answer:

x=2 y=-1

Step-by-step explanation:

2x+x-3=3

3x-3=3

3x=6

x=2

y=2-3

y=-1

So, the formula that we're going to use is:

Compound interest

A = P(1 + r/n)ⁿˣ

Where

A = amount

P = principal = $100

r = rate = 8% = 8 / 100 = 0.08

x = time

In this case, n=4 since it is being compounded quarterly, so:

The answer is $126.82.

the given expression is,

= (-2-8) - (-6x + 5)

= -10 + 6x - 4

= 6x - 14

so the simpl

Answer:

\(\huge\boxed{x=65}\)

Step-by-step explanation:

Look at the picture.

\(m\angle VUP=180^o-2\cdot60^o=180^o-120^o=\boxed{60^o}\\\\m\angle UPQ=360^o-2\cdot60^o=360^o-120^o=\boxed{240^o}\\\\m\angle QRM=180^o-2\cdot60^o=180^o-120^o=60^o\\\\m\angle RMQ=180^o-135^o=45^o\\\\in\ \triangle QRM:\ m\angle MQR=180^o-(60^o+45^o)=180^o-105^o=75^o\\\\m\angle PQW=360^o-(2\cdot60^o+75^o+80^o)=360^o-(120^o+155^o)\\\\=360^o-275^o=\boxed{85^o}\\\\m\angle UVW=\boxed{90^o}\)

VWQPU is the pentagon. The sum of interior angles in a pentagon is equal 540°. Therefore:

\(x^o=540^o-(90^o+60^o+240^o+85^o)=540^o-475^o=65^o\)

Answer:

In explanation below.

Step-by-step explanation:

Presumably, the proof you have in mind is to use a=b=2–√a=b=2 if 2–√2√22 is rational, and otherwise use a=2–√2√a=22 and b=2–√b=2. The non-constructivity here is that, unless you know some deeper number theory than just irrationality of 2–√2, you won't know which of the two cases in the proof actually occurs, so you won't be able to give aa explicitly, say by writing a decimal approximation.

Answer:

1,2,4 and 5

Step-by-step explanation:

those are correct

Answer: 12.9

Step-by-step explanation:

5.1 + 2 + 5.8 = 12.9

The perimeter of the triangle ABC is 12.9 units.

The perimeter of a triangle is the sum of length of three sides of the triangle.

Given, the triangle ABC is rotated by 135 degrees about point S to create the triangle A'B'C'.

We know, after rotation of a figure, the whole dimension of the figure remains the same.

Therefore, AB = A'B' = 5.1 units

BC = B'C' = 2 units

CA = C'A' = 5.8 units

Hence, the perimeter of the triangle ABC

= (5.1 + 2 + 5.8) units

= 12.9 units

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Answer:

40=5×8

Step-by-step explanation:

I think, hope this helps

The equation for the number 40 is 5 times as many as 8 is; 40=5×8

An equation represents equality of two or more mathematical expression.

A mathematical equation is a statement with two equal sides and an equal sign in between. An equation is, for instance, 4 + 6 = 10. Both 4 + 6 and 10 can be seen on the left and right sides of the equal sign, respectively.

Solutions to an equation are those values of the variables involved in that equation for which the equation is true.

We are given that the number 40 is 5 times as many as 8

40 is 5 times as many as 8 means

5×8 = 40

Then the equation will be;

40=5×8

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\(~~25x^2-16=0\\\\\implies (5x)^2 -4^2 =0\\\\\implies (5x+4)(5x-4)=0\\\\\implies x = -\dfrac{4}5, ~~~x=\dfrac 45\)

The expressions that are equivalent to 2(4f + 2g)², (4, f, plus, 2, g) are:(a) 8f² + 8g² + 16fg (8, f, squared, plus, 8, g, squared, plus, 16, f, g) (b) 2f(4f + 2g) · 2f(4f + 2g) (2, f, times, the quantity, 4, f, plus, 2, g,

end the quantity, times, 2, f, times, the quantity, 4, f,

plus, 2, g) (d) 4(2f + 2g)² (4,

times, the quantity, 2, f, plus, 2, g,

end the quantity, squared)In summary,

the three correct expressions equivalent to

2(4f + 2g)², (4, f, plus, 2, g) are:

8f² + 8g² + 16fg, 2f(4f + 2g) ·

2f(4f + 2g), and 4(2f + 2g)².

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a) The graph of h(t) from t = 0 to t = 0.8 seconds is a sinusoidal curve that starts at 10 mm, reaches a maximum of 20 mm at t = 0.4 seconds, and returns to 10 mm at t = 0.8 seconds.

b) The sinusoidal function for the height h(t) is \(h(t) = 5 \times sin(5\pit) + 15.\)

c) The height of the mass is above 17 mm for approximately 0.133 seconds.

a) To sketch a graph of h from t = 0 to t = 0.8 seconds, we need to consider the given information:

The maximum and minimum heights of the mass and the time for one complete cycle.

The maximum height is 20 mm, the minimum height is 10 mm, and the time for one cycle is 0.8 seconds.

We can start by plotting the maximum and minimum heights on the y-axis and marking the time for one cycle on the x-axis.

At t = 0 seconds (when the mass is released), the height is 10mm.

At t = 0.4 seconds, halfway through the cycle, the height is the average of the maximum and minimum heights, which is (20 + 10) / 2 = 15 mm.  

At t = 0.8 seconds, the height is 10 mm again, completing one full cycle.

Using this information, we can sketch a sinusoidal graph that starts at the minimum height, increases to the maximum height, and then decreases back to the minimum height within one cycle.

b) To find a sinusoidal function for the height h(t), we can use the general form: \(h(t) = A \times sin(2\pi / T \times t + \phi) + C,\)

where A is the amplitude, T is the period, φ is the phase shift, and C is the vertical shift.

In this case, the amplitude is (20 - 10) / 2 = 5 mm (half the difference between the maximum and minimum heights).

The period is 0.8 seconds, which is the time for one complete cycle.

Since the mass starts at the minimum height at t = 0, there is no phase shift (φ = 0). The vertical shift (C) is 10 mm, the minimum height.

Therefore, the sinusoidal function for the height h(t) is: \(h(t) = 5 \times sin(2\pi / 0.8 \times t) + 10\)

c) To find for how many seconds the height of the mass is above 17 mm, we can set up an inequality using the sinusoidal function h(t) and solve for t.

h(t) > 17

5 \(\times\) sin(2π / 0.8 \(\times\) t) + 10 > 17

5 \(\times\) sin(2.5π \(\times\) t) + 10 > 17

Subtracting 10 from both sides:

\(5 \times sin(2.5\pi \times t) > 7\)

Now, we can solve for t by taking the inverse sine (arcsine) of both sides:

2.5π \(\times\) t > arcsin(7/5)

Dividing both sides by 2.5π:

t > (1 / (2.5π)) \(\times\) arcsin(7/5)

Approximately, t > 0.285 seconds.

Therefore, the height of the mass is above 17 mm for more than 0.285 seconds.

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True, Frustrated Freight is any freight or shipment that it has been halted from moving.

Generally speaking, the elements that make a bundle quit moving or being shipped to the normal beneficiary are issues to do with marking, bundling, as well as making.

Frustrated Freight alludes to products that can't be conveyed to the expected beneficiary because of different reasons like an erroneous location, beneficiary inaccessibility, or harm during travel.

This can prompt the shipment never arrive at its expected objective, causing disillusionment for the shipper and beneficiary, and possibly bringing about monetary misfortunes.

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Answer:

equation of the line passing through two points

we can chose two points from the graph

I took (5,0) and (0,5)

x-x1/x2-x1=y-y1/y2-y1

x-5/0-5=y-0/5-0

(x-5)*5=y*(-5)

x-5=-y

y+x=5

Step-by-step explanation:

range of the function is [- infinite, infinite]

if x is determined as R number then even range of the function will be R ,real numbers.

Step-by-step explanation:

Let the other 4 friends arrange themselves first. There are 4! ways to do so.

Now there are 3 spaces between the 4 friends, and 1 at each end (total 5 spaces)

Example: _F_F_F_F_

Alice can choose to stand in any of the 5 spaces. David can choose to stand in any of the 4 remaining spaces (now Alice and David are not next to each other)

Total number of ways = 4! * 5 * 4 = 480.

Answer:

t=-105

Step-by-step explanation:

divide 7 by -1/15 to get t separated