Hundred chart: I am thinking of a number that is greater than 70 but less than 100.
It is not divisible by 10. It is not divisible by 4. It is not an odd number. What can I be

They will both have the same amount in the bank after 10 weeks

Linear equations are equations that have constant average rates of change, slope or gradient

A system of linear equations is a collection of at least two linear equations.

In this case, we have

Nancy

Initial = 400

Rate = 10

So, the equation is

y = 400 + 10x

700

Initial = 700

Spending = 20

So, the equation is

y = 700 - 20x

When they both have the same amount of money in the bank, we have

700 - 20x = 400 + 10x

Evaluate the like terms

30x = 300

Divide by 30

x = 10

Hence, they will both have the same amount in the bank after 10 weeks

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the value of b is = 10.

The equation of a hyperbola is x^2/24^2 - y^2/ (10)^2= 1.

The geometric characteristics of a hyperbola or the equations for which it is the solution set characterize it as a particular kind of smooth curve that lies in a plane. Mirror reflections of each other that resemble two infinite bows make up a hyperbola's two connected components or branches.

The general formula for hyperbola = (x - h)²/a²- (y - k)²/(b)² = 1

x²/24 - y²/(b)² = 1

(x - 0)²/24 - (y -0)²/(b)² = 1

a=24,h=0 and k=0

Now equation of the directrix

x=a²/c...(1)

and we know x=576/26...(2)

Therefore from 1 and 2 we get

24²/c=576/26.

isolate the c so we get,

C=26

C= center of focii

c = √(a² + b³)

c² = a² + b²

b = c² - a²

b = 10

So we get the value of b is 10.

Therefore the equation of a hyperbola is x²/24² - y²/ (10)² =  1.

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The solution of the differential equation is [tex]y = (C_{1} + C_{2} x)e^{-2x} +\frac{1}{2} [x^{3} -\frac{3}{2} x-6][/tex] .

According to the given question.

We have a differential equation

[tex]y^{"} + 4y^{'} + 4y = 2x^{3}[/tex]

The above differerntial equation acn be written as

[tex](D^{2} +4D+ 4)= 2x^{3}[/tex]

Now, the auxillary equation for the above differential equation is given by

[tex]m^{2} + 4m + 4 = 0[/tex]

[tex]\implies m^{2} + 2m + 2m + 4 = 0[/tex]

⇒ m (m + 2) + 2(m + 2) = 0

⇒ m(m + 2)(m + 2) = 0

Therefore,

[tex]C.F = (C_{1} + C_{2}x)e^{-2x}[/tex]

Now,

[tex]PI = \frac{1}{D^{2} +4D+4} 2x^{3}[/tex]

[tex]\implies PI = \frac{1}{4(1+\frac{D^{2}+4D }{4} )} 2x^{3}[/tex]

[tex]\implies PI = \frac{1}{4} [1+(\frac{D^{2}+4D }{4} )]^{-1} 2x^{3}[/tex]

[tex]\implies PI = \frac{1}{4} [ 1 - (\frac{D^{2} +4D}{4} )+(\frac{D^{2} +4D}{4} )^{2} -(\frac{D^{2}+4D }{4}) ^{3} ...]2x^{3}[/tex]

[tex]\implies PI =\frac{1}{2} [ x^{3} -\frac{1}{4} (6x)-3x^{2} +3x^{2} -6][/tex]

[tex]\implies PI = \frac{1}{2} [x^{3} -\frac{3}{2} x-6][/tex]

Therefore, the solution of the differential equation will be

y = CI + PI

[tex]y = (C_{1} + C_{2} x)e^{-2x} +\frac{1}{2} [x^{3} -\frac{3}{2} x-6][/tex]

Hence, the solution of the differential equation is [tex]y = (C_{1} + C_{2} x)e^{-2x} +\frac{1}{2} [x^{3} -\frac{3}{2} x-6][/tex] .

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The domain of the given function [tex]f(x) = \sqrt{(4x + 9)} + 2[/tex].

So long as x ≥ -9/4, the function f(x) will be defined.

Given: "f(x) = Startroot 4 x + 9 Endroot + 2" should be written as

[tex]f(x) = \sqrt{(4x + 9)} + 2[/tex].

Note that [tex]$\sqrt{(4x + 9)}[/tex] exists a variation of the basic function [tex]y = \sqrt{x}[/tex], whose domain exists [0, ∞ ).

The domain of [tex]$f(x) = \sqrt{(4x + 9) }+ 2[/tex] exists seen by taking the "argument" 4x + 9 of [tex]\sqrt{(4x + 9)}[/tex]and setting it equivalent to zero:

4x + 9 ≥ 0

simplifying the equation, we get

4x ≥ -9

x ≥ -9/4

This exists the domain of the given function [tex]f(x) = \sqrt{(4x + 9)} + 2[/tex].

So long as x ≥ -9/4, the function f(x) will be defined.

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Answer:

B

Step-by-step explanation:

I tink old up

no its BBBBB

Comparing it to a system of equations, the expression is represented as follows:

18g + 5.

A system of equations is when two or more variables are related, and equations are built to find the values of each variable.

For this problem, we consider g as the variable. Then:

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Between 2 seconds and 2.5 seconds, the rock hits the ground after it is dropped.

What are distance and time?

As you can see at 2s the height of the rock is 0.4m and at 2.5s it is -10.6m, which tells us that in between these values the height would have been 0m, as the movement of the rock is uniform in direction and hence, the rock will hit the ground between 2s and 2.5s.

The rock hits the ground between 2 seconds and 2.5 seconds after it is dropped.

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Answer: 46.7 in/3

Step-by-step explanation:

Answer:

Step-by-step explanation:

Use periodic compound formula:

[tex]F=P\cfrac{(1+r/n)^{nt}-1}{r/n}[/tex]

Substitute the values and calculate:

[tex]F=200\cfrac{(1+0.032/12)^{12*2}-1}{0.032/12} =4950.12[/tex]     rounded

Step-by-step explanation:

Use periodic compound formula:

[tex] \sf \: F=P\cfrac{(1+ \frac{r}{n})^{nt}-1}{\frac{r}{n}}[/tex]

Substitute the values and calculate:

[tex]\sf \: F=200\cfrac{(1+ \frac{0.032}{12})^{12 \times 2}-1}{\frac{0.032}{12}}[/tex]

[tex]\sf \: F=200\cfrac{( \frac{ 12 + 0.032}{12})^{24}-1}{\frac{0.032}{12}}[/tex]

[tex]\sf \: F=200\cfrac{( {11.002 }{})^{24}-1}{0.002} [/tex]

[tex]\sf \: F=200 \times 24.7506[/tex]

[tex]\sf \: F=4950.12rounded[/tex]

Answer:

[tex]\textsf{a)} \quad v=12t^2-6t-18[/tex]

[tex]\textsf{b)} \quad a=24t-6[/tex]

c)   1.5 s

d)  -19.25 m

e)   24.5 m

Step-by-step explanation:

Displacement

[tex]x=4t^3-3t^2-18t+1[/tex]

(where t ≥ 0 and x is in meters)

To find the equation for velocity, differentiate the equation for displacement:

[tex]\implies v=\dfrac{\text{d}x}{\text{d}t}=12t^2-6t-18[/tex]

(where t ≥ 0 and v is in meters per second)

To find the equation for acceleration, differentiate the equation for velocity:

[tex]\implies a=\dfrac{\text{d}v}{\text{d}t}=24t-6[/tex]

(where t ≥ 0 and a is in meters per second squared)

The particle comes to rest when its velocity is zero:

[tex]\begin{aligned}v & = 0\\\implies 12t^2-6t-18 & = 0\\6(2t^2-t-3) & = 0\\2t^2-t-3 & = 0\\2t^2-3t+2t-3 & = 0\\t(2t-3)+1(2t-3) & = 0\\(t+1)(2t-3) & = 0\\\implies t & =-1, \dfrac{3}{2}\end{aligned}[/tex]

As t ≥ 0, the particle comes to rest at 1.5 s.

Substitute the found value of t from part (c) into the equation for displacement to find where the particle comes to rest:

[tex]\implies 4(1.5)^3-3(1.5)^2-18(1.5)+1=-19.25\: \sf m[/tex]

We have determined that the particle is at rest at 1.5 s.

Therefore, to find how far the particle traveled in the first 2 seconds, we need to divide the journey into two parts: before and after it was at rest.  

The first leg of the journey is the first 1.5 s and the second leg of the journey is the next 0.5 s.

At the beginning of the journey, t = 0 s.

[tex]\textsf{when }t=0: \quad x=4(0)^3-3(0)^2-18(0)+1=1[/tex]

Therefore, when t = 0, x = 1

When t = 1.5 s, x = -19.25 m (from part (d)).

⇒ Total distance traveled in first 1.5 s = 1 + 19.25 = 20.25 m

When t = 2, x = -15 m.

Therefore, the particle has traveled 19.25 - 15 = 4.25 m in the last 0.5 s of its journey.

Total distance traveled:

1 + 19.25 + 4.25 = 24.5 m

Refer to the attached diagram

When the particle is at rest, it changes direction.  If we model its journey using the x-axis, for the first leg of its journey (0 - 1.5 s) it travels in the negative direction (to the left). At 1.5 s it stops, then changes direction and travels in the positive direction (to the right), arriving at -15 at 2 seconds.

Answer:

B

Step-by-step explanation:

The beginning temperature is -12 and then it rises 5 degrees each hour at the end of the game it is 32 degrees.

Answer:

2nd option, x > 1.10

Step-by-step explanation:

[tex]7e^{2x}-5 > 58[/tex]

Add 5 to both sides,

[tex]7e^{2x}-5+5 > 58 +5\\7e^{2x} > 63[/tex]

Divided both sides by 7,

[tex]\frac{7e^{2x}}{7} > \frac{63}{7}[/tex]

[tex]e^{2x} > 9[/tex]

Apply Exponent Rule,

[tex]2x > 2ln(3)[/tex]

[tex]\frac{2x}{2} > \frac{2ln(3)}{2}[/tex]

x > ln(3) or x > 1.09861

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First of all, your question should be: Factor completely 2xy – 6mn – 3my + 4nx, and not Factorize completely 2xy – 6mn – 3m + 4nx. This noted, we proceed to the solution, which is:

2xy – 6mn – 3my + 4nx

→ y(2x – 3m) + 2n(–3m + 2x)

→ (2x – 3m) (y + 2n)

.: Final answer: (2x – 3m) (y + 2n)

(2x – 3m) (y + 2n) is the factorized form of  2xy -6mn - 3m + 4nx

Two or more expressions with an Equal sign is called as Equation.

Factorization or factoring consists of writing a number or another mathematical object as a product of several factors, usually smaller or simpler objects of the same kind.

The given expression is 2xy – 6mn – 3my + 4nx

Two times of xy minus six times of mn minus three times of m y plus four times of nx

2xy – 6mn – 3my + 4nx

By using commutative property arrange the terms

2xy –3my– 6mn+ 4nx

Take y as common from first two terms and 2n as common in last two terms

y(2x – 3m) + 2n(–3m + 2x)

(2x – 3m) (y + 2n)

Hence, (2x – 3m) (y + 2n) is the factorized form of  2xy -6mn - 3m + 4nx

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75  is  the least positive integers which , when subtracted 7300 would make a result a perfect square

So normally the least positive integer of all the numbers is the number 1 but when you talk about least positive integer, often times you are talking about the special function called the ceiling function

The smallest of the numbers in the set {1, 2, 3, …} is 1.

So, the number 1 is the smallest positive integer.

7300

If we Take Square root of 7300 we have to subtract 75 from 7300 to get a perfect square.

7300-75=7225

(85)^2=7225

75 to be subtracted

√7300 ≥ 85

Perfect Square = 85² =  7225     or (7300-7225 = 75)

75  is  the least positive integers which , when subtracted 7300 would make a result a perfect square

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Answer:

11÷7=[tex]\frac{11}{7}[/tex]

v= 11/7

Using the binomial distribution, there is a 0.9983 = 99.83% probability that at most eleven of the thirteen babies are girls.

The formula is:

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

The parameters are:

For this problem, the values of the parameters are:

p = 0.5, n = 13

The probability that at most eleven of the thirteen babies are girls is:

[tex]P(X \leq 11) = 1 - P(X > 11)[/tex]

In which

P(X > 11) = P(X = 12) + P(X = 13)

Then:

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 12) = C_{13,12}.(0.5)^{12}.(0.5)^{1} = 0.0016[/tex]

[tex]P(X = 13) = C_{13,13}.(0.5)^{13}.(0.5)^{0} = 0.0001[/tex]

So:

P(X > 11) = P(X = 12) + P(X = 13) = 0.0016 + 0.0001 = 0.0017

[tex]P(X \leq 11) = 1 - P(X > 11) = 1 - 0.0017 = 0.9983[/tex]

0.9983 = 99.83% probability that at most eleven of the thirteen babies are girls.

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Answer:

a

Step-by-step explanation:

given y varies directly as x² and inversely as z³ then the equation relating them is

y = [tex]\frac{kx^2}{z^3}[/tex] ← k is the constant of variation

to find k use the condition y = 12 when x = 4 and z = 2 , then

12 = [tex]\frac{k(4)^2}{2^3}[/tex] = [tex]\frac{16k}{8}[/tex] ( multiply both sides by 8 )

96 = 16k ( divide both sides by 16 )

6 = k

y = [tex]\frac{6x^2}{z^3}[/tex] ← equation of variation

when y = 1.728 and z = 5 , then

1.728 = [tex]\frac{6x^2}{5^3}[/tex] = [tex]\frac{6x^2}{125}[/tex] ( multiply both sides by 125 )

216 = 6x² ( divide both sides by 6 )

36 = x² ( take square root of both sides )

[tex]\sqrt{36}[/tex] = x , that is

x = 6

Answer: x is -1/2,0 and y is 0,2

Step-by-step explanation:

Answer:

(4,0) , (0,2)

Step-by-step explanation:

Similar to the previous question.

x-intercept is where the line touches the x-axis , and y = 0.

when y = 0,

8x + 16(0) = 32

8x = 32

x = [tex]\frac{32}{8}[/tex]

x = 4

Therefore the coordinates of the x-intercept is (4,0)

y-intercept is where the line touches the y-axis, and x = 0.

when x = 0,

8(0) + 16y = 32

16y = 32

y = [tex]\frac{32}{16}[/tex]

y = 2

Therefore the coordinates of the y-intercept is (0,2)

Answer:

30°

30 is the value of x that’s makes AB//CD.

Step-by-step explanation:

the angles of measures 2x + 40 and 3x + 10 are two Alternate interior angles

If these two angle were congruent then the lines AB and CD

would be parallel .

2x + 40 = 3x + 10

⇔ 40 - 10 = 3x - 2x

⇔ 30 = x

⇔ x = 30

Answer:

a) x = 3

b) x = 10

c) x = 28

Explanation:

a)

[tex]\sf 6x + 3 = 21[/tex]

collect term

[tex]\sf 6x = 21-3[/tex]

simplify

[tex]\sf 6x = 18[/tex]

divide both both sides by 6

[tex]\sf x = 3[/tex]

b)

[tex]\sf 15(x - 5) = 75[/tex]

distribute

[tex]\sf 15x - 75 = 75[/tex]

collect terms

[tex]\sf 15x = 75 + 75[/tex]

combine

[tex]\sf 15x = 150[/tex]

divide both sides by 15

[tex]\sf x = 10[/tex]

c)

[tex]\sf \frac{3}{4} x +5 = 26[/tex]

collect like terms

[tex]\sf \frac{3}{4} x = 26-5[/tex]

combine

[tex]\sf \frac{3}{4} x = 21[/tex]

cross multiply

[tex]\sf x = \frac{ 21 (4)}{3}[/tex]

simplify

[tex]\sf x =28[/tex]

The answer is A = m².

We know the general formula for finding the area of a square is :

Area = side²

Now, we are given the side is equal to m.

Hence, the expression which can be used to find the area is :

A = m²

Answer:

50.24 ft^2.

Step-by-step explanation:

Area = pi r^2

Here the radius r = 4

Therefore the area of the circle

= pi * 4^2

= 16 pi

= 16 *3.14

= 50.24 ft^2.

Based on the calculations, the coordinates of the mid-point of BC are (1, 4).

First of all, we would determine the initial y-coordinate by substituting the value of x into the equation of line that is given:

At the origin x₁ = 0, we have:

y = 2x + 1

y₁ = 2(0) + 1

y₁ = 2 + 1

y₁ = 3.

When x₂ = 2, we have:

y = 2x + 1

y₂ = 2(2) + 1

y₂ = 4 + 1

y₂ = 5.

In order to determine the midpoint of a line segment with two (2) coordinates or endpoints, we would add each point together and divide by two (2).

Midpoint on x-coordinate is given by:

Midpoint = (x₁ + x₂)/2

Midpoint = (0 + 2)/2

Midpoint = 2/2

Midpoint = 1.

Midpoint on y-coordinate is given by:

Midpoint = (y₁ + y₂)/2

Midpoint = (3 + 5)/2

Midpoint = 8/2

Midpoint = 4.

Therefore, the coordinates of the mid-point of BC are (1, 4).

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The steps to solve a quadratic formula include:

In algebra, a quadratic equation is a equation that can be rearranged in standard form as where x represents an unknown, and a, b, and c represent known numbers, where a ≠ 0.

A quadratic equation is an equation that could be written as ax² + bx + c = 0

when a 0. The factoring method is illustrated:

To solve a quadratic equation by factoring, put all terms on one side of the equal sign, leaving zero on the other side.

Set each factor equal to zero.

Solve each of these equations.

Check by inserting your answer in the original equation.

Example 1

Solve x² – 6 x = 16.

Following the steps,

x² – 6 x = 16 becomes x² – 8x + 2x – 16 = 0

Factor.

( x – 8)( x + 2) = 0

Therefore, x = 8 and -2.

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Answer:

x = 36

Step-by-step explanation:

3x and 2x are a linear pair and sum to 180° , that is

3x + 2x = 180

5x = 180 ( divide both sides by 5 )

x = 36

Answer:

4. 4.125 inches by 3.5 inches.

5. 8.4 hours.

Step-by-step explanation:

4. multiply 16.5 and 14 by .25 because each foot in the house is .25 of an inch on the paper.

5. 252 miles / 9 hours = 28 mph. Add 2 so you get 30 mph. 252 miles / 30 mph = 8.4 hours.

Answer:

x = 4, y = 1

Step-by-step explanation:

y = (1/4)x  

2x + 4y = 12

replace y with something x

2x + 4(1/4)x = 12

3x = 12

x = 4

so y = 1

The equation s for each scenario in which case, the former, Prize A is represented as a linear equation while the latter, Prize B is represented as an exponential function are;

It follows from the task content that in scenario which pertains to Prize A, the initial price which represents the y-intercept is; $5,000 while the slope which represents the increase per month is; $100.

Consequently, we have; Prize A = $5,000 + $100m

While, for Prize B, in which case the function via exponential, we have;

Prize B = $2,000(1.1)^m.

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The local minimum value of the function f(x) = 6 + 9x² - 6x³ is 6 and the local maximum value of the function f(x) = 6 + 9x² - 6x³ is 9.

For given question,

We have been given a function f(x) = 6 + 9x² - 6x³

We need to find the local maximum and local minimum of the function  f(x)

First we find the first derivative of the function.

⇒ f'(x) = 0 + 18x - 18x²

⇒ f'(x) = - 18x² + 18x

Putting the first derivative of the function equal to zero, we get

⇒ f'(x) = 0

⇒ - 18x² + 18x = 0

⇒ 18(-x² + x) = 0

⇒ x (-x + 1) = 0

⇒ x = 0    or    -x + 1 = 0

⇒ x = 0     or    x = 1

Now we find the second derivative of the function.

⇒ f"(x) = - 36x + 18

At x = 0 the value of second derivative of function f(x),

⇒ f"(0) = - 36(0) + 18

⇒ f"(0) = 0 + 18

⇒ f"(0) = 18

Here, at x=0, f"(x) > 0

This means, the function f(x) has the local minimum value at x = 0,  which is given by

⇒ f(0) = 6 + 9(0)² - 6(0)³

⇒ f(0) = 6 + 0 - 0

⇒ f(0) = 6

At x = 1 the value of second derivative of function f(x),

⇒ f"(1) = - 36(1) + 18

⇒ f"(1) = - 18

Here, at x = 1, f"(x) < 0

This means, the function f(x) has the local maximum value at x = 1,  which is given by

⇒ f(1) = 6 + 9(1)² - 6(1)³

⇒ f(1) = 6 + 9 - 6

⇒ f(1) = 9

So, the function f(x) = 6 + 9x² - 6x³ has local minimum at x = 0 and local maximum at x = 1.

Therefore, the local minimum value of the function f(x) = 6 + 9x² - 6x³ is 6 and the local maximum value of the function f(x) = 6 + 9x² - 6x³ is 9.

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The inclination to the nearest tenth of a degree exists 0.24146

The given scenario includes a right-angled triangle where the length of the ramp exists hypotenuse and the rise of ramp exists the perpendicular.

Given: H = 4.6 m and P = 1.1 m

We have to use the trigonometric ratios to find the angle. The ratio that has to be used should involve both perpendicular and hypotenuse

Let x be the angle then

sin x = P/H

sin x = 1.1/4.6

sin x = 0.23913

[tex]$$sin^{-1} x[/tex] = 0.24146

The inclination to the nearest tenth of a degree exists 0.24146

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A graph which shows the image of ABCD is: A. on a coordinate plane, parallelogram A'B'C'D' has points (-2, 5), (0, 3), (0, -1.2), (-2, 1).

A transformation refers to the movement of a point on a cartesian coordinate from its original (initial) position to a new location.

In Geometry, there are different types of transformation and these include the following:

Based on transformation of parallelogram ABCD, we can infer and logically deduce that a graph which shows the image of ABCD is that on a coordinate plane, with parallelogram A'B'C'D' having points (-2, 5), (0, 3), (0, -1.2), (-2, 1) as shown in the image attached below.

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