In 2005, there were 18,200 students at college A with a projected enrolment increase of 750 students per year in the same year there were 34.450 students at college with a procede decline of 500 students per year According to these projections when will the colleges have the same enrollment? What will be the enrolment in each college at that me? ment In the year , the enrollment at both colleges will be the same The total enrolment at each college will be students

Answer:

[tex]\hat{n}\ =\ \ \dfrac{\dfrac{x}{a^2}\hat{i}+\ \dfrac{y}{b^2}\hat{j}+\ \dfrac{z}{c^2}\hat{k}}{\sqrt{(\dfrac{x}{a^2})^2+(\dfrac{y}{b^2})^2+(\dfrac{z}{c^2})^2}}[/tex]

Step-by-step explanation:

Given equation of ellipsoids,

[tex]u\ =\ \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}[/tex]

The vector normal to the given equation of ellipsoid will be given by

[tex]\vec{n}\ =\textrm{gradient of u}[/tex]

            [tex]=\bigtriangledown u[/tex]

[tex]=\ (\dfrac{\partial{}}{\partial{x}}\hat{i}+ \dfrac{\partial{}}{\partial{y}}\hat{j}+ \dfrac{\partial{}}{\partial{z}}\hat{k})(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2})[/tex]

[tex]=\ \dfrac{\partial{(\dfrac{x^2}{a^2})}}{\partial{x}}\hat{i}+\dfrac{\partial{(\dfrac{y^2}{b^2})}}{\partial{y}}\hat{j}+\dfrac{\partial{(\dfrac{z^2}{c^2})}}{\partial{z}}\hat{k}[/tex]

[tex]=\ \dfrac{2x}{a^2}\hat{i}+\ \dfrac{2y}{b^2}\hat{j}+\ \dfrac{2z}{c^2}\hat{k}[/tex]

Hence, the unit normal vector can be given by,

[tex]\hat{n}\ =\ \dfrac{\vec{n}}{\left|\vec{n}\right|}[/tex]

             [tex]=\ \dfrac{\dfrac{2x}{a^2}\hat{i}+\ \dfrac{2y}{b^2}\hat{j}+\ \dfrac{2z}{c^2}\hat{k}}{\sqrt{(\dfrac{2x}{a^2})^2+(\dfrac{2y}{b^2})^2+(\dfrac{2z}{c^2})^2}}[/tex]

[tex]=\ \dfrac{\dfrac{x}{a^2}\hat{i}+\ \dfrac{y}{b^2}\hat{j}+\ \dfrac{z}{c^2}\hat{k}}{\sqrt{(\dfrac{x}{a^2})^2+(\dfrac{y}{b^2})^2+(\dfrac{z}{c^2})^2}}[/tex]

Hence, the unit vector normal to each point of the given ellipsoid surface is

[tex]\hat{n}\ =\ \ \dfrac{\dfrac{x}{a^2}\hat{i}+\ \dfrac{y}{b^2}\hat{j}+\ \dfrac{z}{c^2}\hat{k}}{\sqrt{(\dfrac{x}{a^2})^2+(\dfrac{y}{b^2})^2+(\dfrac{z}{c^2})^2}}[/tex]

Answer:

Step-by-step explanation:

Sample size of 8 infants were taken and their birth lengths in inches recorded and also 6 months lengths.

If x is length at birth time, and y 6 month length

we have as per table below.

x y x^2 y^2 xy

1 19.75 25.5 390.0625 650.25 503.625

2 20.5 26.25 420.25 689.0625 538.125

3 19 25 361 625 475

4 21 26.75 441 715.5625 561.75

5 19 25.75 361 663.0625 489.25

6 18.5 25.25 342.25 637.5625 467.125

7 20.25 27 410.0625 729 546.75

8 20 26.5 400 702.25 530

Total 158 208 3125.625 5411.75 4111.625

[tex]∑ x,     ∑ x2 , ∑ y,           ∑ xy,           ∑ y2\\158 208 3125.625 5411.75 4111.625[/tex]

SSxx = 3125.625 and SSyy = 5411.75

Answer:

-½

Step-by-step explanation:

Simplify \frac{25}{100}

100

25

to \frac{1}{4}

4

1

.

-\sqrt{\frac{1}{4}}−√

4

1

2 Simplify \sqrt{\frac{1}{4}}√

4

1

to \frac{\sqrt{1}}{\sqrt{4}}

√

4

√

1

.

-\frac{\sqrt{1}}{\sqrt{4}}−

√

4

√

1

3 Simplify \sqrt{1}√

1

to 11.

-\frac{1}{\sqrt{4}}−

√

4

1

4 Since 2\times 2=42×2=4, the square root of 44 is 22.

-\frac{1}{2}−

2

1

Done

Decimal Form: -0.5

Answer: D.)

Step-by-step explanation:

The first step would be to reduce the fraction in the square root, so divide both the numerator and denominator by 25 to get 1/4.

Then calculate the root, any root of one equals one so that stays as is. The exponential form of 4 would be 2^2.

Then reduce the index of the radical and exponent with 2 and you get answer D

Answer:

110,000 base 2

Step-by-step explanation:

column 1 [the first position in the number]:

1+1=0, (carry 1)

column 2:

0+1 +1 (carried)=0, (carry 1)

column 3:

1+0+1 (carried)=0, (carry 1)

column 4:

0+1+1 (carried)=0, (carry 1)

column 5:

1+1+1=1, (carry 1)

then you write the last 1 'cause there is n number to add with:

[tex]10,101_{2}+11,011_{2}=110,000_{2}[/tex]

In binary system the highest number to write is 1, if you add 1+1, it jumps to 0, and you have to carry 1 to the next position.

If you are not sure about the sum, you can convert the numbers in base 2, to base 10, so you can know if it is correct:

[tex]10,101_{2}=21_{10}\\11,011_{2}=27_{10}\\110,000_{2}=48_{10}[/tex]

So 21+27=48.

In decimal system when you add 9+1, it jumps to 0 and then you have to carry 1 to the next position, because the the highest number you can write is 9.

Answer:

The first two coins are quarters, and the one on the right is a nickle.

the two quarters [0.25+0.25] is 0.50 cents. Add the nickle [0.5] and you have 0.55 cents!

ex:     0.25+0.25+0.5

             0.50+0.5

                 =0.55 (cents)

Step-by-step explanation:

Answer:

[tex]y(t)\ =\ C_1e^{-2t}+C_2e^t-t\dfrac{e^{-2t}}{3}-\dfrac{1}{3}[/tex]

Step-by-step explanation:

As given in question, we have to find the solution of differential equation

[tex]y"+y'-2y=1[/tex]

by using the variation in parameter method.

From the above equation, the characteristics equation can be given by

[tex]D^2+D-2\ =\ 0[/tex]

[tex]=>D=\ \dfrac{-1+\sqrt{1^2+4\times 2\times 1}}{2\times 1}\ or\ \dfrac{-1-\sqrt{1^2+4\times 2\times 1}}{2\times 1}[/tex]

[tex]=>\ D=\ -2\ or\ 1[/tex]

Since, the roots of characteristics equation are real and distinct, so the complementary function of the differential equation can be by

[tex]y_c(t)\ =\ C_1e^{-2t}+C_2e^t[/tex]

Let's assume that

     [tex]y_1(t)=e^{-2t}[/tex]          [tex]y_2(t)=e^t[/tex]

[tex]=>\ y'_1(t)=-2e^{-2t}[/tex]        [tex]y'_2(t)=e^t[/tex]

   and g(t)=1

Now, the Wronskian can be given by

[tex]W=y_1(t).y'_2(t)-y'_1(t).y_2(t)[/tex]

   [tex]=e^{-2t}.e^t-e^t(-e^{-2t})[/tex]

   [tex]=e^{-t}+2e^{-t}[/tex]

   [tex]=3e^{-t}[/tex]

Now, the particular solution can be given by

[tex]y_p(t)\ =\ -y_1(t)\int{\dfrac{y_2(t).g(t)}{W}dt}+y_2(t)\int{\dfrac{y_1(t).g(t)}{W}dt}[/tex]

[tex]=\ -e^{-2t}\int{\dfrac{e^t.1}{3.e^{-t}}dt}+e^{t}\int{\dfrac{e^{-2t}.1}{3.e^{-t}}dt}[/tex]

[tex]=\ -e^{-2t}\int{\dfrac{1}{3}dt}+\dfrac{e^t}{3}\int{e^{-t}dt}[/tex]

[tex]=\dfrac{-e^{-2t}}{3}.t-\dfrac{1}{3}[/tex]

[tex]=-t\dfrac{e^{-2t}}{3}-\dfrac{1}{3}[/tex]

Now, the complete solution of the given differential equation can be given by

[tex]y(t)\ =\ y_c(t)+y_p(t)[/tex]

      [tex]=C_1e^{-2t}+C_2e^t-t\dfrac{e^{-2t}}{3}-\dfrac{1}{3}[/tex]

Answer:

(a) The probability is 0.4696

(b) The probability is 0.5304

(c) The probability is 0.0929

Step-by-step explanation:

The total number of ways in which we can select k elements from a group n elements is calculate as:

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

So, the number of ways in which we can select four students from a group of 19 students is:

[tex]19C4=\frac{19!}{4!(19-4)!}=3,876[/tex]

On the other hand, the number of ways in which we can select four students with no math majors is:

[tex](16C4)*(3C0)=(\frac{16!}{4!(16-4)!})*(\frac{3!}{0!(3-0)!})=1820[/tex]

Because, we are going to select 4 students form the 16 students that aren't math majors and select 0 students from the 3 students that are majors.

At the same way, the number of ways in which we can select four students with one, two and three math majors are 1680, 360 and 16 respectively, and they are calculated as:

[tex](16C3)*(3C1)=(\frac{16!}{3!(16-3)!})*(\frac{3!}{1!(3-1)!})=1680[/tex]

[tex](16C2)*(3C2)=(\frac{16!}{2!(16-2)!})*(\frac{3!}{2!(3-1)!})=360[/tex]

[tex](16C1)*(3C3)=(\frac{16!}{1!(16-1)!})*(\frac{3!}{3!(3-3)!})=16[/tex]

Then, the probability that the group has no math majors is:

[tex]P=\frac{1820}{3876} =0.4696[/tex]

The probability that the group has at least one math major is:

[tex]P=\frac{1680+360+16}{3876} =0.5304[/tex]

The probability that the group has exactly two math majors is:

[tex]P=\frac{360}{3876} =0.0929[/tex]

In short, to calculate the probability of certain events in a group selection, you would identify the total possible groups, and then calculate how many of these groups satisfy your desired conditions. The probability is then calculated as the favorable events over the total possibilities.

This problem is a classic example of combinatorics and probability. The total number of ways to select four students from a total of 19 is given by the combination function: 19 choose 4. The denominator for all our probability calculations will be this total number of possible groups.

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Answer:

2500

Step-by-step explanation:

We have to find the largest product of two numbers whose sum is 100.

Let the two numbers be x and y.

Thus, we can write x+y=100

We can calculate the value of y as:

y = 100 - x

The product of these number can be written as: (x)(y) = (x)(100-x) = 100x - x²

Let f(x) = 100x - x²

Now, the first derivative of this function with respect to x is

[tex]\frac{df(x)}{dx}[/tex] = 100-2x

Equating [tex]\frac{df(x)}{dx}[/tex] = 0, we get,

100-2x = 0

⇒ x = 50

Now, we find the second derivative of the the function f(x) with respect to x

[tex]\frac{d^2f(x)}{dx^2}[/tex] = -2

Since, [tex]\frac{d^2f(x)}{dx^2}[/tex] < 0, then by double derivative test the function have a local maxima at x = 50

This, x = 50 and y = 100-50 =50

Largest product = (50)(50) = 2500

The sum of 7.25 L and 875 cL, reduced to milliliters, is 16,000 mL as per the concept of addition.

To add 7.25 L and 875 cL, we need to convert the centiliters to liters before performing the addition.

1. Convert 875 cL to liters:

Since there are 100 centiliters in a liter, we divide 875 by 100 to get the equivalent in liters:

875 cL ÷ 100 = 8.75 L

2. Now that both quantities are in liters, we can add them together:

7.25 L + 8.75 L = 16 L

3. Finally, to convert the result to milliliters, we multiply by 1000 since there are 1000 milliliters in a liter:

16 L × 1000 = 16,000 mL

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Final answer:

To add 7.25 L and 875 cL and reduce the result to milliliters, convert 875 cL to liters to get 8.75 L, then add it to 7.25 L to total 16 L. Finally, convert 16 L to milliliters by multiplying by 1,000, resulting in 16,000 mL.

Explanation:

To add 7.25 L and 875 cL and reduce the result to milliliters, you first need to understand the conversion between liters, centiliters, and milliliters. Remember, there are 1000 milliliters (mL) in a liter (L) and 10 milliliters in a centiliter (cL).

First, let's convert 875 cL to liters to simplify the addition. Since there are 10 mL in a cL, and 1000 mL in a L, you would convert as follows:

875 cL = 875 / 10 = 87.5 mL

However, we need to recognize the proper conversion to liters in the step above. Correctly, it should state: 875 cL = 8.75 L (since 100 cL = 1 L).

Once we have both measurements in liters, we can easily add them:

7.25 L + 8.75 L = 16.0 L

To convert the total liters to milliliters, multiply by 1,000 (since there are 1,000 mL in 1 L).

16.0 L × 1,000 = 16,000 mL

Answer:

The linear problem is to maximize [tex]Z = C_ {1} X_ {1} + C_ {2}X_ {2} = 60X_ {1} + 50X_ {2}[/tex], s.a.

subject to

[tex]\frac {1} {5} X_ {1} + \frac {1} {4} X_ {2} \leq 20\\\\\frac {1} {3} X_ {1} + \frac {1} {6} X_ {2} \leq 24\\\\X_ {2} \geq 2\\\\X_ {1}, X_ {2} \geq 0[/tex]

Step-by-step explanation:

Let the decision variables be:

[tex] X_ {1} [/tex]: number of units of product 1 to produce.

[tex] X_ {2} [/tex]: number of units of product 2 to produce.

Let the contributions be:

[tex]C_ {1} = 60\\\\C_ {2} = 50[/tex]

The objective function is:

[tex]Z = C_{1} X_{1}+ C_{2}X_{2} = 60X_ {1} + 50X_ {2}[/tex]

The restrictions are:

[tex]\frac {1} {5} X_ {1} + \frac {1} {4} X_ {2} \leq 20\\\\\frac {1} {3} X_ {1} + \frac {1} {6} X_ {2} \leq 24\\\\X_ {2} \geq 2\\\\X_ {1}, X{2} \geq 2\\\\[/tex]

The linear problem is to maximize [tex]Z = C_ {1} X_ {1} + C_ {2}X_ {2} = 60X_ {1} + 50X_ {2}[/tex], s.a.

subject to

[tex]\frac {1} {5} X_ {1} + \frac {1} {4} X_ {2} \leq 20\\\\\frac {1} {3} X_ {1} + \frac {1} {6} X_ {2} \leq 24\\\\X_ {2} \geq 2\\\\X_ {1}, X_ {2} \geq 0[/tex]

Answer:

Ans. the annual rate of return, in order to turn $5,000 into $24,000 in 15 years is 11.02% annual.

Step-by-step explanation:

Hi, well, in order to find the value of the interest rate of return, we need to solve for "r" the following equation,

[tex]Future Value=PresentValue(1+r)^{n}[/tex]

Where:

n= years (time that the money was invested)

r=annual rate of return (Decimal)

So, let´s see the math of this.

[tex]24,000=5,000(1+r)^{15}[/tex]

[tex]\frac{24,000}{5,000} =(1+r)^{15}[/tex]

[tex]\sqrt[15]{\frac{24,000}{5,000} } =1+r[/tex]

[tex]\sqrt[15]{\frac{24,000}{5,000} } -1=r[/tex]

[tex]r=0.11023[/tex]

So the annual rate of return that turns $5,000 into $24,000 in 15 years is 11.02%.

N=15; PV=5,000; FV=24,000; PMT=N.A; I/Y=11.02% P/Y=N.A

Best of Luck.

Answer:

There is a 47.50% probability that the chosen senator is a Democrat.

Step-by-step explanation:

This can be formulated as the following problem:

What is the probability of B happening, knowing that A has happened.

It can be calculated by the following formula:

[tex]P = \frac{P(B).P(A/B)}{P(A)}[/tex]

Where P(B) is the probability of B happening, P(A/B) is the probability of A happening knowing that B happened and P(A) is the probability of A happening.

In your problem we have that:

A(what happened) is the probability of a gun owner being chosen:

There are 100 people in the survay(53 Democrats, 45 Republicans ans 2 Independents), and 40 of them have guns(19 Democrats, 21 Republicans). So, the probability of a gun owner being chosen is:

[tex]P(A) = \frac{40}{100} = 0.4[/tex]

[tex]P(A/B)[/tex] is the probability of a senator owning a gun, given that he is a Democrat. 19 of 53 Democrats own guns, so the probability of a democrat owning a gun is:

[tex]P(A/B) = \frac{19}{53} = 0.3585[/tex]

[tex]P(B)[/tex] is the probability that the chosen senators is a Democrat. There are 100 total senators, 53 of which are Democrats, so:

[tex]P(B) = \frac{53}{100} = 0.53[/tex]

If a senator participating in that survey was picked at random and turned out to be a gun owner, what was the probability that he or she was a Democrat?

[tex]P = \frac{P(B).P(A/B)}{P(A)} = \frac{(0.53)*(0.3585)}{(0.40)} = 0.4750[/tex]

There is a 47.50% probability that the chosen senator is a Democrat.

Answer:

The answer is no: the set of all 2x2 matrices such that det(A)=0 is not a subspace of the vector space of all 2x2 matrices.

Step-by-step explanation:

In order for a set of matrices to be a subspace of all 2x2 matrices, three conditions must be satisfied:

1) That the set is not empty.

2) If A and B are both 2x2 matrices with zero determinant then the matrix A+B should also be a matrix with zero determinant.

3) The determinant of c*A, where "c" is any complex number and A is any matrix of the set, should be zero.

1)

The first condition is satisfied by the set of all 2x2 matrices such that det(A)=0, since there are plenty 2x2 matrices with zero determinant.

2)

The second condition is not satisfied, since from the determinant properties, we know that:

[tex]det(A+B)\geq det(a)+det(B)[/tex]

The equality might hold, but it is not a general characteristic. For example, if we consider the following matrices:

[tex]A = \left[\begin{array}{cc}1&0\\0&0\end{array}\right], \quad B = \left[\begin{array}{cc}0&0\\0&1\end{array}\right][/tex]

We can easily check that the determinant of both matrices is zero, nevertheless the determinant of the sum is different than zero.

Therefore, the set of all 2x2 matrices such that det(A)=0 is not a subspace of the vector space of all 2x2 matrices.

The set of all 2x2 matrices such that det(A) = 0 is not a subspace of the vector space of all 2x2 matrices.

To determine if a set is a subspace of a vector space, it must satisfy three conditions:

1. The set must contain the zero vector (in this case, the zero matrix).

2. The set must be closed under addition, meaning that if A and B are matrices in the set, then A + B must also be in the set.

3. The set must be closed under scalar multiplication, meaning that if A is a matrix in the set and c is a scalar, then cA must also be in the set.

Let's examine each condition:

1. The zero matrix, denoted by O, has a determinant of det(O) = 0, so it is included in the set. This condition is satisfied.

2. Consider two 2x2 matrices A and B with determinant 0:

[tex]\[ A = \begin{pmatrix} a b \\ c d \end{pmatrix}, \quad \text{det}(A) = ad - bc = 0 \][/tex]

 [tex]\[ B = \begin{pmatrix} e f \\ g h \end{pmatrix}, \quad \text{det}(B) = eh - fg = 0 \][/tex]

 The sum of A and B is:

 [tex]\[ A + B = \begin{pmatrix} a+e b+f \\ c+g d+h \end{pmatrix} \][/tex]

 The determinant of A + B is:

 [tex]\[ \text{det}(A + B) = (a+e)(d+h) - (b+f)(c+g) \][/tex]

 [tex]\[ \text{det}(A + B) = ad + ah + ea + eh - bc - bg - cf - fg \][/tex]

 Since det(A) = 0 and det(B) = 0, we have:

 [tex]\[ ad - bc = 0 \][/tex]

 [tex]\[ eh - fg = 0 \][/tex]

 However, this does not imply that det(A + B) = 0. The cross terms ah, ea, bg, and cf may not sum to zero, and thus det(A + B) may not be zero. Therefore, the set is not closed under addition, and this condition is not satisfied.

3. Consider a scalar c and a matrix A with determinant 0:

 [tex]\[ cA = c \begin{pmatrix} a b \\ c d \end{pmatrix} = \begin{pmatrix} ca cb \\ cc cd \end{pmatrix} \][/tex]

 The determinant of cA is:

 [tex]\[ \text{det}(cA) = (ca)(cd) - (cb)(cc) = c^2(ad - bc) \][/tex]

 Since det(A) = 0, we have ad - bc = 0, and thus det[tex](cA) = c^2(0) = 0[/tex]for any scalar c. This means that the set is closed under scalar multiplication, and this condition is satisfied.

Answer:

[tex]\frac{20,410 \text{ grams}}{254\text{ mm}}[/tex]

Step-by-step explanation:

We have been given that adhesive tape made from fabric has a tensile strength of not less than 20.41 kg/2.54 cm of width. We are asked to reduce these quantities to grams and millimeters.

We know 1 kg equals 1000 grams and 1 cm equals 10 mm.

[tex]\frac{20.41\text{ kg}}{\text{2.54 cm}}[/tex]

[tex]\frac{20.41\text{ kg}}{\text{2.54 cm}}\times \frac{\text{1 cm}}{\text{10 mm}}[/tex]

[tex]\frac{20.41\text{ kg}}{2.54\times\text{10 mm}}[/tex]

[tex]\frac{20.41\text{ kg}}{254\text{ mm}}[/tex]

[tex]\frac{20.41\text{ kg}}{254\text{ mm}}\times \frac{\text{1000 grams}}{\text{1 kg}}[/tex]

[tex]\frac{20.41\times \text{1000 grams}}{254\text{ mm}}[/tex]

[tex]\frac{20,410 \text{ grams}}{254\text{ mm}}[/tex]

Therefore, our required measurement would be [tex]\frac{20,410 \text{ grams}}{254\text{ mm}}[/tex].

Answer:

a. [tex]p \wedge q[/tex]

b. [tex]\neg p \wedge q[/tex]

c. [tex]p\Rightarrow q[/tex]

d. [tex]p \Leftrightarrow q[/tex]

e. [tex]\neg p \Rightarrow \neg q[/tex]

Step-by-step explanation:

a. 4 is an even integer and -5 is a negative prime number, can be represented by: [tex]p \wedge q[/tex]

b. 4 is not an even integer and-5 is a negative prime number, can be represented by: [tex]\neg p \wedge q[/tex]

c. If 4 is an even integer, then-5 is a negative prime number, can be represented by: [tex]p\Rightarrow q[/tex]

d. 4 is an even integer if and only if-5 is a negative prime number, can be represented by: [tex]p \Leftrightarrow q[/tex]

e. If 4 is not an even integer, then-5 is not a negative prime number,  can be represented by: [tex]\neg p \Rightarrow \neg q[/tex]

The statements a-e are translated into the language of logic as p ∧ q, ¬p ∧ q, p → q, p ↔ q, and ¬p → ¬q, respectively. These represent different logical relationships between the statements '4 is an even integer' and '-5 is a negative prime number'.

To rewrite the provided statements using logical connectors and the given identifiers (p: 4 is an even integer, q: -5 is a negative prime number), we proceed as follows:

a. p ∧ q: This reads "p and q", representing the statement "4 is an even integer and -5 is a negative prime number".

b. ¬p ∧ q: The symbol ¬ stands for "not", so this reads "not p and q", representing "4 is not an even integer and -5 is a negative prime number".

c. p → q: This reads "p implies q", representing "If 4 is an even integer, then -5 is a negative prime number".

d. p ↔ q: This reads "p if and only if q", representing "4 is an even integer if and only if -5 is a negative prime number".

e. ¬p → ¬q: This reads "not p implies not q", representing "If 4 is not an even integer, then -5 is not a negative prime number".

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Answer:

The value of x is [tex]\frac{(3c+b)}{3-c}[/tex].

Step-by-step explanation:

The given equation is

[tex]cx+b=3(x-c)[/tex]

Using distributive property we get

[tex]cx+b=3(x)+3(-c)[/tex]

[tex]cx+b=3x-3c[/tex]

To solve the above equation isolate variable terms.

Subtract 3x and b from both sides.

[tex]cx-3x=-3c-b[/tex]

Taking out common factors.

[tex]x(c-3)=-(3c+b)[/tex]

Divide both sides by (c-3).

[tex]x=-\frac{(3c+b)}{c-3}[/tex]

[tex]x=\frac{(3c+b)}{3-c}[/tex]

Therefore the value of x is [tex]\frac{(3c+b)}{3-c}[/tex].

Answer:

The statement is true for every n between 0 and 77 and it is false for [tex]n\geq 78[/tex]

Step-by-step explanation:

First, observe that, for n=0 and n=1 the statement is true:

For n=0: [tex]\sum^{n}_{i=0} (2i)^4=0 \leq 0=(4n)^4[/tex]

For n=1: [tex]\sum^{n}_{i=0} (2i)^4=16 \leq 256=(4n)^4[/tex]

From this point we will assume that [tex]n\geq 2[/tex]

As we can see, [tex]\sum^{n}_{i=0} (2i)^4=\sum^{n}_{i=0} 16i^4=16\sum^{n}_{i=0} i^4[/tex] and [tex](4n)^4=256n^4[/tex]. Then,

[tex]\sum^{n}_{i=0} (2i)^4 \leq(4n)^4 \iff \sum^{n}_{i=0} i^4 \leq 16n^4[/tex]

Now, we will use the formula for the sum of the first 4th powers:

[tex]\sum^{n}_{i=0} i^4=\frac{n^5}{5} +\frac{n^4}{2} +\frac{n^3}{3}-\frac{n}{30}=\frac{6n^5+15n^4+10n^3-n}{30}[/tex]

Therefore:

[tex]\sum^{n}_{i=0} i^4 \leq 16n^4 \iff \frac{6n^5+15n^4+10n^3-n}{30} \leq 16n^4 \\\\ \iff 6n^5+10n^3-n \leq 465n^4 \iff 465n^4-6n^5-10n^3+n\geq 0[/tex]

and, because [tex]n \geq 0[/tex],

[tex]465n^4-6n^5-10n^3+n\geq 0 \iff n(465n^3-6n^4-10n^2+1)\geq 0 \\\iff 465n^3-6n^4-10n^2+1\geq 0 \iff 465n^3-6n^4-10n^2\geq -1\\\iff n^2(465n-6n^2-10)\geq -1[/tex]

Observe that, because [tex]n \geq 2[/tex] and is an integer,

[tex]n^2(465n-6n^2-10)\geq -1 \iff 465n-6n^2-10 \geq 0 \iff n(465-6n) \geq 10\\\iff 465-6n \geq 0 \iff n \leq \frac{465}{6}=\frac{155}{2}=77.5[/tex]

In concusion, the statement is true if and only if n is a non negative integer such that [tex]n\leq 77[/tex]

So, 78 is the smallest value of n that does not satisfy the inequality.

Note: If you compute  [tex](4n)^4- \sum^{n}_{i=0} (2i)^4[/tex] for 77 and 78 you will obtain:

[tex](4n)^4- \sum^{n}_{i=0} (2i)^4=53810064[/tex]

[tex](4n)^4- \sum^{n}_{i=0} (2i)^4=-61754992[/tex]

We will use mathematical induction to prove this inequality.

We will prove this inequality using mathematical induction. First, let's check the base case when n = 0. The left-hand side (LHS) of the inequality is 0 and the right-hand side (RHS) is (4*0)^4 = 0. So, the inequality holds for n = 0.

Next, assume that the inequality holds for some positive integer k, i.e.,

∑([2i]^4) ≤ (4k)^4 (where the sum is taken from i = 0 to k)

We will prove that it also holds for k + 1. Adding the (k+1)th term to both sides of the inequality:

∑([2i]^4) + ([2(k+1)]^4) ≤ (4k)^4 + ([2(k+1)]^4)

Now, simplifying the LHS and RHS:

(∑([2i]^4)) + ([2(k+1)]^4) ≤ (4k)^4 + ([2(k+1)]^4)

Answer:

Step-by-step explanation:

Given that the solution of a certain differential equation is of the form

[tex]y(t) = ae^{7t} +be^{11t}[/tex]

Use the initial conditions

i) y(0) =1

[tex]1=a(1)+b(1)\\a+b=1[/tex] ... I

ii) y'(0) = 4

Find derivative of y first and then substitute

[tex]y'(t) = 7ae^{7t} +11be^{11t}\\y'(0) =7a+11b \\7a+11b =4 ...II[/tex]

Now using I and II we solve for a and b

Substitute b = 1-a in II

[tex]7a+11(1-a) = 4\\-4a+11 =4\\-4a =-7\\a = 1.75 \\b = -0.75[/tex]

Hence solution is

[tex]y(t) = 1.75e^{7t} -0.75e^{11t}[/tex]

Answer:

y(t) = a exp(3t) + b exp(4t) conditions, y(0) = 3 y'(0) = 3 y(0) = a exp(3 x 0) + b exp(4 x 0) = a exp(0) + b exp(0) = (a x 1) + (b x 1) = a + b y'(0) = 0 so the linear equation is, a + b = 3

Step-by-step explanation:

Answer:

$114

Step-by-step explanation:

make a calender and count every 4 days for dogs and every 6 days for the lawn. Then add all the money up.

The problem is a step-by-step calculation with several possible combinations of type A and type B vehicles when a total of 101 counts are registered. A systematic approach is required to find all possible whole number solutions.

This problem is an example of a diophantine problem or a linear equation in two variables. If we denote the number of type A vehicles by 'a', and the number of type B vehicles by 'b', the problem can be represented by the equation 2a + 9b = 101.

As you are looking for all possible solutions, you have to do a systematic search. You will find that:

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Answer:

Solution: [tex]B=5+25e^{4-4t}[/tex]

Step-by-step explanation:

Given: [tex]\dfrac{dB}{dt}+4B=20[/tex]  

with B(1)=30

The differential equation in form of linear differential equation,

[tex]\dfrac{dy}{dt}+Py=Q[/tex]

Integral factor, IF: [tex]e^{\int Pdt}[/tex]

General Solution:

[tex]y\cdot IF=\int Q\cdot IFdt[/tex]

[tex]\dfrac{dB}{dt}+4B=20[/tex]  

P=4, Q=20

IF= [tex]e^{\int 4dt}=e^{4t}[/tex]

Solution:

[tex]Be^{4t}=\int 20e^{4t}dt[/tex]

[tex]Be^{4t}=5e^{4t}+C[/tex]

[tex]B=5+Ce^{-4t}[/tex]

B(1)=30 , Put t=1, B=30

[tex]30=5+Ce^{-4}[/tex]

[tex]C=25e^4[/tex]

[tex]B=5+25e^{4-4t}[/tex]

Answer: The number is 40.

Step-by-step explanation:

Since we have given that

Let the number be 'x'.

If a number is decreased by 40%.

So, number becomes,

[tex]\dfrac{100-40}{100}\times x\\\\=\dfrac{60}{100}\times x\\\\=0.6x[/tex]

According to question, the result becomes 24.

So, our equation becomes,

[tex]0.6x=24\\\\x=\dfrac{24}{0.6}\\\\x=\dfrac{240}{6}\\\\x=40[/tex]

Hence, the number is 40.

Answer:

5

Step-by-step explanation:

-7x-3x+2= -8x-8;

-10x+2= -8x-8;

-10x+2+8x= -8;

-10x+8x= -8-2;

-2x= -10;

x=(-10)/(-2);

x=5.

During the flood season, the water rises 26.4 feet up the side of the lake.

The angle of elevation is given as 15 degrees 30 minutes,

Now, it can be converted to decimal degrees as 15.5 degrees.

Let's denote the distance up the side of the lake as x feet.

Now set up a trigonometric equation using the tangent function:

tan(15.5°) = (7.3 feet) / x

We can solve for x by rearranging the equation:

x = (7.3 feet) / tan(15.5°)

Evaluating this expression gives us:

x = (7.3 feet) / 0.277

x = 26.4 feet

Therefore, the water rises 26.4 feet up the side of the lake.

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Answer:

the rate is 0.208

Answer:

[tex]f'(0)=0[/tex]

Step-by-step explanation:

Applying the chain rule

[tex]\frac{d}{dx} (f(-x))=-\frac{df}{dx}[/tex]

Then it becomes

[tex]\frac{df}{dx} =-\frac{df}{dx}[/tex]

In x=0

[tex]\frac{d[tex]f'(0)=-f'(0)\\f'(0)+f'(0)=0\\2f'(0)=0\\[/tex]f}{dx} =-\frac{df}{dx}[/tex]

Then

[tex]f'(0)=0[/tex]

Answer:

25

Step-by-step explanation:

Given:

Volume of Prochlorperazine injection available = 10 mL

Dose per vial = 5 mg/mL

Now,

The total mass of dose present in 10 mL = Volume × Dose

or

The total mass of dose present in 10 mL = 10 × 5 = 50 mg

Thus,

The number of 2 mg dose that can be withdrawn = [tex]\frac{\textup{50 mg}}{\textup{2 mg}}[/tex]

or

The number of 2 mg dose that can be withdrawn = 25

Answer: 25 doses of 2 mg each from the 10-mL vial

Step-by-step explanation:

To determine how many 2-mg doses can be withdrawn from a 10-mL vial containing Prochlorperazine at a concentration of 5 mg/mL, you can use the following calculation:

1. Calculate the total amount of Prochlorperazine in the vial:

  Total amount = Concentration × Volume

  Total amount = 5 mg/mL × 10 mL

  Total amount = 50 mg

2. Now, calculate how many 2-mg doses can be withdrawn:

  Number of 2-mg doses = Total amount / Dose per patient

  Number of 2-mg doses = 50 mg / 2 mg/dose

  Number of 2-mg doses = 25 doses

So, you can withdraw 25 doses of 2 mg each from the 10-mL vial of Prochlorperazine.

Answer:

Pigs = 20 and Chickens = 50

Step-by-step explanation:

Let the number of pigs be x

and let the number of chicken be y

Thus, x + y = 70

Since Chicken has 2 legs and pigs has 4 legs.

⇒ 4x + 2y = 180

Solving both equations,

We get, x = 20 and y = 50

Thus number of pigs = 20

and, number of chickens = 50.

Answer:

1) yes

2) no, (fog)(x)=3-x/2

3) no, (f o g)(x)= -2x+6/3x-5

The first and second pair of functions are not inverses of each other, while the third pair of functions are inverses.

The first pair of functions, f(x) = 2x + 3/(4x - 3) and g(x) = 3x + 3/(4x - 2), are not inverses of each other. To determine this algebraically, we need to calculate (f o g)(x) and (g o f)(x) and check if they equal to x. In this case, (f o g)(x) is x + 2/3, which is not equal to x, therefore f and g are not inverses of each other.

The second pair of functions, f(x) = -x^3 + 2 and g(x) = 3(cubedroot)x - 2/(2), are also not inverses. By calculating (f o g)(x) and (g o f)(x), we found that (f o g)(x) = x - 14/8, which is not equal to x.

The third pair of functions, f(x) = -2x + 4/(2 - 5x) and g(x) = (4 - 2x)/(5 - 2x), is indeed inverses of each other. By calculating (f o g)(x) and (g o f)(x), we found that (f o g)(x) = x and (g o f)(x) = x, which means f and g are inverses of each other.

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Answer:

The answers are given below:

Step-by-step explanation:

1. a. (─13)³⁵

1. b. The product will be negative. The expanded form shows 34 negative factors plus one more negative factor. Any even number of negative factor yields a positive product. The remaining 35th negative factor negates the resulting product.

2. 4 times.

3. Arnie is not correct. The base, ─3.1, should be in parentheses to prevent ambiguity. At present the notation is not correct.