Which type of arc requires another entity to specify its start point?

The table that shows a function that is decreasing only over the interval (–1, ∞) is:

A 2-column table with 6 rows. The first column is labeled x with entries negative 3, negative 2, negative 1, 0, 1, 2. The second column is labeled f of x with entries negative 1, negative 3, negative 5, negative 2, negative 1, 2.

A function is decreasing if when we increase the value of x, we decrease the value of y, and vice versa.

Decreasing on the (–1, ∞) means that as x increases on the interval, i.e. x = -100, then x = -10, then x = -2, the value of y decreases. The function that follows this pattern only on this interval is:

A 2-column table with 6 rows. The first column is labeled x with entries negative 3, negative 2, negative 1, 0, 1, 2. The second column is labeled f of x with entries negative 1, negative 3, negative 5, negative 2, negative 1, 2.

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Answer:

A

Step-by-step explanation:

Answer:

[tex]2 \tan (6x)+2 \sec (6x)+\text{C}[/tex]

Step-by-step explanation:

Fundamental Theorem of Calculus

[tex]\displaystyle \int \text{f}(x)\:\text{d}x=\text{F}(x)+\text{C} \iff \text{f}(x)=\dfrac{\text{d}}{\text{d}x}(\text{F}(x))[/tex]

If differentiating takes you from one function to another, then integrating the second function will take you back to the first with a constant of integration.

Given indefinite integral:

[tex]\displaystyle \int \dfrac{12}{1-\sin (6x)}\:\:\text{d}x[/tex]

[tex]\boxed{\begin{minipage}{5 cm}\underline{Terms multiplied by constants}\\\\$\displaystyle \int a\:\text{f}(x)\:\text{d}x=a \int \text{f}(x) \:\text{d}x$\end{minipage}}[/tex]

If the terms are multiplied by constants, take them outside the integral:

[tex]\implies 12\displaystyle \int \dfrac{1}{1-\sin (6x)}\:\:\text{d}x[/tex]

Multiply by the conjugate of 1 - sin(6x) :

[tex]\implies 12\displaystyle \int \dfrac{1}{1-\sin (6x)} \cdot \dfrac{1+\sin(6x)}{1+\sin(6x)}\:\:\text{d}x[/tex]

[tex]\implies 12\displaystyle \int \dfrac{1+\sin(6x)}{1-\sin^2(6x)} \:\:\text{d}x[/tex]

[tex]\textsf{Use the identity} \quad \sin^2 x+ \cos^2 x=1:[/tex]

[tex]\implies \sin^2 (6x) + \cos^2 (6x)=1[/tex]

[tex]\implies \cos^2 (6x)=1- \sin^2 (6x)[/tex]

[tex]\implies 12\displaystyle \int \dfrac{1+\sin(6x)}{\cos^2(6x)} \:\:\text{d}x[/tex]

Expand:

[tex]\implies 12\displaystyle \int \dfrac{1}{\cos^2(6x)}+\dfrac{\sin(6x)}{\cos^2(6x)} \:\:\text{d}x[/tex]

[tex]\textsf{Use the identities }\:\: \sec \theta=\dfrac{1}{\cos \theta} \textsf{ and } \tan\theta=\dfrac{\sin \theta}{\cos \theta}:[/tex]

[tex]\implies 12\displaystyle \int \sec^2(6x)+\dfrac{\tan(6x)}{\cos(6x)} \:\:\text{d}x[/tex]

[tex]\implies 12\displaystyle \int \sec^2(6x)+\tan(6x)\sec(6x) \:\:\text{d}x[/tex]

[tex]\boxed{\begin{minipage}{5 cm}\underline{Integrating $\sec^2 kx$}\\\\$\displaystyle \int \sec^2 kx\:\text{d}x=\dfrac{1}{k} \tan kx\:\:(+\text{C})$\end{minipage}}[/tex]

[tex]\boxed{\begin{minipage}{6 cm}\underline{Integrating $ \sec kx \tan kx$}\\\\$\displaystyle \int \sec kx \tan kx\:\text{d}x= \dfrac{1}{k}\sec kx\:\:(+\text{C})$\end{minipage}}[/tex]

[tex]\implies 12 \left[\dfrac{1}{6} \tan (6x)+\dfrac{1}{6} \sec (6x) \right]+\text{C}[/tex]

Simplify:

[tex]\implies \dfrac{12}{6} \tan (6x)+\dfrac{12}{6} \sec (6x)+\text{C}[/tex]

[tex]\implies 2 \tan (6x)+2 \sec (6x)+\text{C}[/tex]

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Substitute [tex]y=6x[/tex] and [tex]dy=6\,dx[/tex] to transform the integral to

[tex]\displaystyle \int \frac{12}{1-\sin(6x)} \, dx = 2 \int \frac{dy}{1 - \sin(y)}[/tex]

Now substitute [tex]t=\tan\left(\frac y2\right)[/tex] and [tex]dt=\frac12 \sec^2\left(\frac y2\right) \, dy[/tex] to transform this to

[tex]\displaystyle 2 \int \frac{dy}{1 - \sin(y)} = 2 \int \frac1{1-\frac{2t}{1+t^2}}\cdot\frac{2\,dt}{1+t^2} = 4 \int \frac{dt}{(t-1)^2}[/tex]

Finally, substitute [tex]s = t-1[/tex] and [tex]ds=dt[/tex] to get

[tex]\displaystyle 4 \int \frac{dt}{(t-1)^2} = 4 \int \frac{ds}{s^2} = -\dfrac4s + C[/tex]

Now recover the antiderivative in terms of [tex]x[/tex].

[tex]\displaystyle \int \frac{12}{1-\sin(6x)} \, dx = -\frac4s + C \\\\ ~~~~~~~~ = -\frac4{t-1} + C \\\\ ~~~~~~~~ = -\frac4{\tan\left(\frac y2\right) - 1} + C \\\\ ~~~~~~~~ = \boxed{-\frac4{\tan(3x) - 1} + C}[/tex]

The correct option for the matrix will be:

a) If an n x n matrix A has fewer than n distinct eigenvalues, then A is not diagonalizable.

It could have repeated eigenvalues as long as the basis of each eigenspace is equal to the multiplicity of that eigenvalue.

b) If A is diagonalizable the A2 is diagonalizable

If A is diagonalizable then there exists an invertible matrix

c) If Rn has a basis of eigenvectors of A, then A is diagonalizable.

d) A is diagonalizable if and only if A has n eigenvalues, counting multiplicity.

e) If A is diagonalizable, then A is invertible.

It’s invertible if it doesn’t have a zero as eigenvalue but this doesn’t affect diagonalizable.

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Answer:

54 degrees

Step-by-step explanation:

The actual angles of the triangle are not changing since this is just a reflection. The angles inside a triangle all add up to 180, so we can do 180-59-67 to get 54.

Mathematical logic is important as it's a way to learn new experience through continuous self assessment.

Logic is important as it enables us to form sound judgements and beliefs.

The study of logic can help as it can help us to understand disagreement and ambiguity.

It also helps us in making a reasonable emotional life.

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Answer:

8.048 liters

Step-by-step explanation:

Convert 8542 ml to liters:  (8542 ml)(1 liter/1000 ml) = 8.542 liters

Now subtract 0.458 liters to find the amount that a fish tank can hold:

 8.542 liters

-0.458  liters

8.048 liters is the fish tank capacity

Answer:

50 minutes

Step-by-step explanation:

3:20 pm - 4:50 = 90 minutes

90 = 2a + 15 + a

90-15= 75

75= 2a + a

75 = 3a

75/3 = 25

25 x 2 = 50 minutes

The portion of the mixture that is fruit to the nearest hundredth is 10.00 ounces

12 ounce juice:

= 70/100 × 12

= 0.7 × 12

= 8.4 ounces

Water = 30%

16 ounces;

= 10/100 × 16

= 0.1 × 16

= 1.6 ounces

Portion of mixture that is fruits = 8.4 ounces + 1.6 ounces

= 10.00 ounces

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Answer:

96 is the number of original women investors.

Step-by-step explanation:

Let X be the number of young women in the initial group.  They raised 480000, so the average payment per person was (480000/X).  

When 4 pull out, the new average is (480000/(X-4)).  We are told that this new average required the remaining women (X-4) to add another 20000.

The four women therefore had contributed 20000 in total, making their average 20000/4 = 5000 each.  

This would have been the same amount contributed by all X women.  Thus, we can set the average (480000/X) equal to 5000

(480000/X) = 5000

(480000) = 5000X

(480000)/5000 = X

X = 96

The original number of women was 96.

===

Check

(96 Women)(5000/Woman) = 480000  CHECKS

(4 Women pull out)*(5000) = 20000 that needs to be added to stay at 480000.  CHECKS

[tex]P(R \: then \: Y) = \frac{3}{10} \times \frac{1}{10} = \frac{3}{100} [/tex]

[tex]note \: that = \\ order \: matters \: (no \: factorial) \\ replacement \: (equal \: sample \: space)[/tex]

Answer:

c. 3/100

Step-by-step explanation:

there are 10 marbles

First draw (there are 3 red marbles)

[tex]P=\frac{3}{10}[/tex]

Second draw (there is one yellow marble)

[tex]P=\frac{1}{10}[/tex]

probability of the event:

[tex]P=(\frac{3}{10} )(\frac{1}{10} )=\frac{3}{100}[/tex]

Hope this helps

Answer:

Θ ≈ 37° , x ≈ 13 cm

Step-by-step explanation:

(a)

using the cosine ratio in the right triangle

cosΘ = [tex]\frac{adjacent}{hypotenuse}[/tex] = [tex]\frac{BC}{AC}[/tex] = [tex]\frac{8}{10}[/tex] , then

Θ = [tex]cos^{-1}[/tex] ([tex]\frac{8}{10}[/tex] ) ≈ 37° ( to the nearest whole number )

(b)

using the sine ratio in the right triangle and the exact value

sin60° = [tex]\frac{\sqrt{3} }{2}[/tex] , then

sin60° = [tex]\frac{opposite}{hypotenuse}[/tex] = [tex]\frac{x}{15}[/tex] = [tex]\frac{\sqrt{3} }{2}[/tex] ( cross- multiply )

2x = 15[tex]\sqrt{3}[/tex] ( divide both sides by 2 )

x = 7.5[tex]\sqrt{3}[/tex] ≈ 13 cm ( to the nearest whole number )

It is determined while using the binomial distribution that there is still a 1.145=114.5% chance that she produces no more than 11 of them.

There are just two possible results for each throw. Either she succeeds or she fails. The binomial probability distribution is employed to answer this issue since the probability of completing a shot is regardless of all other throws.

Binomial probability distribution-

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

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

where,

the no. of success= x

the no. of trials = n

the probability of a success on one trial = p

The probability of throwing not more than 11 will be:

P(X<11) = P(X=0) + P(X=1) +P(X=2) + P(X=3) +P(X=4) +P(X=5)+P(X=6)+P(X=7)+P(X=8)+P(X=9)+P(X=10)+P(X=11)

Where,

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

[tex]P(X=0) = C_{18,0} .(0.689)^{0}(0.311)^{18}[/tex]≈0

[tex]P(X=1) = C_{18,1} .(0.689)^{1}(0.311)^{17}[/tex]≈0

[tex]P(X=2) = C_{18,2} .(0.689)^{2}(0.311)^{16}[/tex]≈0

[tex]P(X=3) = C_{18,3} .(0.689)^{3}(0.311)^{15}[/tex]≈0

[tex]P(X=4) = C_{18,4} .(0.689)^{4}(0.311)^{14}[/tex]≈0

[tex]P(X=5) = C_{18,5} .(0.689)^{5}(0.311)^{13}[/tex]= 0.0003

[tex]P(X=6) = C_{18,4} .(0.689)^{6}(0.311)^{12}[/tex]=0.0016

[tex]P(X=7) = C_{18,7} .(0.689)^{7}(0.311)^{11}[/tex]=0.0062

[tex]P(X=8) = C_{18,8} .(0.689)^{8}(0.311)^{10}[/tex]=0.0188

[tex]P(X=9) = C_{18,9} .(0.689)^{9}(0.311)^{9}[/tex]=0.0463

[tex]P(X=10) = C_{18,10} .(0.689)^{10}(0.311)^{8}[/tex]=0.9232

[tex]P(X=11) = C_{18,11} .(0.689)^{11}(0.311)^{7}[/tex]=0.1488

So,

P(X<11) = P(X=0) + P(X=1) +P(X=2) + P(X=3) +P(X=4) +P(X=5)+P(X=6)+P(X=7)+P(X=8)+P(X=9)+P(X=10)+P(X=11)

=0+0+0+0+0+0.0003+0.0016+0.0062+0.0188+0.0463+0.9232+0.1488 =1.145

Therefore, she makes 1.145=114.5% probability, no more than 11 of them.

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The solution of the equation are as follows:

x = 6 and y = -5

4x + 5y = -1

-5x - 8y = 10

Therefore,

20x + 25y = -5

-20x - 32y = 40

-7y = 35

y = -5

Hence,

4x + 5(-5) = -1

4x - 25 = -1

4x = -1 + 25

4x = 24

x = 24 / 4

x = 6

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Let [tex]n=abcba[/tex] be such a number. If 18 divides [tex]n[/tex], then both 2 and 9 divide [tex]n[/tex].

To be divisible by 2, we must have [tex]a\in\{2,4,6,8\}[/tex]. Meanwhile we can have [tex](b,c)\in\{0,1,2,\ldots,9\}^2[/tex].

To be divisible by 9, we the sum of the digits of [tex]n[/tex] must itself be divisible by 9, or

[tex]2a+2b+c=9k[/tex]

for some integer [tex]k[/tex].

The largest value of [tex]2a+2b+c[/tex] is 2•8 + 2•9 + 9 = 43, so we must have [tex]k\in\{1,2,3,4\}[/tex].

I'm not sure what the best way to get the final count may be, but there are 44 such numbers. It's rather tedious to do by hand.

• If [tex]k=1[/tex], then [tex]2a+2b+c=9[/tex], and we can do this in 4 ways.

For example,

2•2 + 2•0 + 5 = 9 [tex](n = 20502)[/tex]

• If [tex]k=2[/tex], then [tex]2a+2b+c=18[/tex] and can be done in 16 ways.

2•2 + 2•3 + 8 = 18 [tex](n = 23832)[/tex]

• If [tex]k=3[/tex], then [tex]2a+2b+c=27[/tex] and can be done in 18 ways.

2•2 + 2•7 + 9 = 27 [tex](n = 27972)[/tex]

• If [tex]k=4[/tex], then [tex]2a+2b+c=36[/tex] and can be done in 6 ways.

2•6 + 2•8 + 8 = 36 [tex](n = 68886)[/tex]

Answer:

2061.9[tex]mm^{3}[/tex]

Step-by-step explanation:

We can see the original shape of the gold block is a cuboid.

Volume of Cuboid = Length x Width x Height

Volume of Original Gold Block = 27mm x 8mm x 12mm = [tex]2592mm^{3}[/tex]

Next, we can see the shape of the cut-out hole is a cylinder.

Volume of Cylinder = [tex]\pi r^{2} h[/tex]

We know that r = 0.5 x diameter = 0.5 x 5 = 2.5mm

Volume of cut-out hole = [tex]\pi (2.5)^{2} (27)\\[/tex] = [tex]168.75\pi mm^{3}[/tex]

Volume of Remaining Gold Block = Volume of Original Gold block - Volume of cut-out hole

= [tex]2592mm^{3} -168.75\pi mm^{3}[/tex]

= 2061.9[tex]mm^{3}[/tex]

Answer:

See attached image

Step-by-step explanation:

The expression that is not equivalent to (1 - sin²(x)) tan(-x) is D. (cos²x - 1)(cot -x).

It should be noted that (cos²x - 1)(cot -x). is not equivalent to the given expression.

This is illustrated as:

(cos²x - 1)(cot -x)

= (-sin²x) × (-cot x)

= sin²x × cosx/sin x = sinxcosx

In conclusion, the correct option is D.

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The expression _______ is not equivalent to (1 − sin2(x)) tan(-x).

a. (1 - cos^2(x)) cot(-x)

b. (cos^2(x) - 1) cot(x)

c. (sin^(x) - 1) tan(x)

d. (cos^2(x) - 1) cot(-x)

Answer:

Ralph Warren purchased 27 shares of stock at 16 3/8 per share. He paid a $27.50 brokerage fee. He later sold all 27 shares at 17 5/8 and paid a $28.75 brokerage fee. (36) What was his total cost for the stock including his brokerage fee? (37) What did he receive from the sale of the stock after he paid the brokerage fee? (38) Did he have a capital gain or loss? (39) How much was the gain or loss? (40) What was the net change from 16 3/8 to 17 5/8?

Step-by-step explanation:

The discriminant of the quadratic equation is -20 and there are two non real solutions.

The discriminant of a quadratic equation uses the equation:

[tex]b^{2} -4ac[/tex]

Where the value of this calculation can tell you what solutions there are, plug known values in:

[tex]x^{2} +6x+14[/tex]

a=1

b=6

c=14

[tex]b^{2} -4ac[/tex]

which is equal to -20 on putting the values of a,b and c in the equation

As -20 < 0, this means that there is not a real solution, resulting in the first option being correct.

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For a proportional relationship, the constant is found dividing all values of y by each respective value of x.

A proportional relationship is a function in which the output variable is given by the input variable multiplied by a constant of proportionality, that is:

y = kx

In which k is the constant of proportionality.

The constant can be represented as follows:

[tex]k = \frac{y}{x}[/tex]

Hence the constant is found dividing all values of y by each respective value of x.

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Step-by-step explanation:

Look, you're taking these kinds of questions too serious and that's why you believe they're hard, now pay close attention:

the areas are defined in cm² but the radius is defined in cm, so you need find the positive root of cm².

[tex] \sqrt{ \frac{5500 {cm}^{2} }{220 {cm}^{2} } } = \frac{r}{5cm} = = > \\ 5 = \frac{r}{5} = = = > r = 25[/tex]

and the question wants the base area:

[tex]s = \pi {r}^{2} = = = > s = 3.14 \times {(25cm)}^{2} = = > 1963.5 {cm}^{2} [/tex]

Hello,

2x² - 8x = 0

2x × x - 2x × 4 = 0

2x(x - 4) = 0

2x = 0 or x - 4 = 0

x = 0 or x = 4

The numbers : 4,-10 and -4 are integers.

A number without a decimal or fractional element is known as an integer, which can be both positive and negative, including zero.

The Latin word "integer" signifies "whole" or "intact." Thus, fractions and decimals are not included in integers.

All whole numbers and negative numbers are considered integers. This means that if we combine negative numbers with whole numbers, a collection of integers results.

"Negative 1 and one-third" includes a fractional part, so it is not an integer.

"2.5" and "0.ModifyingAbove 13 with Bar" contain a decimal, so it is also not an integer.

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Based on the calculations, the correct value of the test statistic is equal to 3.2.

For samples A and B, the hypothesis is given by:

H₀: μ₁ ≤ μ₂

H₁: μ₁ > μ₂

Since both samples have a normal distribution, we would use a pooled z-test to determine the value of the test statistic:

[tex]z = \frac{\bar{x_1} - \bar{x_2} -(\mu_1 - \mu_2) }{\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_1^2}{n_1}} }[/tex]

Substituting the given parameters into the formula, we have;

[tex]z = \frac{33 - 25 -(0) }{\sqrt{\frac{3^2}{4} + \frac{4^2}{4}} }\\\\z = \frac{8 }{\sqrt{\frac{9}{4} + \frac{16}{4}} }\\\\z = \frac{8 }{\sqrt{\frac{25}{4} } }\\\\z = \frac{8 }{\frac{5}{2} }}\\\\z = 8 \times \frac{2}{5}[/tex]

z = 3.2.

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Answer:

  7

Step-by-step explanation:

A 2-way table can be useful for recording the given information and for finding the missing numbers.

The attached table shows the numbers of families booking in July and August, and also counted by destination. Totals are at the right and bottom, and the grand total is the number 100 at the lower right.

Underlined numbers are those in the problem statement. The remaining numbers are computed so as to make the totals be correct.

Overall, 30+35 = 65 went to Portugal or France, so 100-65 = 35 went to Spain. Of those, 20 booked in August, so 15 booked to Spain in July.

59 booked in August, so a total of 41 booked in July.

Now we know 19 booked for France and 15 booked for Spain in July, so the remaining 7 booked for Portugal in July.

The dosage at which the resulting blood pressure is maximized is x at 0.2 and the maximum dosage is 0.0008.

In the question,

The function is [tex]B(x) = 0. 06x^2 - 0. 2x^3[/tex]

To find the maximum or minimum, take the derivative and set it equal to zero.

⇒ [tex]B'(x) = 2(0. 06)x - 3(0. 2)x^{2}[/tex]

Setting it equal to zero, we get

⇒ [tex]0.12x - 0.6x^{2}=0[/tex]

⇒ 0.6x (0.2-x) = 0

⇒ x = 0 or x = 0.2

Now substitute x = 0.2 in B(x), we get

⇒ [tex]B(0.2) = 0. 06(0.2)^{2} - 0. 2(0.2)^{3}[/tex]

⇒ B(0.2) = 0.0024 - 0.0016

⇒ B(0.2) = 0.0008

To know B(0.2) is maximum, let us find the values for x = 1 and x = 0.01.

For x = 1,

⇒ [tex]B(1) = 0. 06(1)^{2} - 0. 2(1)^{3}[/tex]

⇒ B(1) = 0.06-0.2

⇒ B(1) = -1.04

For x = 0.01,

⇒ [tex]B(0.01) = 0. 06(0.01)^{2} - 0. 2(0.01)^{3}[/tex]

⇒ B(0.01) = 0.000006 - 0.0000002

⇒ B(0.01) = 0.0000058

Thus, x = 0.2 is the maximum.

Hence we can conclude that the dosage at which the resulting blood pressure is maximized is x at 0.2 and the maximum dosage is 0.0008.

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The solution to the inequality is (–2.5, ∞)

The inequality is given as:

5.1(3 + 2.2x) > –14.25 – 6(1.7x + 4)

Open the brackets

15.3 + 11.22x > –14.25 – 10.2x - 24

Collect the like terms

11.22x + 10.2x > -15.3 - 14.25- 24

Evaluate the like terms

21.42x > -53.55

Divide both sides by 21.42

x > -2.5

This can also be represented as (–2.5, ∞)

Hence, the solution to the inequality is (–2.5, ∞)

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Answer:

c

Step-by-step explanation:

on edge