Simplify to a single power of 3:

Answer:

[tex]fg x5 \\ 5fg[/tex]

[tex] \large \bf \implies \angle{BAC} = 40 \degree[/tex]

[tex] \bf \implies \angle{ABC} + \angle{BAC} = 90\degree[/tex] [The acute angles of a right triangle are complementary]

[tex] \bf \implies 50x + 40x = 90\degree[/tex]

Substitute :

[tex]\angle{BAC} = 40x \: \: , \: \: \angle{ABC} = 50x \: into \: \angle{ABC} + \angle{BAC} = 90\degree[/tex]

Calculate 50x + 40x = 90° ↑

Substitute x = 1 into [tex]\bf{\angle{BAC} = 40x}[/tex] ↑

[tex] \boxed{ \bold{\angle{BAC} = 40} }\: \mathfrak{ans.}[/tex]

The most appropriate form of equation of a line in 3D will be given by-

[tex]\frac{x + 1}{1} = \frac{y-3}{-5} = \frac{z-5}{0}[/tex]

is the required equation of the line.

What is equation of line in 3D?

Suppose a line passes through two points ([tex]x_1, y_1, z_1[/tex]) and ([tex]x_2, y_2, z_2[/tex]).

Equation of line is given by

[tex]\frac{x-x_1}{l} = \frac{y - y_1}{m} = \frac{z - z_1}{n}[/tex]

where [tex]l, m, n[/tex] are the Direction ratios

[tex]l = x_2 - x_1,\\ m = y_2 - y_1, \\n = z_2 - z_1[/tex]

Here,

[tex]x_{1} =-1, y_{1} = 3, z_{1} = 5, x_2 = 0, y_2 = -2, z_2 = 5[/tex]

l = 0-(-1)=1

m = -2-3 = -5

n= 5-5=0

Equation of line =

[tex]\frac{x -(- 1)}{1} = \frac{y-3}{-5} = \frac{z-5}{0}[/tex]

[tex]\frac{x + 1}{1} = \frac{y-3}{-5} = \frac{z-5}{0}[/tex]

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Step-by-step explanation:

So, the expression “three times the variable x” can be written in a number of ways: 3x, 3(x), or 3 · x. Use the distributive property to expand the expression 9(4 + x).

Answer:

3 < x < 5

Step-by-step explanation:

Expression is |2x − 8| < 2

The absolute rule says that if |u| < a then -a < u < a

Here u = 2x-8

So we get -2 < 2x - 8 < 2

This means

2x - 8 > -2            

==> 2x > -2 + 8       (add 8 to both sides)

==> 2x > 6               (simplify)

==> x > 3                  (divide by 2 both sides)

and,

2x - 8 < 2 gives

==> 2x < 2 + 8          (add 8 to both sides)

==> 2x < 10              (simplify)

==> x < 5                  (divide by 2 both sides)

So the solution to |2x − 8| < 2

is 3 < x < 5

The dimensions of the rug would be 27 feet long and 19 feet wide with a 6 foot space around the carpet uniformly.

She wants to buy a rug for a room that is 25 ft wide and 33 ft long.

She wants  to leave a uniform strip of floor around the rug.

she can afford to buy 513 square feet of carpeting.

Cynthia room area = lw

where

Therefore,

Cynthia room area = 25 × 33

Cynthia room area = 825 ft²

She wants to leave a uniform strip of floor all around the rug and has affordability to buy 513 square feet of carpet.

Therefore, the dimension the rug should have is as follows:

(25 - 6) (33 - 6) = 513

19 × 27 = 513 ft²

Therefore, the dimensions of the rug would be 27 feet long and 19 feet wide with a 6 foot space around the carpet uniformly.

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Answer:

a)

[tex]2.35 \times {10}^{ - 3} [/tex]

b)

[tex]2.5 \times {10}^{1} [/tex]

c)

[tex]4 \times {10}^{3} [/tex]

Step-by-step explanation:

above I think

the chances are always 50 50

Answer:

The number of books in the owner's office are 4.

Step-by-step explanation:

It is given that the book store had a new bookshelf that could hold 62 books the owner put 5 books on each of the shelves and the rest back in his office.

Now, let us assume the there are 'n' number of total books that the owner have with him;

=> total books = n

and the total books the shelf could hold is;

=> total books the shelf could hold = 62

Further, it is provided that the owner put 5 books on each of the shelves and the rest back in his office;

=> n = 62/5

=> n = 12.4

Therefore, the number of books in the owner's office are 4.

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Answer:

[tex]\textsf{Rewrite the original equation as $x^2+\dfrac{1}{3}x=\boxed{\dfrac{2}{9}}$}[/tex]

[tex]\textsf{Add appropriate number to make the left side a perfect square trinomial}[/tex]

[tex]x^2+\dfrac{1}{3}x+\boxed{\dfrac{1}{36}}=\dfrac{2}{9}+\boxed{\dfrac{1}{36}}[/tex]

[tex]\textsf{Factor the left side as a perfect square and combine the right hand side into one number}[/tex][tex]\left(x+\boxed{\dfrac{1}{6}}\:\right)^2=\boxed{\dfrac{1}{4}}[/tex]

[tex]\textsf{Final answers $x=\boxed{\dfrac{1}{3}, - \dfrac{2}{3}}$}[/tex]

Step-by-step explanation:

Given equation:

[tex]18x^2+6x-4=0[/tex]

Add 4 to both sides:

[tex]\implies 18x^2+6x-4+4=0+4[/tex]

[tex]\implies 18x^2+6x=4[/tex]

Divide both sides by 18:

[tex]\implies \dfrac{18x^2}{18}+\dfrac{6x}{18}=\dfrac{4}{18}[/tex]

[tex]\implies x^2+\dfrac{1}{3}x=\dfrac{2}{9}[/tex]

Add the square of half the coefficient of x to both sides:

[tex]\implies x^2+\dfrac{1}{3}x+\left(\dfrac{\frac{1}{3}}{2}\right)^2=\dfrac{2}{9}+\left(\dfrac{\frac{1}{3}}{2}\right)^2[/tex]

[tex]\implies x^2+\dfrac{1}{3}x+\left(\dfrac{1}{6}}\right)^2=\dfrac{2}{9}+\left(\dfrac{1}{6}\right)^2[/tex]

[tex]\implies x^2+\dfrac{1}{3}x+\dfrac{1}{36}=\dfrac{2}{9}+\dfrac{1}{36}[/tex]

Factor the perfect square trinomial on the left side and combine the numbers on the right side:

[tex]\implies \left(x+\dfrac{1}{6}\right)^2=\dfrac{1}{4}[/tex]

Square root both sides:

[tex]\implies \sqrt{\left(x+\dfrac{1}{6}\right)^2}=\sqrt{\dfrac{1}{4}}[/tex]

[tex]\implies x+\dfrac{1}{6}=\pm \dfrac{\sqrt{1}}{\sqrt{4}}[/tex]

[tex]\implies x+\dfrac{1}{6}=\pm \dfrac{1}{2}[/tex]

Subtract 1/6 from both sides:

[tex]\implies x+\dfrac{1}{6}-\dfrac{1}{6}=\pm\dfrac{1}{2}-\dfrac{1}{6}[/tex]

[tex]\implies x=-\dfrac{1}{6}\pm\dfrac{1}{2}[/tex]

Therefore:

[tex]\implies x=-\dfrac{1}{6}+\dfrac{1}{2}=\dfrac{1}{3}[/tex]

[tex]\implies x=-\dfrac{1}{6}-\dfrac{1}{2}=-\dfrac{2}{3}[/tex]

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Answer:

between 4 and 5

Step-by-step explanation:

The function g (x) is called an inner function and the function f (x) is called an outer function. Hence, we can also read f [g (x)] as “the function g is the inner function of the outer function f”.

Given that,

f(x) = 2x²+2  and

g(x)=x2-1

So find the (f- g)(x).

(f- g)(x) means,

Multiply x into f and g functions

Then, (f- g)(x) = f(x)-g(x)

Replace with f(x) and g(x) values in this equation

So,

(f- g)(x)= f(x)-g(x)

          = (2x²+2) - (x2-1)

Distribute the subtraction to all three of the last terms.          

          = 2x²+2-2x+1

          = 2x²-2x+2+1

Combine like terms

  (f- g)(x) = 2x²-2x+3

Therefore,

f(x) = 2x²+2 and

g(x)=x2-1,    

So,

       (f- g)(x) = 2x²-2x+3.

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Answer:

11

Step-by-step explanation:

If the school earns $264 for 8 senior tickets and 12 child tickets and $237 for 11 senior tickets and 6 child tickets then the price of 1 senior ticket be $15 and the price of child ticket be $12.

Given that on the first day of ticket sales, the school sold 8 senior citizen tickets and 12 child tickets for a total of $264 and the school earns $237 on IInd day by selling 11 senior citizen tickets and 6 child tickets.

We are required to define the variables and solve the system of the equations.

Suppose the price of 1 ticket of senior citizen be x.

Suppose the price of 1 ticket of child be y.

The equations will be:

8x+12y=264--------1

11x+6y=237--------2

Multiply equation 1 by 11 and multiply equation 2 by 8 and then subtract equation 2 from equation 1.

88x+132y-88x-48y=2904-1896

84y=1008

y=1008/84

y=12

Use the value of y in equation 1 to get the value of x.

8x+12y=264

8x+12*12=264

8x+144=264

8x=264-144

8x=120

x=120/8

x=15

Hence if the school earns $264 for 8 senior tickets and 12 child tickets and $237 for 11 senior tickets and 6 child tickets then the price of 1 senior ticket be $15 and the price of child ticket be $12.

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The cost of a Senior citizen ticket is $15, while a children's ticket cost $12.

On the first day:

The school sold 12 child tickets and 8 senior citizen tickets for a total amount of $264.

On the second day:

The school sold 11 senior citizen tickets and 6 child tickets for a total of $237.

Let A be the price of a senior citizen ticket and B be the price of a child ticket.

So, the equation for the first day:

8A + 12B = 264

The equation for the second day,

11A + 6B = 237

Multiplying the equation for the second day by 2 and subtracting the equation for the first day.

We get,

22A + 12B - 8A - 12B = 474 - 264

14A = 210

A = 15

Substituting the value of A in the equation 8A + 12B = 264,

8A + 12B = 264

8(15) + 12B = 264

120 + 12B = 264

12B = 144

B = 12

Therefore, the price of a senior citizen ticket is $15 and the price of a child ticket is $12.

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9.9 and 9.95 are the result of rounding off 9.949 to the nearest tenth and hundredth.

So, rounding off:

Therefore, rounding off of 9.949 to the nearest tenth and hundredth is 9.9 and 9.95.

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Answer: 1.083333

Step-by-step explanation:

Answer:

The circumference is 153.93804... or 153.94...

Step-by-step explanation:

Hope it helps! =D

Answer:

153.94 meters.

Step-by-step explanation:

The formula for finding the circumference of a circle is C = πd, where d is the diameter of the circle. With a diameter of 49m, the circumference can be calculated as C = π(49) ≈ 153.94m. Therefore, the circumference of the circle is approximately 153.94 meters.

The greatest number of binders Eula can buy is 5

The greatest number of notebooks Eula can buy is 10

The greatest number of binders she can buy  if Eula buys 7 notebooks is 3/2

It is an order relationship that is greater than, greater than, or equal to, less than, or less than or equal to—between two numbers or algebraic expressions.

We have,

4x + 2y ≤ 20

x = number of binders

y = number of notebooks

The greatest number of binders Eula can buy:

Put y = 0.

4x + 2 x 0 ≤ 20

4x ≤ 20

x ≤ 20/4

x ≤ 5

The greatest number of notebooks Eula can buy:

Put x = 0.

4x + 2y ≤ 20

4 x 0 + 2y ≤ 20

2y ≤ 20

y ≤ 10

Eula buys 7 notebooks then, the greatest number of binders she can buy:

4x + 2y ≤ 20

4x + 2 x 7 ≤ 20

4x + 14 ≤ 20

4x ≤ 20 - 14

4x ≤ 6

x ≤ 6/4

x ≤ 3/2

Thus,

The greatest number of binders Eula can buy is 5

The greatest number of notebooks Eula can buy is 10

The greatest number of binders she can buy  if Eula buys 7 notebooks is 3/2

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Answer:

Eula needs to buy binders that cost $4 each and notebooks that cost $2 each. She has $20. The graph of the inequality 4x + 2y ≤ 20, which represents the situation, is shown.What is the greatest number of binders Eula can buy? What is the greatest number of notebooks Eula can buy? If Eula buys 7 notebooks, what is the greatest number of binders she can buy?  ⇒ 1

Step-by-step explanation:

The slope of line AB with points A(4, 5) and B(9, 7) is 2/5.

The slope or gradient of a line is a number that describes both the direction and the steepness of the line.

We have,

A(4, 5) and B(9, 7)

The slope of a line with points A and B is given by:

= d - b / c - a

Where A(a, b) and B(c, d) are the coordinates of the points.

We have the points:

A(4, 5) = (a, b)

B(9, 7) = (c, d)

The slope of the line AB:

= (7 - 5) / (9 - 4) = 2 / 5

Therefore the slope of line AB with points A(4, 5) and B(9, 7) is 2/5.

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The value of the expression 3m^3n^2(8mn^3) is 24m^4n^5

The expression is given as:

3m^3n^2(8mn^3)

Rewrite properly as

3m^3n^2(8mn^3) = 3m^3n^2 * (8mn^3)

Remove the bracket

So, we have

3m^3n^2(8mn^3) = 3m^3n^2 * 8mn^3

Apply the law of indices in evaluating the product

So, we have

3m^3n^2(8mn^3) = 24m^4n^5

Hence, the value of the expression 3m^3n^2(8mn^3) is 24m^4n^5

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Answer:

f'(x) = 5

Step-by-step explanation:

differentiate using the power rule

[tex]\frac{d}{dx}[/tex] (a[tex]x^{n}[/tex] ) = na[tex]x^{n-1}[/tex] and [tex]\frac{d}{dx}[/tex] (constant) = 0

given

f(x) = 5x + 5 , then

f'(x) = 5[tex]x^{(1-1)}[/tex] + 0

     = 5[tex]x^{0}[/tex] + 0

    = 5

Answer:

a. 2x + 11

b. (5/x) - 2

c. (x + 8) * 4

Step-by-step explanation:

I may be wrong on b and c because the wording is a bit weird. Here is a list of common wording and their respective things. Have a great day!

If Leslie is about to type 56 words per minute, she would be able to type 8 pages in one hour

How many words can Leslie type in one hour?

The fact that Leslie can type 56 words per minute means that he is able, means that the number of words she is able to type in one hour is determined as 56 words multiplied 60 minutes which make an hour

number of words in one hour=56*60

number of words in one hour=3360

The number of pages typed is determined as the 3360 words typed in one hour divided by the number of words in a page

number of pages type=3360/420

number of pages typed=8 pages

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The number of cars in the parking lot is 105.

The first step is to determine the total number of cars that the parking lot can accommodate. This can be determined by multiplying the number of rows by the total number of cars in each row.

Multiplication is the mathematical operation that is used to determine the product of two or more numbers.

Total number of cars that can be in the parking lot  = number of rows x number of cars in each row

6 x 20 = 120

The next step is to subtract the number of empty spaces from the capacity of the parking lot . Subtraction is the operation that is used to determine the difference between two or more numbers.

Number of cars = capacity of the parking lot - empty spaces

120 - 15 = 105

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The direction cosines of the line are:

x = 8 / √116

y = 6 / √116

z = -4 / √116

Direction ratio helps in knowing the components of a line or a vector with reference to the three-axis, the x-axis, y-axis, and z-axis respectively.

We have,

Let the direction ratio of the line be (x, y, z)

The line is perpendicular to the lines whose direction cosines are proportional to 3,-2,3 and 1,-2,-1.

Now we have,

3x - 2y + 3z = 0 _____(1)

x - 2y - z = 0 _____(2)

We will find the value of x, y, and z using cross product.

x / 2 + 6 = y / 3 + 3 = z / -6 + 2

x / 8 = y / 6 = z / -4

We get,

x = 8

y = 6

z = -4

The direction cosines of the line are:

x = 8 / √8² + 6² + (-4)²

x = 8 / √64 + 36 + 16

x = 8 / √116

y = 6 / √116

z = -4 / √116

Thus the direction cosines of the line are:

x = 8 / √116

y = 6 / √116

z = -4 / √116

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Answer:

108

Step-by-step explanation:

3²(2³+4) calculate to the power of 2 and get 9.

9 (2³+4) calculate 2 to the power of 3 and get 8.

9 (8+4) add 8 and 4 to get 12.

9 · 12 = 108 multiply 9 and 12 to get 108.

Answer:

The reference angle is π/6.

Cosine is -(√3)/2.

Step-by-step explanation:

To find the reference angle, find the acute angle in quadrant I and use it as a reference for the given expression.

For the cosine, the cosine is the sine of the complementary angle. The complementary angle is the given angle beside it minus a right angle, which is exactly 90 degrees. If the angle is 25 degrees, its complementary angle will be double its amount, 50 degrees. Then, for angle angle measured "theta", the cosine is equal to the sine's right-angle subtracted by theta.

Answer:

[tex]2(35)+4=74[/tex]

∠SQR = 74

Step-by-step explanation:

[tex](2m+4)+(3m+1)=180\\5m+5=180\\ -5\\5m=175\\/5\\175/5=35[/tex]

Answer:

SQR = 74

Step-by-step explanation:

2m + 4 + 3m + 1 = 180

5m + 5 = 180

5m = 175

m = 35

35 * 2 + 4

70 + 4 = 74

Angle SQR = 74

There are 40 rows with 53 seats in each row.

This auditorium is divided into rows of seats, and because it is rectangular, each row has the same number of seats.

So,

Therefore, there are 40 rows with 53 seats in each row because we can't have a negative number of rows.

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In the differential equation

[tex]\dfrac{dy}{dx} + 12y = \cos\left(\dfrac{\pi x}{12}\right)[/tex]

multiply on both sides by the integrating factor

[tex]\mu = \exp\left(\displaystyle\int12\,dx\right) = e^{12x}[/tex]

Then the left side condenses to the derivative of a product.

[tex]e^{12x} \dfrac{dy}{dx} + 12 e^{12x} y = e^{12x} \cos\left(\dfrac{\pi x}{12}\right)[/tex]

[tex]\dfrac{d}{dx}\left[e^{12x}y\right] = e^{12x}\cos\left(\dfrac{\pi x}{12}\right)[/tex]

Integrate both sides with respect to [tex]x[/tex], and use the initial condition [tex]y(0)=0[/tex] to solve for the constant [tex]C[/tex].

[tex]\displaystyle \int \frac{d}{dx} \left[e^{12x}y\right] \, dx = \int e^{12x} \cos\left(\dfrac{\pi x}{12}\right) \, dx[/tex]

As an alternative to integration by parts, recall

[tex]e^{ix} = \cos(x) + i \sin(x)[/tex]

Now

[tex]e^{12x} \cos\left(\dfrac{\pi x}{12}\right) = e^{12x} \mathrm{Re}\left(e^{i\pi x/12}\right) = \mathrm{Re}\left(e^{(12+i\pi/12)x}\right)[/tex]

[tex]\displaystyle \int \mathrm{Re}\left(e^{(12+i\pi/12)x}\right) \, dx = \mathrm{Re}\left(\int e^{(12+i\pi/12)x} \, dx\right)[/tex]

[tex]\displaystyle. ~~~~~~~~ = \mathrm{Re}\left(\frac1{12+i\frac\pi{12}} e^{(12+i\pi/12)x}\right) + C[/tex]

[tex]\displaystyle. ~~~~~~~~ = \mathrm{Re}\left(\frac{12 - i\frac\pi{12}}{12^2 + \frac{\pi^2}{12^2}} e^{12x} \left(\cos\left(\frac{\pi x}{12}\right) + i \sin\left(\frac{\pi x}{12}\right)\right)\right) + C[/tex]

[tex]\displaystyle. ~~~~~~~~ = \frac{12}{12^2 + \frac{\pi^2}{12^2}} e^{12x} \cos\left(\frac{\pi x}{12}\right) + \frac\pi{12} e^{12x} \sin\left(\frac{\pi x}{12}\right) + C[/tex]

[tex]\displaystyle. ~~~~~~~~ = \frac1{12(12^4+\pi^2)} e^{12x} \left(12^4 \cos\left(\frac{\pi x}{12}\right) + \pi (12^4+\pi^2) \sin\left(\frac{\pi x}{12}\right)\right) + C[/tex]

Solve for [tex]y[/tex].

[tex]\displaystyle e^{12x} y = \frac1{12(12^4+\pi^2)} e^{12x} \left(12^4 \cos\left(\frac{\pi x}{12}\right) + \pi (12^4+\pi^2) \sin\left(\frac{\pi x}{12}\right)\right) + C[/tex]

[tex]\displaystyle y = \frac1{12(12^4+\pi^2)} \left(12^4 \cos\left(\frac{\pi x}{12}\right) + \pi (12^4+\pi^2) \sin\left(\frac{\pi x}{12}\right)\right) + C[/tex]

Solve for [tex]C[/tex].

[tex]y(0)=0 \implies 0 = \dfrac1{12(12^4+\pi^2)} \left(12^4 + 0\right) + C \implies C = -\dfrac{12^3}{12^4+\pi^2}[/tex]

So, the particular solution to the initial value problem is

[tex]\displaystyle y = \frac1{12(12^4+\pi^2)} \left(12^4 \cos\left(\frac{\pi x}{12}\right) + \pi (12^4+\pi^2) \sin\left(\frac{\pi x}{12}\right)\right) - \frac{12^3}{12^4+\pi^2}[/tex]

Recall that

[tex]R\cos(\alpha-\beta) = R\cos(\alpha)\cos(\beta) + R\sin(\alpha)\sin(\beta)[/tex]

Let [tex]\alpha=\frac{\pi x}{12}[/tex]. Then

[tex]\begin{cases} R\cos(\beta) = 12^4 \\ R\sin(\beta) = 12^4\pi+\pi^3 \end{cases} \\\\ \implies \begin{cases} (R\cos(\beta))^2 + (R\sin(\beta))^2 = R^2 = 12^8 + (12^4\pi + \pi^3)^2 \\ \frac{R\sin(\beta)}{R\cos(\beta)}=\tan(\beta)=\pi+\frac{\pi^3}{12^4}\end{cases}[/tex]

Whatever [tex]R[/tex] and [tex]\beta[/tex] may actually be, the point here is that we can condense [tex]y[/tex] into a single cosine expression, so choice (D) is correct, since [tex]\cos(x)[/tex] is periodic. This also means choice (C) is also correct, since [tex]\beta=\cos(x)\implies\beta=\cos(x+2n\pi)[/tex] for infinitely many integers [tex]n[/tex]. This simultaneously eliminates (A) and (B).