. Given that r satisfies the inequality 20.x > 33,
find the smallest value of x if .r is a prime number.​

The value of f(5) is positive

At f(x) = 0, the value of x is 1

For the interval f(x) ≤ 0, the values of x are [-2, 1]

From the question, we have the following parameters that can be used in our computation:

The graph

On the graph, we have

f(5) = 1

This means that f(5) is positive

Also, we have

When f(x) = 0, the value of x is 1

For the interval f(x) ≤ 0, we have the values of x to be

-2 ≤ x ≤ 1

When represented as an interval, we have

[-2, 1]

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Answer:

40%

Step-by-step explanation:

to change a fraction to a percentage multiply by 100% , that is

\(\frac{20}{50}\) × 100% = 0.4 × 100% = 40%

Answer:

40%

Step-by-step explanation:

Step-by-step explanation:

I've skipped some steps. Hope you get this.

Answer:

(a) \(\displaystyle{\int \left(x^2-2x+\dfrac{1}{x^2}\right) \, dx} = \dfrac{x^3}{3} - x^2 - \dfrac{1}{x} + \text{C}}\)

(b) \(\displaystyle{\int x\left(x+3\right)^2 \, dx = \dfrac{x^4}{4} + 2x^3 + \dfrac{9x^2}{2} + \text{C}}\)

Step-by-step explanation:

Part A

Consider the indefinite integral with respect to x:

\(\displaystyle{\int \left(x^2-2x+\dfrac{1}{x^2}\right) \, dx}\)

We can separately integrate each terms as they are in addition and subtraction:

\(\displaystyle{\int x^2 \, dx - \int 2x \, dx + \int \dfrac{1}{x^2} \, dx}\)

For the last term, we can rewrite in negative exponent as \(x^{-2}\) so we can apply power rules to all these terms (making it easier):

\(\displaystyle{\int x^2 \, dx - \int 2x \, dx + \int x^{-2} \, dx}\)

The middle term -- we can separate the constant out of integral:

\(\displaystyle{\int x^2 \, dx - 2\int x \, dx + \int x^{-2} \, dx}\)

Apply the power integration rule (for n ≠ 1):
\(\displaystyle{\int x^n \, dx = \dfrac{x^{n+1}}{n+1} + \text{C} \ \ \left(\text{where C is a constant}\right)}\)

But we can add C later after we integrate all these terms, therefore:

\(\displaystyle{\dfrac{x^{2+1}}{2+1} - 2\cdot \dfrac{x^{1+1}}{1+1} + \dfrac{x^{-2+1}}{-2+1}}\\\\\displaystyle{\dfrac{x^{3}}{3} - 2\cdot \dfrac{x^{2}}{2} + \dfrac{x^{-1}}{-1}}\\\\\displaystyle{\dfrac{x^3}{3} - x^2 - x^{-1}}\)

Rewrite the negative exponent back to fraction form:

\(\displaystyle{\dfrac{x^3}{3} - x^2 - \dfrac{1}{x}}\)

After integrating all terms, we add + C every time. Hence:

\(\displaystyle{\int \left(x^2-2x+\dfrac{1}{x^2}\right) \, dx} = \dfrac{x^3}{3} - x^2 - \dfrac{1}{x} + \text{C}}\)

Part B

Consider the indefinite integration with respect to x of:

\(\displaystyle{\int x\left(x+3\right)^2 \, dx}\)

First, expand (x + 3)² to x² + 6x + 9:

\(\displaystyle{\int x\left(x^2+6x+9\right) \, dx}\)

Then disribute the x inside the brackets:

\(\displaystyle{\int \left(x^3+6x^2+9x\right) \, dx}\)

Integrate separately:

\(\displaystyle{\int x^3 \, dx + \int 6x^2 \, dx + \int 9x \, dx}\)

Separate the constants:

\(\displaystyle{\int x^3 \, dx + 6\int x^2 \, dx + 9\int x \, dx}\)

For the power integration formula, recall above. Thus:

\(\displaystyle{\dfrac{x^{3+1}}{3+1} + 6 \cdot \dfrac{x^{2+1}}{2+1} + 9 \cdot \dfrac{x^{1+1}}{1+1}}\\\\\displaystyle{\dfrac{x^{4}}{4} + 6 \cdot \dfrac{x^{3}}{3} + 9 \cdot \dfrac{x^{2}}{2}}\\\\\displaystyle{\dfrac{x^4}{4} + 2x^3 + \dfrac{9x^2}{2}}\)

Add + C after finishing the integration process, hence:

\(\displaystyle{\int x\left(x+3\right)^2 \, dx = \dfrac{x^4}{4} + 2x^3 + \dfrac{9x^2}{2} + \text{C}}\)

Let me know if you have any questions!

Answer:

D

Step-by-step explanation:

since we have -5 and 3, in order to translate f(g) into f(x), we subtract 8 from f(g)

The graph of g(x) is the same as the graph of f(x) translated up 8 units. Therefore, option B is the correct answer.

A translation in math moves a shape left or right and/or up or down. The translated shapes look exactly the same size as the original shape, and hence the shapes are congruent to each other. They just have been shifted in one or more directions.

Given that, f(x)= 6x-5 and g(x)= 6x+3.

When the shape is moved up by k units, then replace y with y + k.

Here, g(x)=6x-5+8

g(x)= 6x+3

The graph of g(x) is the same as the graph of f(x) translated up 8 units. Therefore, option B is the correct answer.

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Answer:

C

Step-by-step explanation:

i just guessed but it was correct

Answer:

3

Step-by-step explanation:

Answer:

2x bigger

Step-by-step explanation:

because if you split the rectange in half, it would be 2x?

Answer:

1. A      2. A? I think.      3. C

Step-by-step explanation:

The radius of the semicircle protractor is approximately 4.693 cm.

Given,Perimeter of a semicircle protractor = 14.8 cm.

To find:The radius of a semicircle protractor.Solution:We know that the perimeter of a semicircle protractor is the sum of the straight edge of a protractor and half of the circumference of the circle whose radius is the radius of the protractor.

Circumference of a circle = 2πrWhere, r is the radius of the circle.If the radius of the semicircle protractor is r, then Perimeter of a semicircle protractor = r + πr [∵ half of the circumference of a circle =\((1/2) × 2πr = πr]14.8 = r + πr14.8 = r(1 + π) r = 14.8 / (1 + π)r ≈ 4.693\) cm.

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Answer:

area of the rectangle =

\( {2x}^{2} {cm}^{2} \)

Step-by-step explanation:

length = 2x

width = x

area of the rectangle = length × width

\(a = l \times w \\ a = 2x \times x \\ a = {2x}^{2} \)

The number of students who just passed and didn't get any division is  195.

We can start by finding the number of students who got first or second division:

Next, we can find the total number of students who got a division:

Total number of students who got a division = Number of students who got first division + Number of students who got second division

= 125 + 175 = 300

Now, we can find the number of students who failed:

Number of students who failed = 5

Finally, we can subtract the number of students who got a division and the number of students who failed from the total number of students to find the number of students who just passed and didn't get any division:

Number of students who just passed and didn't get any division = Total number of students - Number of students who got a division - Number of students who failed

= 500 - 300 - 5

= 195

Therefore, the number of students who just passed and didn't get any division is 195.

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Answer:

The answer is C

Step-by-step explanation:

A trend line should attempt to be in the middle of the data points not above them. If it does not do as staed above, it is NOT a correct trend line.

Have an amazing day!

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Answer: The answer is 8 1/3

I hope this helps

Step-by-step explanation:

Answer:

the answer for your question is 8 1/3

Answer:

x=-3

Step-by-step explanation:

-15x=45

x=-45/15

x=-3

Answer:

okay x=-3

Step-by-step explanation:

-15x=45

-15/-15=45/15

x=-3

Cotangent is the inverse of tangent:

cotΘ = 1/tanΘ

cotΘ = \(1/\frac{\sqrt{6} }{2}\)

         =\(\frac{2}{\sqrt{6}}\)

         =\(\frac{2}{\sqrt{6}}*\frac{\sqrt{6}}{\sqrt{6}}=\frac{2\sqrt{6}}{6}\) Rationalize the denominator

cotΘ = \(\frac{\sqrt{6}}{3}\)

Plug and solve.

\(\frac{\frac{\sqrt{6}}{2}+\frac{\sqrt{6}}{3} }{\frac{\sqrt{6}}{2}-\frac{\sqrt{6}}{3}}\)

We'll break the problem into 2, simplify the denominator, then simplify the numerator.

Simplify the denominator with the common denominator 6, of 2 and 3:

\({\frac{\sqrt{6}}{2}-\frac{\sqrt{6}}{3} = {\frac{3\sqrt{6}}{6}-\frac{2\sqrt{6}}{6} = {\frac{3\sqrt{6}-2\sqrt{6}}{6} = {\frac{\sqrt{6}}{6}\)

Simplify the Numerator with the common denominator 6, of 2 and 3:

\({\frac{\sqrt{6}}{2}+\frac{\sqrt{6}}{3} = {\frac{3\sqrt{6}}{6}+\frac{2\sqrt{6}}{6} = {\frac{3\sqrt{6}+2\sqrt{6}}{6} = {\frac{5\sqrt{6}}{6}\)

Back in original equation with simplified numerator and denominator:

\(\frac{\frac{5\sqrt{6}}{6}}{\frac{\sqrt{6}}{6}}\)

Multiply by reciprocal and simplify:

\(\frac{5\sqrt{6}}{6}*\frac{6}{\sqrt{6}} = \frac{5\sqrt{6} * 6}{6\sqrt{6}} = 5*\frac{6\sqrt{6}}{6\sqrt{6}} = 5*1 = 5\)

Answer:

y = -2x +17

Step-by-step explanation:

line equation y - y1 = m (x - x1)

y - 7 = -2 (x - 5)

y - 7 = -2x + 10

y = -2x + 10 + 7

y = -2x +17

Answer:

1000 in

Step-by-step explanation:

Given

The dimension of the small cube is 1\ in.1 in.

No of the cubes is 10^6106

If the small cubes are arranged on the floor

The area of the cubes is

\Rightarrow 10^6\times 1^2\ in.^2⇒106×12 in.2

Answer:

60

Step-by-step explanation:

40%=0.40

0.4x=24

x=24/0.4

x=60

Answer:

19.99%

Step-by-step explanation:

hope this helps

here you go

A triangle must have a third side that is bigger than the sum of any two of its sides. There cannot be a triangle with these side lengths because in this instance, 5 + 15 = 20 is not greater than 250.

To check whether the given lengths can form the sides of a right triangle, we need to check if they satisfy the Pythagorean theorem, which states that in a right triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.

Let's label the sides of the triangle as a, b, and c, where c is the hypotenuse. Then, the Pythagorean theorem can be written as:

a^2 + b^2 = c^2

Plugging in the given values, we get:

5^2 + 15^2 = (√250)^2

Simplifying the left-hand side, we get:

25 + 225 = 250

This is not true, since 25 + 225 = 250 does not hold. Therefore, the given lengths cannot form the sides of a right triangle.

In fact, we can see that the given lengths violate the triangle inequality, which states that the sum of any two sides of a triangle must be greater than the third side. In this case, 5 + 15 = 20 is not greater than √250, so a triangle with these side lengths cannot exist.

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Answer: 12.12 units

Let us create the figure to make it easier to understand how to solve this question.

Each side measures 14 units. Since the triangle is an equilateral triangle, the height of the triangle is directly at the middle of the triangle, hence dividing it into two equal parts. This is why the base of the triangle is divided into 7s.

As we can see, the height of the triangle will create a right triangle on both sides of the triangle. Since we now have a right triangle, we can solve for the height of the triangle using the Pythagorean theorem.

The Pythagorean theorem is noted as:

Where:

a and b are the shorter sides of the triangle while c is the longest side of the triangle.

We have:

a = 7

c = 14

We look for b:

Therefore, the height of the triangle is 12.12 units.

Answer:

e-5=2f

take '-5' to the other side where '2f' is

e=2f+5

The lengths of the other two sides of the right triangle are 24m and 25m, respectively.

Let's assume the consecutive integers representing the lengths of the other two sides of the right triangle are x and x + 1, where x is the smaller integer. We are given that the shortest side measures 7m. Now, we can use the Pythagorean theorem to solve for the lengths of the other two sides.

According to the Pythagorean theorem, in a right triangle, the square of the length of the hypotenuse (the longest side) is equal to the sum of the squares of the lengths of the other two sides.

Using this theorem, we have the equation:

\(7^2 + x^2 = (x + 1)^2\)

Expanding and simplifying this equation, we get:

\(49 + x^2 = x^2 + 2x + 1\)

Now, we can cancel out \(x^2\) from both sides of the equation:

49 = 2x + 1

Next, we can isolate 2x:

2x = 49 - 1

2x = 48

Dividing both sides by 2, we find:

x = 24

Therefore, the smaller integer representing the length of one side is 24, and the consecutive integer representing the length of the other side is 24 + 1 = 25.

Hence, the lengths of the other two sides of the right triangle are 24m and 25m, respectively.

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The correct answer is: O Yes, distance and angle measure are preserved.

When a triangle ABC is reflected across the line y = x, the pre-image and image are congruent.

This is because the line y = x is the perpendicular bisector of the segment joining each corresponding point of the pre-image and image.

Reflection across the line y = x is a type of transformation known as an isometry, which preserves both distance and angle measure.

Here's why:

Distance preservation:

When a point is reflected across the line y = x, the distance between the original point and its reflection remains the same.

This holds true for all corresponding points of the triangle.

Therefore, the distance between any two corresponding points in the pre-image and image triangle will be equal, resulting in distance preservation.

Angle preservation: When a line segment is reflected across the line y = x, the angle between the line segment and the line y = x is preserved. This means that the corresponding angles in the pre-image and image triangle will be congruent.

Since both distance and angle measure are preserved during reflection across the line y = x, the pre-image and image triangles are congruent.

It's important to note that congruence under reflection across a line holds only when the line of reflection is the same for both the pre-image and image.

If the line of reflection were different, the triangles would not be congruent.

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Answer:

Step-by-step explanation:

To prove sin(a+b)*sin(a-b)=cos^2b-cos^2a

we simplify the left side  sin(a+b)*sin(a-b) first

sin(a+b) = sin a cos b + cos a sin b

sin(a-b) = sin a cos b - cos a sin b

sin(a+b)*sin(a-b) = (sin a cos b + cos a sin b) x (sin a cos b -cos a sin b)

sin a cos b((sin a cos b + cos a sin b) - cos a sin b (sin a cos b + cos a sin b)

open the bracket

sin a cos b(sin a cos b) + sin a cos b(cos a sin b) -cos a sin b (sin a cos b)+ cos a sin b ( cos a sin b)

sin²a cos²b + sin a cos b cos a sin b - cos a sin b sin a cos b + cos²a sin²b

sin²a cos²b  + 0 + cos²a sin²b

sin²a cos²b + cos²a sin²b

sin²a  = 1-cos² a

sin²b  = 1-cos² b

(1-cos² a)cos² b - cos² a(1-cos² b)

= cos² b - cos² a cos² b - cos² a +cos² a cos² b

choose like terms

cos² b - cos² a - cos² a cos² b + cos² a cos² b = cos² b - cos² a  + 0

cos² b - cos² a

left hand side equals right hand side

Answer: (6,11π/6).

Step-by-step explanation:We need to find the radius r and the angle θ. Remember that r2=x2+y2, so

because of the signs of x and y, our angle is in quadrant IV. Therefore, we find that θ=11π/6.

So the final answer is (6,11π6).