Find the Least Common Multiple of of these two monomials:

If we run the following line:<br> y = int(3 * '4')<br><br> what is the value of y?

Hitman42 [59]

The value of y in the code segment is 444

<h3>How to determine the value of y?</h3>

The code segment is given as:

y = int(3 * '4')

Evaluate the expression in the bracket

y =int('444')

Remove the bracket

y = 444

Hence, the value of y in the code segment is 444

Read more about code segments at:

brainly.com/question/26683418

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3 0

2 years ago

Please help i need the answer asap ill give brainliest

timofeeve [1]

80.1 years, im not 100% sure of my answer though

6 0

2 years ago

Jaycee draws a sequence of squares on the coordinate plane. First she draws a square in the center with corners at 2,2 spots and

Anon25 [30]

This is based on understanding what dilation means in a graph transformation.

<em />

<em>The dilation from first square directly to sixth square will be; (x,y) -> (243, 243)</em>

<em />

In transformations, dilation of an object involves producing an image of the object that is the same shape but not the same size.

This means that if we want to dilate a square, we will produce a bigger square of a different size.

We are told one corner of the first square she drew is (2, 2). This means that one side of the square is 2 units as the four sides of a square are equal.

For the second square, she dilates the first one using (x, y) -> (3x, 3y).

This means the corner that was (2, 2) will now be (3 × 2), (3 × 2) = (6, 6)

For the third square, it will be; (3 × 6), (3 × 6) = (18, 18)

For the fourth square, it will be; (3 × 18), (3 × 18) = (54, 54)

For the fifth square, it will be; (3 × 54), (3 × 54) = (162, 162)

For the sixth square, it will be; (3 × 162), (3 × 162) = (486, 486)

Since first square was (2, 2), then it means dilation from first square directly to sixth square will be; (x,y) -> (486/2, 486/2)

⇒ (x,y) -> (243, 243)

Read more at; brainly.com/question/2523916

4 0

1 year ago

Please solve i will give brainiest 100 point question ****** do the whole page please need to pass or i will fail its my final t

dmitriy555 [2]

Answer:

1. Find the difference between the areas.

<u>Area of the small rectangle</u>: $(x+2)(x+7)=x^2+7x+2x+14=x^2+9x+14$

<u>Area of the big rectangle</u>: $(x+9)(x+11)=x^2+11x+9x+99=x^2+20x+99$

The difference is: $11x+85$

$( x^2+20x+99)- (x^2+9x+14)=x^2+20x+99-x^2-9x-14=11x+85$

2.

You can solve this question just by looking at the graph.

a) The height is 4 meters.

$f(d)=h=-2d^2+7d+4$

To find the height of the bleachers, we should consider the moment before the shoot, when the distance is equal to 0.

$f(0)=h=-2(0)^2+7(0)+4$

$h=4$

The height is 4 meters.

b) 9 meters.

For $d=1$

$f(1)=h=-2(1)^2+7(1)+4$

$f(1)=h=-2+7+4$

$h=9$

b) The ball travels 4 meters.

But to calculate it, it is when $h=0$

$0=-2d^2+7d+4$

Using the quadratic formula:

$d=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$d=\frac{-7 \pm \sqrt{7^2-4\left(-2\right)4}}{2\left(-2\right)}$

$d=\frac{-7\pm\sqrt{81}}{-4}$

$d=\frac{-7\pm9}{-4}$

It will give us to solutions, once it is a quadratic equation, but we are talking about a positive distance.

$d=-\frac{1}{2} \text{ or }d=4$

3.

In this question, we have to find the area of the cylinder and the sphere.

From the information given, we have

a = 5mm and d = 5mm, therefore the radius is 2.5 mm.

The volume of a cylinder:

$V=\pi r^2h$

$V=\pi (2.5)^2 \cdot 5$

$V=31.25 \pi$

$V_{c} \approx 98.17 \text{ m}^3$

The volume of the sphere:

$V=\frac{4}{3} \pi r^2$

$V_{s} \approx 65.4 \text{ m}^3$

The volume of the capsule is approximately $163.57 \text{ m}^3$

3 0

3 years ago

The scores on a test for a sample of 42 statistics students are summarized in the following table. 9, 90, 18, 80, 15, 70. Find t

igomit [66]

<h3>Answer: 78.6</h3>

Work Shown:

$\begin{array}{|c|c|} \cline{1-2}\text{Number of People} & \text{Score}\\\cline{1-2}9 & 90\\\cline{1-2}18 & 80\\\cline{1-2}15 & 70\\\cline{1-2}\end{array}$

Multiply across the rows to form a third column. Example: 9*90 = 810 in the first row.

$\begin{array}{|c|c|c|} \cline{1-3}\text{Number of People} & \text{Score} & \text{Number*Score}\\\cline{1-3}9 & 90 & 810\\\cline{1-3}18 & 80 & 1440\\\cline{1-3}15 & 70 & 1050\\\cline{1-3}\end{array}$

The values in the third column add to 810+1440+1050 = 3300

Divide this over the total number of people which is 9+18+15 = 42

The mean is 3300/42 =78.571 approximately which rounds to 78.6

5 0

1 year ago