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3 years ago

Three students, Carson, Andrew, and London, line up one behind the other. How

Mathematics

1 answer:

bezimeni [28]3 years ago

8 0

They can stand 9 different ways

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Please help if you can thank you.

jok3333 [9.3K]

Answer:

113.04

Step-by-step explanation:

The formula is pie × radius × radius pie=3.14 so 3.14 × 6 × 6 = 113.04

7 0

3 years ago

Read 2 more answers

A system of linear equations includes the line that is created by the equation y = x+ 3, graphed below, and the line through the

ad-work [718]

Answer:

(8,11)

Step-by-step explanation:

1st line

y = x+ 3

We need to find the equation of the other line

We have two points (3, 1) and (4, 3)

Slope

m = (y2-y1)/(x2-x1)

= ( 3-1)/(4-3)

= 2/1 = 2

The slope intercept form is y = mx+b where m is the slope and b is the y intercept

y =2x+b

Using the point (4,3)

3 = 2(4)+b

3 = 8+b

3-8=b

-5 =b

y = 2x-5

We have 2 lines

y =x+3 and y = 2x-5

Setting them equal to each other

x+3 = 2x-5

Subtract x from each side

x+3-x = 2x-5-x

3 = x-5

Add 5 to each side

3+5 =x-5+5

8=x

Now we can find y

y =x+3

y = 8+3

y=11

7 0

3 years ago

The points (5, 3), (3, -4), (10, 3), and (8, -4) are the vertices of a polygon. What type of polygon is formed by these points?

zhuklara [117]

Opposite sides are parallel. An appropriate choice is ...

B) Parallelogram

3 0

3 years ago

Read 2 more answers

Given the equation y=14x2−5, the minimum is ___ and the range is ___

mel-nik [20]

Answer:

The minimum is -5

The range is $y \ge -5$

f(14) = 2739

Step-by-step explanation:

Given

$y = 14x^2 - 5$

Solving (a): The minimum

A quadratic function is represented as:

$y = ax^2 + bx + c$

If a > 0, then the function has a minimum

By comparison

$a = 14$ --- the function has a minimum

$b = 0$

$c = -5$

To calculate the minimum, we first calculate the following is calculated as:

$m = -\frac{b}{2a}$

So, we have:

$m = -\frac{0}{2*14}$

$m = 0$

So, the minimum is at f(m)

We have: $y = 14x^2 - 5$

$f(0) = 14 *0^2 - 5$

$f(0) = - 5$

Solving (b): The range

In (a), we have:

$f(0) = - 5$ --- the minimum

This implies that the smallest value of y on the graph is -5.

So, the range is:

$r = \{y|y\ge -5\}$

Solving (c): f(14)

We have:

$y = 14x^2 - 5$

So:

$f(14) = 14 * 14^2 - 5$

$f(14) = 2739$

4 0

3 years ago

How would i solve this? Thank you

WITCHER [35]

6 because 2 divided by 12 is 6

8 0

3 years ago