Can somebody explain to me stem and leaf diagrams please?

1 answer:

5 0

They are a method for showing the frequency of a data plot in a much more compact and numerical way.

1[2 3
2[1 7
3[3 4 5 7
4[ 0 0 1

Key 1[2= 12

meaning that 3[5 is 35 and 4[0 is 40 and so and so forth

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A rectangle has an area of x square units. The rectangle is dilated to create a new rectangle with an area of 9x square units. D

Alika [10]

The reactangle has an area of x.

So that means two numbers multiplied to each other equal to x. We can just think of itwo numbers that when multiplied equal to x as a and b.

So that means the side lengths are a and b. And ab = x (will be using this right next).

And it was multiplied by some scale factor that made the new area 9x. So if the scale factor is y, then:

$(y\times a)\times (y\times b)=9x$
$y^2ab=9x$
$y^2x=9x$
$y=3$

y can not equal to -3, because the length of a side of any polygon can never be negative.

So the dilation is 3.

And so if the dimension is 5 by 8.

Then the area is $5 \times 8=40$

And so then $x=40$

And then after the dilation, it would be $9x=9(40) = 360$

We could check it another way too. Since we said the scale factor is 3, then:

$( 5 \times 3) \times (8 \times 3) = 15 \times 24 = 360$

So basically 15 by 24 is the new dimensions :)

6 0

3 years ago

What is the value of x/3y , when x = 18 and y = 3? 3/18 2 3 6

Anestetic [448]

x = 18

y = 3

x/3y becomes 18/3(3) = 18/9 = 2

4 0

3 years ago

Read 2 more answers

Find the x- and y- intercepts of parabola y=5x^2-16x+10

____ [38]

Y-INTERCEPT

$y = 5x^2 - 16x + 10$

The y-intercept is where the equation/curve/parabola cosses the y-axis.

The y-axis is where x = 0. (The x-axis is where y = 0)

To find the y-intercept:

$\text{y-axis} \rightarrow \text{x = 0} \rightarrow y = 5(0)^2 -16(0) + 10 = 10$

The y-intercept must be at (0, 10)

X-INTERCEPT (ROOTS/SOLUTIONS)

$y = 5x^2 - 16x + 10\\\text{make it equal 0}\\y = 0\\\therefore 5x^2 - 16x + 10 = 0$

We need to use the quadratic formula

The quadratic formula helps us find what values of $x$ make the equation = 0

Quadratic formula: $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

$x=\frac{-(-16) + \sqrt{(-16)^2-4(5)(10)}}{2(5)}\\\\x = \frac{16 + \sqrt{256-200}}{10}\\x = \frac{16 + \sqrt{56}}{10}\\x = \frac{16 + 2\sqrt{14}}{10}\\x = \frac{8 + \sqrt{14}}{5}\\\\\\x=\frac{-(-16) - \sqrt{(-16)^2-4(5)(10)}}{2(5)}\\\text{doing the same thing...}\\\text{end up with...}\\x = \frac{8 - \sqrt{14}}{5}\\$