Plzzz help I don’t understandddddd

Find the percent of increase from 20 to 45.

Aloiza [94]

It is a 125% increase from 20 to 45

4 0

3 years ago

1. A right triangle is shown below. Use the Pythagorean Theorem to determine the lengths of each side. Phythagorean Theorem: a2

erik [133]

This may not be very pretty.
(2x + 5) * (2x + 5) expanded is (use foil)
F = 4x^2
O = 2x * 5 = 10x
I = 2x + 5 = 10x
L = 25
Total = 4x^2 + 20x + 25

(x + 3)(x + 3) = x^2 + 6x + 9 by the same method

(2x + 7)(2x + 7) = 4x^2 + 28x + 49 Same method.

4x^2 + 20x + 25 + x^2 + 6x + 9 = 4x^2 + 28x + 49
5x^2 + 26x + 34 = 4x^2 + 28x + 49 Collect everything on the left.
x^2 - 2x - 15 = 0
(x - 5)(x + 3) = 0
x + 3 will have no meaning.
x - 5 = 0
x = 5

2x + 5 = 2*5 + 5 = 15
x + 3 = 5 + 3 = 8
2x+ 7 = 2*5 +7 = 17

Check
15^2 + 8^2 = 17^2
225 + 64 = ? 289
289 = 289

a = 15 <<<<< answer
b = 8 <<<<< answer
c = 17 <<<<< answer

5 0

3 years ago

Help, and if you can give me a step by step explanation

Llana [10]

Answer:

the first option

Step-by-step explanation:

variability !

what does that word tell us ?

it means that there are more individuals differences.

you could also use "accuracy" as the opposite - we are aiming for the mean value ...

imagine some bow and arrow tournament.

who wins ?

the person with the highest accuracy across all the attempts (and that means the lowest variability in the results across all attempts relatively to the target center representing the predefined mean value).

now look at the graphic for neighborhood A.

and then for neighborhood B.

which one has the data points more clustered around the center (where the mean value is going to be) ? this one has lower variability than the one where the data points are having more than one cluster or are even all over the place.

remember, for the variability you have to add all the differences to the mean value. the smaller the differences to the mean value, the smaller the variability.

in neighborhood B almost all data points have a larger difference to the mean value.

so, the variability will be higher here.

5 0

2 years ago

Rita drank the following ounces of milk during the last week

Andre45 [30]

Answer:

Mean = 14.14

Mode = 14

Median = 14

Range = 16

Step-by-step explanation:

Given Data Set : 18 , 14 , 10 , 22, 14 , 15 and 6

Identifying the mean

Mean = Sum of given data values / total # of given values

Here sum of given data values = 18 + 14 + 10 + 22 + 14 + 15 + 6 = 99

And there are 7 given data values

So mean = 99 / 7 =

14.14

( approximately )

Identifying the mode

The mode is the number that appears most often

14 appears twice, which is the most times out of any other value

Hence, the mode is

14

Identifying the median

The median is the middle number in terms of value

To make this easier, its best to order the given values from least to greatest
Doing so we acquire 6 , 10 , 14 , 14 , 15 , 18 , 22

Then you go to the middle of the listed terms.

Doing so we would end up with 14 being in the middle

Hence the median is

14

Identifying the range

Range = largest given value - smallest given value

Here the most rita drank was 22 ounces and the least amount of milk she drank was 6 ounces

So the range would be 22 - 6 =

16

ounces

And we are done!

3 0

2 years ago

A surveyor leaves her base camp and drives 42km on a bearing of 032degree she then drives 28km on a bearing of 154degree,how far

ValentinkaMS [17]

Answer:

The surveyor is 36.076 kilometers far from her camp and her bearing is 16.840° (standard form).

Step-by-step explanation:

The final position of the surveyor is represented by the following vectorial sum:

$\vec r = \vec r_{1} + \vec r_{2} + \vec r_{3}$ (1)

And this formula is expanded by definition of vectors in rectangular and polar form:

$(x,y) = r_{1}\cdot (\cos \theta_{1}, \sin \theta_{1}) + r_{2}\cdot (\cos \theta_{2}, \sin \theta_{2})$ (1b)

Where:

$x, y$ - Resulting coordinates of the final position of the surveyor with respect to origin, in kilometers.

$r_{1}, r_{2}$ - Length of each vector, in kilometers.

$\theta_{1}, \theta_{2}$ - Bearing of each vector in standard position, in sexagesimal degrees.

If we know that $r_{1} = 42\,km$ , $r_{2} = 28\,km$ , $\theta_{1} = 32^{\circ}$ and $\theta_{2} = 154^{\circ}$ , then the resulting coordinates of the final position of the surveyor is: