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MrRissso [65]
4 years ago
10

Helppppppppppppppppppppppppppp

Mathematics
2 answers:
zvonat [6]4 years ago
5 0
I'm almost 50% positive that it is B
ElenaW [278]4 years ago
3 0
Look up the answers :) its how i aced my finals ;) jk dont do that u will be caught, it is cheating, i never did it tbh, im just messing, but yeah im also pretty positive its B
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Carlos performed a transformation on trapezoid EFGH to create E′F′G′H′, as shown in the figure below:
Nataliya [291]

Answer:

90° clockwise rotation or 270° anticlockwise rotation

Step-by-step explanation:

(x,y)=(y,-x)

<em>E(-6,-4)=E'(-4,6)</em>

<em>F(-4,-4)=F'(-4,4)</em>

<em>G(-2,-6)=G'(-6,2)</em>

<em>H(-7,-7)=H'(-7,7)</em>

Therefore, E prime is (-4,6), F prime is (-4,4), G prime is (-6,2) and H prime is (-7,7).

4 0
2 years ago
What is the product?
victus00 [196]

Answer:

\begin{bmatrix}-2\\0 \end{bmatrix}

Step-by-step explanation:

We have the product, 2\times \begin{bmatrix}-1\\0 \end{bmatrix}.

It is known that,

'When we multiply a scalar with a matrix, the scalar is multiplied by each element of the matrix'.

So, we get,

2\times \begin{bmatrix}-1\\0 \end{bmatrix}.

⇒ \begin{bmatrix}-1\times 2\\0\times 2 \end{bmatrix}

⇒ \begin{bmatrix}-2\\0 \end{bmatrix}

So, the resulting product is \begin{bmatrix}-2\\0 \end{bmatrix}.

8 0
3 years ago
Read 2 more answers
Does anyone know the answer to this one?
polet [3.4K]

Answer:

\large\boxed{a(x+4)(x-5)=0,\ a\in\mathbb{R}-\{0\}}\\\\\text{for}\ a=1\to\boxed{x^2-x-20=0}

Step-by-step explanation:

\text{If}\ x_1\ \text{and}\ x_2\ \text{are the roots of the quadratic equation}\ ax^2+bx+c=0,\\\text{then}\ ax^2+bx+c=a(x-x_1)(x-x_2).\\\\\text{If}\ x_1\ \text{and}\ x_2\ \text{, then they are the roots of the quadratic equation.}\\\\\text{We have}\ x_1=-4\ \text{and}\ x_2=5.\ \text{Therefore we have the equation:}\\\\a(x-(-4))(x-5)=0\\\\a(x+4)(x-5)=0\qquad\text{for any value of}\ a\ \text{except 0}.

7 0
3 years ago
Please help!
quester [9]

Answer:

The third one

Step-by-step explanation:

Median - Middle number in terms of value

Mode - outcome that appears the most

Mean - average ( add the numbers together and divide that by the amount of numbers )

Data without Kevins score

13, 21, 28, 9, 31

First let's find the mean

To find the mean we simply add the numbers up and divide that by the total amount of numbers ( there are 5 numbers )

13 + 21 + 28 + 9 + 31 = 102

102/5=20.4

So the mean without Kevins score is 20.4

Next let's find the mode

The mode is the number that appears the most

Looking at the data no number appears more than once so there is no mode

Finally let's find the median.

To make it easier to find let's list the numbers in order from least to greatest

9, 13 , 21 , 28, 31

We then go to the middle number ( which would be 21 )

The mode is 21

So for data without Kevins score the mean would be 20.4, the median would be 21 and the mode would be none

Now let's add in Kevin's score and repeat this process

Data with Kevin's score

13, 21, 28, 9, 31, 9

First let's find the mean

( Note that there is 6 numbers this time )

13 + 21 + 28 + 9 + 31 + 9 = 111

111/6 = 18.5

So mean with Kevin's score = 18.5

Next let's find the mode

The number 9 appears the most so the mode is 9

Finally let's find the mean

Once again let's list the numbers from least to greatest

9, 9, 13 , 21 , 28, 31

Because there is an even amount of numbers we will take the sum of the two middle numbers and divide that by 2

The two middle numbers would be 21 and 13

21 + 13 = 34

34/2=17

So median = 17

In the end we get the following

Without Kevins score

Mean = 20.4

Median = 21

Mode = none

With Kevin's score

Mean = 18.5

Median = 17

Mode = 9

This information corresponds with the third table

5 0
3 years ago
What is the answer to <img src="https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B2%7D%20" id="TexFormula1" title=" \frac{1}{2} " alt=
marta [7]
1. 5 -4x =6 +2x 
+4x. +4x
5=6. +6x
-6 -6
-1 = 6x
--- ---
6. 6
-6=x
2. 9 - 2x = 7x
+2x. +2x
9 = 9x
--- ----
9. 9
1 = x
8 0
3 years ago
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