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3 years ago

I did the upgrade what do i do now?

Mathematics

1 answer:

liq [111]3 years ago

3 0

Wdym.? What upgrade are u referring to.?

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Can anyone please help i have no one to help me

adoni [48]

Answer:

<em>Graph below</em>

Step-by-step explanation:

<u>Transformations</u>

Triangle ABC has coordinates A=(-4,2) B=(-2,6) C=(3,2)

We'll use the transformation (x,y) -> (2x, y+3) to map it to the triangle A'B'C'. Let's calculate the coordinates:

A'=(-8,5)

B'=(-4,9)

C'=(6,5)

The image below shows both triangles in the same grid

8 0

3 years ago

Statistics 4. Hol (a) The average age of 5 students is 9 years. Out of them, the ages of 4 students are 5, 7, 8, and 15 years. W

olga nikolaevna [1]

Answer:

10 years

Step-by-step explanation:

the average or mean value is the sum of all data points divided by the number of data points.

x = the unknown age of the 5th student.

average = 9 = (5 + 7 + 8 + 15 + x)/5

45 = 5 + 7 + 8 + 15 + x = 35 + x

x = 45 - 35 = 10

the 5th student is 10 years old.

6 0

2 years ago

Question 5 of 7 -

MatroZZZ [7]

Answer: B and E

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Original Equation:8 - 7x^2 + 3 = 4x^2 - 1Frist step:11 - 7x^2 = -4x^2 - 1Which property justifies the first step?A. Division pro

Tpy6a [65]

8 - 7x^2 + 3 = 4x^2 - 1

8+3-7x^2 = 4x^2-1

11-7x^2 = 4x^2-1

B. Commutative property of addition

We can swap numbers and obtain the same results.

In this case, we swapped 8+3 and added them.

7 0

1 year ago

Find parametric equations and symmetric equations for the line. (Use the parameter t.) The line through (4, −5, 2) and parallel

Nataliya [291]

Answer:

Step-by-step explanation:

From the given information, the symmetric equations for the line pass through(4, -5, 2) i.e ( $x_o, y_o, z_o$ ) and are parallel to $\dfrac{x+5}{1} = \dfrac{y}{2}= \dfrac{z-3}{1}$

The parallel vector to the line i + zj+k = ai + bj + ck

Hence, the equation for the line is :

$x = x_o + at \\ \\ x = y_o + bt \\ \\ x = z_o + ct$

x = 4 + t

y = -5 + 2t

z = 2 + t

Thus, x, y, z = ( 4+t, -5+2t, 2+t )

The symmetric equation can now be as follows:

$\begin {vmatrix} x = 4+ t \\ \\ \dfrac{x-4}{1} = t \begin {vmatirx} \end {vmatrix}$ $\begin {vmatrix} y = - 5+2t \\ \\ \dfrac{y+5}{2} =t \end {vmatrix}$ $\begin {vmatrix} z =2+t \\ \\ \dfrac{z-2}{1} =t \end {vmatrix}$

∴

$\dfrac{x-4}{1}= \dfrac{y+5}{2}=\dfrac{z-2}{1}$

8 0

3 years ago