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Salsk061 [2.6K]
3 years ago
15

Shawna, Dexter, and Tilana are solving the equation Negative 2.5 (5 minus n) + 2 = 15.

Mathematics
2 answers:
skelet666 [1.2K]3 years ago
8 0
All students have the correct answer
mihalych1998 [28]3 years ago
3 0

Answer:

All students are correct

Step-by-step explanation:

Shawna, Dexter, And Tilana

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Factor-
Nastasia [14]

Hello from MrBillDoesMath!

Answer:

See Discussion section below

Discussion:

2x^2 + 5x + 3  = (2x+3)(x+1) which is Choice 3


3x^2+10x+4  =-1/3 (-3 x + sqrt(13) - 5) (3 x + sqrt(13) + 5)

I don't think any of the provided choices are right!


8x^2+10x-3 = (2x+3)(4x-1) which is Choice 2


6x^2-7x-4 = -1/24 (-12 x + sqrt(145) + 7) (12 x + sqrt(145) - 7)

I don't think any of the provided choices are right!


2x^2+x-28  =  (2x-7)(x+4) which is Choice 2


Thank you,

MrB

4 0
3 years ago
Which of the following statements are true
Kobotan [32]

Answer:

Step-by-step explanation:

The answer for this is

See the red line in the graph which says f(X)

See the blue line in the graph which says g(X)

So these are the variables which come under the example of exponential functions

And the answer is option B. Both graphs are exponential functions.

5 0
3 years ago
On a coordinate grid, point R is at (−6, 2) and point S is at (8, 2). The distance (in units) between points R and S is ______.
Talja [164]
You can use the distance formula:
d=sqroot(x2-x1)squared + (y2-y1)squared.
Plug in values: d=sqroot(8-(-6))squared + (2-2)squared
Simplify: d=sqroot(196)+0
Simplify:d=sqroot(196
Square Root: d=14 units
4 0
3 years ago
Finding the tangent of angle E
Pavlova-9 [17]
I don’t know but I need an answer
3 0
2 years ago
Read 2 more answers
Write a possible third degree polynomial with integer coefficient that have zeros: 1 2i, -4. Assume the leading coefficient to b
jasenka [17]

Answer:

The polynomial is:

p(x) = x^3 + 2x^2 - 3x + 20

Step-by-step explanation:

A third degree polynomial can be written in function of it's zeros x_1, x_2, x_3 the following way:

p(x) = a(x - x_1)(x - x_2)(x - x_3)

In which a is the leading coefficient.

Integer coefficient that have zeros: 1+2i, 1-2i, -4

Leading coefficient: 1

So

p(x) = 1(x - (1+2i))(x - (1-2i))(x - (-4))

p(x) = (x - 1 -2i)(x - 1 + 2i)(x + 4)

p(x) = ((x-1)^2 - (2i)^2)(x + 4)

p(x) = (x^2 - 2x + 1 - 4i^2)(x + 4)

Since i^2 = -1

p(x) = (x^2 - 2x + 1 + 4)(x + 4)

p(x) = (x^2 - 2x + 5)(x + 4)

p(x) = x^3 + 4x^2 - 2x^2 - 8x + 5x + 20

p(x) = x^3 + 2x^2 - 3x + 20

3 0
3 years ago
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