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Vladimir79 [104]
3 years ago
5

F(x)=x-2; translation 5 units left

Mathematics
1 answer:
kotykmax [81]3 years ago
3 0

Answer:

nknm

Step-by-step explanation:

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Full explanation Please<br> Will mark brainlist
Katyanochek1 [597]

The sum of 2 sides of a triangle needs to be greater than the length of the 3rd side.

14 + 29 = 43 (this is greater than 41)

14 + 41 = 55 (this is greater than 29)

29 + 41 = 70 (this is greater than 14)


(14, 29,41) can form a triangle.


6 0
2 years ago
Help please please !!!!!!!
lyudmila [28]
In each case, you can use the second equation to create an expression for y that will substitute into the first equation. Then you can write the result in standard form and use any of several means to find the number of solutions.

System A
  x² + (-x/2)² = 17
  x² = 17/(5/4) = 13.6
  x = ±√13.6 . . . . 2 real solutions

System B
  -6x +5 = x² -7x +10
  x² -x +5 = 0
The discriminant is ...
  D = (-1)²-4(1)(5) = -20 . . . . 0 real solutions

System C
  y = 8x +17 = -2x² +9
  2x² +8x +8 = 0
  2(x+2)² = 0
  x = -2 . . . . 1 real solution
7 0
3 years ago
Can someone pls help with this question
serg [7]

Answer:

26

Step-by-step explanation:

Use the Pythagorean Theorem: a^2+b^2=c^2

5 0
3 years ago
Read 2 more answers
PLEASE HELP ME ASAP!!!!!!!!
Svetlanka [38]
Wouldn't it be B ??? Hope this is right:) can you take a look at mine??
5 0
3 years ago
Read 2 more answers
Which of the equations below could be the equation of this parabola?
nirvana33 [79]

Answer:

 y=-4x^2  is the equation of this parabola.

Step-by-step explanation:

Let us consider the equation

y=-4x^2

\mathrm{Domain\:of\:}\:-4x^2\::\quad \begin{bmatrix}\mathrm{Solution:}\:&\:-\infty \:

\mathrm{Range\:of\:}-4x^2:\quad \begin{bmatrix}\mathrm{Solution:}\:&\:f\left(x\right)\le \:0\:\\ \:\mathrm{Interval\:Notation:}&\:(-\infty \:,\:0]\end{bmatrix}

\mathrm{Axis\:interception\:points\:of}\:-4x^2:\quad \mathrm{X\:Intercepts}:\:\left(0,\:0\right),\:\mathrm{Y\:Intercepts}:\:\left(0,\:0\right)

As

\mathrm{The\:vertex\:of\:an\:up-down\:facing\:parabola\:of\:the\:form}\:y=a\left(x-m\right)\left(x-n\right)

\mathrm{is\:the\:average\:of\:the\:zeros}\:x_v=\frac{m+n}{2}

y=-4x^2

\mathrm{The\:parabola\:params\:are:}

a=-4,\:m=0,\:n=0

x_v=\frac{m+n}{2}

x_v=\frac{0+0}{2}

x_v=0

\mathrm{Plug\:in}\:\:x_v=0\:\mathrm{to\:find\:the}\:y_v\:\mathrm{value}

y_v=-4\cdot \:0^2

y_v=0

Therefore, the parabola vertex is

\left(0,\:0\right)

\mathrm{If}\:a

\mathrm{If}\:a>0,\:\mathrm{then\:the\:vertex\:is\:a\:minimum\:value}

a=-4

\mathrm{Maximum}\space\left(0,\:0\right)

so,

\mathrm{Vertex\:of}\:-4x^2:\quad \mathrm{Maximum}\space\left(0,\:0\right)

Therefore,  y=-4x^2  is the equation of this parabola. The graph is also attached.

7 0
3 years ago
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