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Softa [21]
3 years ago
11

Need help please help

Mathematics
1 answer:
Vesna [10]3 years ago
6 0
The vertex is (105,7)

Notice how on the graph, the y value peaks at 7. The y values to the left and right are symmetrical, meaning that the x value for the y value of 7 is the axis of symmetry, and the point (105,7) is the vertex.
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MATH ANYONE PLEASE HELP
BartSMP [9]
3/8 is the answer to this question

8 0
3 years ago
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1. A square lot has an area of 64x² square units. What is the length of each side with full solution​
sergiy2304 [10]

\sqrt{64} = 8

The area of the square can be expressed as

area =  {s}^{2}

So to go back to the side length, you just need to find the square root of the area.

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What is the correct factorisation of?<br>X^2+6x+8​
MaRussiya [10]

Answer:

Step-by-step explanation:

Given that: x  

2

+6x+8

                 =x  

2

+4x+2x+8,        (split middle term)

                 =(x  

2

+4x)+(2x+8),    (group pair of terms)  

                 =x(x+4)+2(x+4),         (factor each binomials)

                 =(x+4)(x+2),              (factor out common factor (x+4)

Hence factors of x  

2

+6x+8 are (x+4)(x+2)

4 0
3 years ago
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How many times does 7 go into 37 without going over
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5 sevens go into 37 withought going over
6 0
3 years ago
The ranger of the function x2+2x-8is allá
allsm [11]

Answer:

\mathrm{Range\:of\:}x^2+2x-8:\quad \begin{bmatrix}\mathrm{Solution:}\:&\:f\left(x\right)\ge \:-9\:\\ \:\mathrm{Interval\:Notation:}&\:[-9,\:\infty \:)\end{bmatrix}

Step-by-step explanation:

As the function is

x^{2} +2x-8

As we know that domain is the set of input values for which the function is defined.

Therefore,

\mathrm{Domain\:of\:}\:x^2+2x-8\::\quad \begin{bmatrix}\mathrm{Solution:}\:&\:-\infty \:

And range is defined as:

The set of the dependent variable for which the function is defined.

As

\mathrm{For\:a\:parabola}\:ax^2+bx+c\:\mathrm{with\:Vertex}\:\left(x_v,\:y_v\right)

\mathrm{If}\:a

\mathrm{If}\:a>0\:\mathrm{the\:range\:is}\:f\left(x\right)\ge \:y_v

a=1,\:\mathrm{Vertex}\:\left(x_v,\:y_v\right)=\left(-1,\:-9\right)

f\left(x\right)\ge \:-9

Therefore,

\mathrm{Range\:of\:}x^2+2x-8:\quad \begin{bmatrix}\mathrm{Solution:}\:&\:f\left(x\right)\ge \:-9\:\\ \:\mathrm{Interval\:Notation:}&\:[-9,\:\infty \:)\end{bmatrix}

The graph is also attached below.

Keywords: graph, domain, range

Learn more about domain and range from brainly.com/question/13856645

#learnwithBrainly

8 0
3 years ago
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