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Rama09
14 days ago
12

Given that a function, g, has a domain of -1 ≤ x ≤ 4 and a range of 0 ≤ g(x) ≤ 18 and that g(-1) = 2 and g(2) = 8, select the st

atement that could be true for g.
Mathematics
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A land surveyor places two stakes 500 ft apart and creates a perpendicular to the line that connects these two stakes. He needs
Svet_ta [12734]

Answer:
We will select the final option as the correct one.
Step-by-step explanation:
A land surveyor positions two stakes 500 ft apart and identifies the midpoint between them.
From that midpoint, he is tasked with placing another stake 100 ft away, maintaining equal distance to the two original stakes.  
To utilize the Perpendicular Bisector Theorem, the land surveyor must identify a line that is "perpendicular to the segment connecting the two stakes and passes through the midpoint of those stakes."
Thus, we will choose the last option as the correct answer.

8 0
3 months ago
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The dot plots show the number of first cousins the boys and girls in Miss Campbell‘s class have. Based on the dot plot, which se
PIT_PIT [12445]

Answer:

The correct choice is the third one.

Step-by-step explanation:

3 0
3 months ago
Part c when sizes of pizzas are quoted in inches, the number quoted is the diameter of the pizza. a restaurant advertises an 8-i
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I believe this situation pertains to "two individuals," though I consider a 16-inch pizza to be more than sufficient. Apologies if I'm mistaken.
8 0
3 months ago
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What is the positive solution to the equation 0 = –x2 + 2x + 1? Quadratic formula: x = StartFraction negative b plus or minus St
zzz [12365]

Answer:

1+\sqrt{2}

Step-by-step explanation:

Consider a quadratic equation given by

ax^2+bx+c=0.... (1)

The quadratic formula then is

x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}

The quadratic equation provided is

0=-x^2+2x+1

This can be expressed as

-x^2+2x+1=0.... (2)

By comparing (1) and (2), we derive:

a=-1,b=2,c=1

Plug these values into the quadratic formula.

x=\dfrac{-2\pm \sqrt{2^2-4(-1)(1)}}{2(-1)}

x=\dfrac{-2\pm \sqrt{4+4}}{-2}

x=\dfrac{-2\pm \sqrt{8}}{-2}

x=\dfrac{-2\pm 2\sqrt{2}}{-2}

Factoring out common elements.

x=\dfrac{-2(1\pm \sqrt{2})}{-2}

x=1\pm \sqrt{2}

The two solutions are

x=1+\sqrt{2} and x=1-\sqrt{2}

We recognize that

\sqrt{2}=1.41

Thus,

1

Consequently, root x=1+\sqrt{2} is positive while x=1-\sqrt{2} is negative.

9 0
2 months ago
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