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nataly862011 [7]
3 years ago
5

Which of the following is the graph of y = negative StartRoot x EndRoot + 1? On a coordinate plane, an absolute value curve curv

es up and to the right in quadrant 4 and starts at y = negative 1. On a coordinate plane, an absolute value curve curves up and to the right in quadrant 4 and starts at y = 1. On a coordinate plane, an absolute value curve opens down and to the left in quadrant 2 and starts at y = 1. On a coordinate plane, an absolute value curve opens down and to the left in quadrant 2 and starts at y = negative 1.
Mathematics
2 answers:
mestny [16]3 years ago
7 0

Answer:

  • B. On a coordinate plane, an absolute value curve curves up and to the right in quadrant 4 and starts at y = 1.

Step-by-step explanation:

<u>Graph of the function:</u>

  • y = -√x + 1

The domain is x ≥ 0, the range y ≤ 1

Correct answer choice is B

  • On a coordinate plane, an absolute value curve curves up and to the right in quadrant 4 and starts at y = 1.

<em>The graph is attached</em>

erica [24]3 years ago
6 0

Answer: it’s B

Step-by-step explanation: got it right on edge 2021

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Phantasy [73]

Answer:

ΔABC≅ΔDEC by AAS

Step-by-step explanation:

You can use the AAS method of congruency.

Since you already have <BAC and <EDC congruent to eachother, and sides BC and EC congruent to each other, you only need that one remaining angle in between. <ACB can be proven congruent to <DCE by the Vertical Angles Theorem, and that gives you the AAS you need to prove that these two triangles are congruent

Hope this helped.

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3 years ago
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NeTakaya

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14

Step-by-step explanation:

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3 years ago
A Population of 50 Fruit Flies is Increased At A Rate of 6% per Day
andriy [413]

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Per how many days?

Step-by-step explanation:


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3 years ago
1. A train at a carnival covers a distance of 4 miles in 20 minutes. How
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16 miles
Since it’s 4 x 4





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3 years ago
What is the distance from P to Q ?A. 7 unitsB. 5 unitsC. 1 unitD. 25 unitshow do I find the distance from P to Q?
Alex17521 [72]

to find the distance between 2 points we should apply the formula

d=\sqrt[]{(x_2-x_1)^2+(y_2-_{}y_1)^2_{}}

call point q as point 1 for reference in the formula and p as point 2

replace the coordinates in the formula

d=\sqrt[]{(3-(-1))^2+(-4-(-1))^2}

simplify the equation

\begin{gathered} d=\sqrt[]{(3+1)^2+(-4+1)^2} \\ d=\sqrt[]{4^2+(-3)^2} \\ d=\sqrt[]{16+9} \\ d=\sqrt[]{25} \\ d=5 \end{gathered}

the distance between the 2 points is 5 units

7 0
1 year ago
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