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vampirchik [111]
3 years ago
15

The water level in Ricky Lake changes at an average of -716 inch every 3 years. a. Based on the rate above, how much will the wa

ter level change after one year? Show your calculations and you can use the vertical number line to help you., using 0 as the original water level.
Mathematics
1 answer:
jolli1 [7]3 years ago
4 0

Answer:

The water level will change at a rate of - 2 7/18 inches per year

Step-by-step explanation:

From the question, we have a change of -7 1/6 inch every 3 years

We now want to calculate the value of the change every year

Mathematically, that will be;

- 7 1/6 divided by 3

= -43/6 divided by 3

= -43/18 = - 2 7/18 inch per year

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Tanya [424]

Answer:

5

Step-by-step explanation:

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2 years ago
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How do I get the ans??
ioda

Answer:

11 sets of forks

Step-by-step explanation:

First subtract 406-278 to find total amount of forks needed which is 128 forks.

Then divide 128 by 12 to see how many sets of forks you need.

You get 10.67 but because you can't buy two thirds of a set, you have to round up to 11 sets of forks.

Hope this helps :)

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3 years ago
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Please help ASAP will give brainlist.
julia-pushkina [17]

Answer:

B

Step-by-step explanation:

Corresponding angles are not always congruent

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2 years ago
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Compare <img src="https://tex.z-dn.net/?f=5%5E%7B149%7D" id="TexFormula1" title="5^{149}" alt="5^{149}" align="absmiddle" class=
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A = 5¹⁴⁹
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Case 1: A>B
Result: True

Case 2: A=B
Result: False

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8 0
3 years ago
What is the range of the function y = x 2?
Leno4ka [110]

Answer:

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

The graph is also attached below.

Step-by-step explanation:

Given the function

y=x^2

  • We know that the range of a function is the set of values of the dependent variable for which a function is defined.

\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(0,\:0\right)

f\left(x\right)\ge \:0

Thus,

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

The graph is also attached below.

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