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KIM [24]
3 years ago
13

How do you graph this helpppp and explain tyty

Mathematics
1 answer:
Ivahew [28]3 years ago
5 0
You graph it by graphing it tbh
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Which answer shows a reflection across the y-axis?
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The answer would be D.
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Estephany has a box that is filled with toys. The box has a length of 3.5 feet, a width of 6 feet, and a height of 3 feet. What
Zigmanuir [339]

Answer:

63

Step-by-step explanation:

The formula for volume of a box is length×width×height = volume

So you would multiply  3.5×6×3=63

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3 years ago
Simplify the following expression as a monomial<br><br> X^2y÷yx^2
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3 years ago
Help with Math PLZ (Questions in screenshot)
-Dominant- [34]

1.) Let's say that the circle on the graph below represents x. The arrow is pointing to all the numbers greater than x, which happens to be on -2. If it points right, this means that x can equal to any number greater than -2. So your answer is x > -2.

2.) For inequalities such as these, you can simplify just like what you do for normal equations. Let's isolate x.

x - 7 > 10

--> x > 10 + 7

---> x > 17

3.) 3x \leq 21

--> \frac{3x}{3} \leq  \frac{21}{3}

---> x \leq 7

4.) 5a - 2 < 18

--> 5a < 18 + 2

---> 5a < 20

----> \frac{5a}{5} < \frac{20}{5}

-----> a < 4

5.) 2t + 8 \geq -4(t + 1)

--> 2t + 8 \geq -4t - 4

---> 2t + 4t \geq -8 - 4

----> 6t \geq -12

-----> t \geq -2

7 0
4 years ago
Solve the following recurrence relation: <br> <img src="https://tex.z-dn.net/?f=A_%7Bn%7D%3Da_%7Bn-1%7D%2Bn%3B%20a_%7B1%7D%20%3D
-Dominant- [34]

By iteratively substituting, we have

a_n = a_{n-1} + n

a_{n-1} = a_{n-2} + (n - 1) \implies a_n = a_{n-2} + n + (n - 1)

a_{n-2} = a_{n-3} + (n - 2) \implies a_n = a_{n-3} + n + (n - 1) + (n - 2)

and the pattern continues down to the first term a_1=0,

a_n = a_{n - (n - 1)} + n + (n - 1) + (n - 2) + \cdots + (n - (n - 2))

\implies a_n = a_1 + \displaystyle \sum_{k=0}^{n-2} (n - k)

\implies a_n = \displaystyle n \sum_{k=0}^{n-2} 1 - \sum_{k=0}^{n-2} k

Recall the formulas

\displaystyle \sum_{n=1}^N 1 = N

\displaystyle \sum_{n=1}^N n = \frac{N(N+1)}2

It follows that

a_n = n (n - 2) - \dfrac{(n-2)(n-1)}2

\implies a_n = \dfrac12 n^2 + \dfrac12 n - 1

\implies \boxed{a_n = \dfrac{(n+2)(n-1)}2}

4 0
3 years ago
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