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Yuri [45]
2 years ago
6

Identify the Asymptotes for the graphs below help pls!

Mathematics
2 answers:
scoray [572]2 years ago
5 0
Asymptotes are lines that continues to approach a given curve but never actually meets it! For the second figure it is the Y axis since the curves slowly appear to approach the Y axis! Not sure about the first figure(probably the X axis)!
Ede4ka [16]2 years ago
3 0
Asymptotes are lines that continues to approach a given curve but never actually meets it! For the second figure it is the Y axis since the curves slowly appear to approach the Y axis! Not sure about the first figure(probably the X axis)!
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34 students in Mr. Jones’s math class passed the final exam. There are 40 students in his class total. What Percent of students
solniwko [45]

Answer:

85%

Step-by-step explanation:

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2 years ago
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Shelby is decorating a rectangular ballroom ceiling with garland. Beginning in a corner, Shelby strings garland along length of
Liono4ka [1.6K]

Answer:

since the room is rectangular and has 4 sides

2 sides have the same length

18*2= 36m

80*2=160m

then add the sums...

the answer is 196 meters

Step-by-step explanation:

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2 years ago
Maya deposits $5000 into a checking account that pays 0.75% annual interest compounded monthly. What will be the balance after 8
Nana76 [90]

~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$5000\\ r=rate\to 0.75\%\to \frac{0.75}{100}\dotfill &0.0075\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{monthly, thus twelve} \end{array}\dotfill &12\\ t=years\dotfill &8 \end{cases} \\\\\\ A=5000\left(1+\frac{0.0075}{12}\right)^{12\cdot 8}\implies A=5000(1.000625)^{96}\implies A\approx 5309.08

5 0
2 years ago
The volume (in cubic inches) of a shipping box is modeled by V =2x³ -19x² +39x, where x is the length (in inches). Determine the
yawa3891 [41]

Answer:

The values of x for which the model is 0 ≤ x ≤ 3

Step-by-step explanation:

The given function for the volume of the shipping box is given as follows;

V = 2·x³ - 19·x² + 39·x

The function will make sense when V ≥ 0, which is given as follows

When V = 0, x = 0

Which gives;

0 = 2·x³ - 19·x² + 39·x

0 = 2·x² - 19·x + 39

0 = x² - 9.5·x + 19.5

From an hint obtained by plotting the function, we have;

0 = (x - 3)·(x - 6.5)

We check for the local maximum as follows;

dV/dx = d(2·x³ - 19·x² + 39·x)/dx = 0

6·x² - 38·x + 39 = 0

x² - 19/3·x + 6.5 = 0

x = (19/3 ±√((19/3)² - 4 × 1 × 6.5))/2

∴ x = 1.288, or 5.045

At x = 1.288, we have;

V = 2·1.288³ - 19·1.288² + 39·1.288 ≈ 22.99

V ≈ 22.99 in.³

When x = 5.045, we have;

V = 2·5.045³ - 19·5.045² + 39·5.045≈ -30.023

Therefore;

V > 0 for 0 < x < 3 and V < 0 for 3 < x < 6.5

The values of x for which the model makes sense and V ≥ 0 is 0 ≤ x ≤ 3.

8 0
3 years ago
Please help! This is timed
gulaghasi [49]
I think the answer is C 1 2/3
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3 years ago
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