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3 years ago

Which of the following answer choices would satisfy the inequality below? -2.74 < x < -2.35

Mathematics

1 answer:

Helga [31]3 years ago

4 0

Answer: -3pi/4

Step-by-step explanation:

-3pi/4 = -2.35619

-2.35619 is closer to 0 making it greater than -2.74 but less than -2.35 bc -2.35 is closer to 0 than -2.35619 is so.... (the greater the negative number, the less value it has)

-2.74 < -3pi/4 < -2.35

hope this helps ^_^

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Use the Pythagorean theorem to calculate the length of RO:

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$A_T=\dfrac{1}{2}|RO||PA|$

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4 years ago

The number of research articles in a prominent journal written by researchers during 1983–2003 can be approximated by U(t) = 4.1

Nimfa-mama [501]

Answer:

Step-by-step explanation:

From the information given:

We are to find; an expression for the total no. of articles written since 1983

∴

The total no. of articles written since 1983 $=\int \limts ^t_0 U(t) dt$

$= \int ^t_0 \limits \dfrac{4.1 e^{0.6\ t}}{0.2+e^{0.6t}} \ dt$

$=\dfrac{4.1}{0.6}\bigg [ \mathtt{In} \bigg | 0.2 + e^{0.6 x} \bigg| \bigg]^t_0$

$=\dfrac{4.1}{0.6}\bigg [ \mathtt{In} \bigg | 0.2 + e^{0.6 x} \bigg| - \mathtt{In} \bigg|0.2+1\bigg|\bigg]$

$=\dfrac{4.1}{0.6}\bigg [ \mathtt{In} \bigg | \dfrac{0.2 + e^{0.6 x}}{1.2} \bigg|\bigg]$

$=\dfrac{41}{6} \mathtt{In} \bigg | {0.2 + e^{0.6 x} \bigg|-\dfrac{41}{6} \mathtt{ In} \bigg | 1.2\bigg |$

$=6.833\times \mathtt{In} \bigg | {0.2 + e^{0.6 x} \bigg|-1.2458$

$=6.833\times \mathtt{In} \bigg | {0.2 + e^{0.6 x} \bigg|-1.25 \ \mathbf{thousand \ articles}$

Therefore, the total number of articles written since 1983 is $=6.833\times \mathtt{In} \bigg | {0.2 + e^{0.6 x} \bigg|-1.25 \ \mathbf{thousand \ articles}$

b. To find how many articles were being written from 1983 to 2003

i.e. t = 2003 - 1983 = 20

∴

Total articles written from 1983 to 2003 is $=\int \limts ^{20}_0 U(t) dt$

$= \int ^{20}_0 \limits \dfrac{4.1 e^{0.6\ t}}{0.2+e^{0.6t}} \ dt$

$=\dfrac{4.1}{0.6}\bigg [ \mathtt{In} \bigg | 0.2 + e^{0.6 x} \bigg| \bigg]^{20}_0$

$=\dfrac{4.1}{0.6}\bigg [ \mathtt{In} \bigg | 0.2 + e^{0.6 *20} \bigg| - \mathtt{In} \bigg|0.2+e^{0.6*0}\bigg|\bigg]$

$=\dfrac{4.1}{0.6}\bigg [ \mathtt{In} \bigg | 0.2 + e^{12} \bigg| - \mathtt{In} \bigg|0.2+e^0\bigg|\bigg]$

$=\dfrac{4.1}{0.6}\bigg [ \mathtt{In} \bigg | 0.2 + e^{12} \bigg| - \mathtt{In} \bigg|0.2+1\bigg|\bigg]$

$=\dfrac{4.1}{0.6}\bigg [ \mathtt{In} \bigg | 0.2 + e^{12} \bigg| - \mathtt{In} \bigg|1.2\bigg|\bigg]$

$=\dfrac{4.1}{0.6}\bigg [ \mathtt{In} \bigg | \dfrac{0.2 + e^{12}}{1.2} \bigg|\bigg]$

= 80.75 thousand articles

3 0

3 years ago

Which of the following answer choices would satisfy the inequality below? -2.74 < x < -2.35​

Which of the following answer choices would satisfy the inequality below? -2.74 < x < -2.35