All categories

3 years ago

Which option please I need help ASAP reply ASAP

Mathematics

2 answers:

IrinaK [193]3 years ago

8 0

Answer:

The answer is probably c

Step-by-step explanation:

its deducted by the x axis

Dafna1 [17]3 years ago

5 0

Answer:

Step-by-step explanation:

You might be interested in

I am not usre what the answer could be

lakkis [162]

Answer:

$x \le 5$

Step-by-step explanation:

Given

Shape: Cube

Dimensions: x, 3x and 2x

$Volume < 900$

Required

Show that $x \le 5$

First, calculate the volume

$Volume = x * 3x * 2x$

$Volume = 6x^3$

$Volume < 900$ becomes

$6x^3 < 900$

Divide both sides by 9

$x^3 < 150$

For $x^3 < 150$ , the value of x as to be: 1, 2, 3, 4 or 5

Hence:

$x \le 5$

3 0

3 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

Help me solve this math problem please!! With an explanation as well. Thank you :)

zlopas [31]

Answer:

hello

Step-by-step explanation: ok look at the years

8 0

3 years ago

Enter the slope-intercept equation of the line shown below. ??!!!!!

Luda [366]

Answer:

y=2x+4

Step-by-step explanation:

2x is the gradient and the y intercept is 4 so we just put them into the equation y=mx+c

5 0

3 years ago

What is fifty four times eighty nine?

o-na [289]

Fifty-Four Multiplied by Eighty-Nine is Four Thousand Eight Hundred and Six. (4,806)

5 0

4 years ago

Read 2 more answers