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3 years ago

Help me please I have a dba and I don’t know the answer!!!

Mathematics

1 answer:

NNADVOKAT [17]3 years ago

7 0

Answer:

9a³b³

how this can be of help

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The temperature during the day in Miami was 92 degrees fahrenheit. During the night, the temperature fell 4 degrees fahrenheit.

BigorU [14]

You need to take away 4 degrees from 92.
"Take away" means subtraction.
92 - 4
The answer is option C.

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3 years ago

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Which of the following,is a factor of 3x^2-9x^2+3x

Savatey [412]

Answer:

-3x(x-1)

Step-by-step explanation:

-6x^2 +3x

-3x(x-1)

Hope this helps!

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3 years ago

Find the solution to the equation 8x + 6 - 2x = 6x +5

Hunter-Best [27]

The answer is no solution.
First you simplify both sides of the equation... Then you subtract 6x from both sides. And lastly you subtract 6 from both sides. Leaving you with no solutions.

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3 years ago

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A manager received an invoice for $86.30 for 30gal ice cream. how much did each gallon cost? a $3.49 b $3.15 c $4.26 d $2.88

Vinil7 [7]

86.3 / 30 = 2.876

Therfore the answer is D $2.88

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3 years ago

In quadratic drag problem, the deceleration is proportional to the square of velocity

Mars2501 [29]

Part A

Given that $a= \frac{dv}{dt} =-kv^2$

Then,

$\int dv= -kv^2\int dt \\ \\ \Rightarrow v(t)=-kv^2t+c$

For $v(0)=v_0$ , then

$v(0)=-kv^2(0)+c=v_0 \\ \\ \Rightarrow c=v_0$

Thus, $v(t)=-kv(t)^2t+v_0$

For $v(t)= \frac{1}{2} v_0$ , we have

$\frac{1}{2} v_0=-k\left( \frac{1}{2} v_0\right)^2t+v_0 \\ \\ \Rightarrow \frac{1}{4} kv_0^2t=v_0- \frac{1}{2} v_0= \frac{1}{2} v_0 \\ \\ \Rightarrow kv_0t=2 \\ \\ \Rightarrow t= \frac{2}{kv_0}$

Part B

Recall that from part A,

$v(t)= \frac{dx}{dt} =-kv^2t+v_0 \\ \\ \Rightarrow dx=-kv^2tdt+v_0dt \\ \\ \Rightarrow\int dx=-kv^2\int tdt+v_0\int dt+a \\ \\ \Rightarrow x=- \frac{1}{2} kv^2t^2+v_0t+a$

Now, at initial position, t = 0 and $v=v_0$ , thus we have

$x=a$

and when the velocity drops to half its value, $v= \frac{1}{2} v_0$ and $t= \frac{2}{kv_0}$

Thus,

$x=- \frac{1}{2} k\left( \frac{1}{2} v_0\right)^2\left( \frac{2}{kv_0} \right)^2+v_0\left( \frac{2}{kv_0} \right)+a \\ \\ =- \frac{1}{2k} + \frac{2}{k} +a$

Thus, the distance the particle moved from its initial position to when its velocity drops to half its initial value is given by

$- \frac{1}{2k} + \frac{2}{k} +a-a \\ \\ = \frac{2}{k} - \frac{1}{2k} = \frac{3}{2k}$

7 0

3 years ago