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2 years ago

Find the perimeter of the square if the length of its diagonal is 12.72 inches. Round to the nearest tenth

Mathematics

1 answer:

Ksju [112]2 years ago

3 0

Answer:

36 in.

Step-by-step explanation:

side = x

x^2 + x^2 = 12.72^2

2x^2 = 161.79

x^2 = 80.899

x = 9

perimeter = 4x = 4 * 9 = 36

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Write this number in scientific notation 0.000000358

Fittoniya [83]

Step-by-step explanation:

0.000000358 written as scientific notation is

$3.58 \times {10}^{ - 7}$

8 0

2 years ago

Find the volume of the triangular prism below 5mm 12mm 3mm

UNO [17]

Answer: 90 mm cubed

Step-by-step explanation:

Formula = (L*W*H) / 2

5 * 12 * 3 = 60 * 3 = 180

180 / 2 = 90

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2 years ago

Profit, P (x), is the difference between revenue, R (x), and cost C (x), so P (x) = R (x) - C (x), Which expression represents P

Ierofanga [76]

I think the answer is C

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2 years ago

6m=234 what does m equal

baherus [9]

Get m by itself so divide each side by 6
m= 234/6
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6 0

3 years ago

Read 2 more answers

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

2 years ago