Find the smallest of two consecutive even integers if the square is 10 more than the larger

Mr.Hoges wants to build a fence around his entire rectangular garden.What is the total distance around his garden

svetoff [14.1K]

Answer:

The total distance of the rectangular garden is the perimeter of the garden.

Step-by-step explanation:

Since the garden is in a rectangular shape. So, the perimeter of the rectangular garden is the distance on which Mr. Hoges can build the fence. However, the perimeter can be determined by adding all the sides. A rectangular shape contains four sides and opposite sides are equal. Thus, it can be determined by using the below formula;

$P=2(l+b)$

Where

l = length of the garden

b is the breadth of the garden

3 0

3 years ago

PLEASE ANSWER ASAP! WILL GIVE BRAINLIEST+12 POINTS!

VikaD [51]

Answer:

a¹⁰

Step-by-step explanation:

$(x^{m} )^{n} =x^{mn}$

So here we have (a⁵)², meaning we need to multiply the exponents to get a¹⁰

7 0

3 years ago

The ratio of angles angles A:B:C is 5:4:3 . find measure of angle A and B And C

mr Goodwill [35]

Answer:

A=75

B=60

C=45

Step-by-step explanation:

5+4+3=12

every triangle is equal to 180 so we divide

180/12=15

then we multiply each angle by 15

5*15

4*15

and 3*15

don't forget to check your work by adding the answers together to get 180

8 0

3 years ago

A polling organization contacts 1657 undergraduates who attend a university and live in the United States and asks whether or no

Dimas [21]

Step-by-step explanation:

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7 0

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