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3 years ago

The perimeter of a square is 36 inches. Find the length of a diagonal. Round your answer to the nearest tenth.

Mathematics

2 answers:

vovangra [49]3 years ago

5 0

Answer:

12.7

Step-by-step explanation:

The diagonal formula is x*sqrt(2) with x being one side length.

To find the side length divide the perimeter by 4.

That gives us 9.

Then, all you have to do is multiply that by the square root of 2 and then round.

Ad libitum [116K]3 years ago

4 0

Answer:

36/4 = 9

9 inch per side lol

9 rad 2 will be diagonal because of the 45 45 90 type triangle

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Beinggreat78 answer fast please :')

Dovator [93]

Answer:

17662.5 sq. feet

Step-by-step explanation:

We know that the formula for Area of a Circle is $A = \pi (r)^2$ .

r = radius; d = diameter, which is r*2

And so, in the question, it gives you that the diameter is 150, first we divide it by 2 to get the radius.

150/2 = 75

The question said to keep $\pi$ =3.14, so then we plug in the equation-

$A = 3.14 * 75^2$

Then we use PEMDAS [Parenthesis; Exponents; Multiply & Divide; Add & Subtract] which would get us-

$A = 3.14 * 5625$ (first doing the Exponent)

And we continue-

$A = 17662.5$ (then multiple)

So our answer is-

17662.5 sq. feet

I hope this helps!
Please give brainliest!

Have a great day

3 0

2 years ago

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Svetach [21]

Answer:

the answer is 54pi cm^3

Step-by-step explanation:

I'm smart

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3 years ago

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A. The function shown on the graph has a greater rate of change, but a lower starting point.

Julli [10]

I think it is A

sorry if it is wrong!

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2 years ago

An airplane is flying at an altitude of 2.7 miles and is 8.3 miles from the runway. Find the angle of depression that the airpla

Art [367]

Answer:

$18.98\°$

Step-by-step explanation:

Let

x-----> the angle of depression

we know that

The sine of angle x is equal to divide the altitude (opposite side to angle x) by the distance from the runway (hypotenuse)

$sin(x)=\frac{2.7}{8.3}$

$x=arcsin(\frac{2.7}{8.3})=18.98\°$

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2 years ago

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1. Consider the population model dP dt = 0.2P 1 − P 135 , where P(t) is the population at time t. (a) For what values of P is th

Kryger [21]

Answer:

$P = 0$ OR $P = 135$

$P > 0$ and $P < 135$

Generally the carrying capacity is can be defined as the highest amount of population and environment can support for an unlimited duration or time period

$P = 67.5$

Step-by-step explanation:

From the question we are told that

The population model is $\frac{dP}{dt} = 0.2P(1 - \frac{P}{135} )$

Generally at equilibrium

$\frac{dP}{dt} = 0$

$0.2P = 0$

=> $P = 0$

$(1 - \frac{P}{135} ) = 0$

=> $P = 135$

Thus at equilibrium P = 0 or P = 135

Generally when the population is increasing we have that

$\frac{dP}{dt} > 0$

$0.2P > 0$

=> $P > 0$

and

$(1 - \frac{P}{135} ) > 0$

$P < 135$

Now when the first value of P i.e $P< 0$ for $\frac{dP}{dt} > 0$

$P_2 > 135$

So when population increasing the values of P are

$P > 0$ and $P < 135$

So to obtain initial values of P where the population converge to the carrying capacity as $t \to [\infty]$

The rate equation can be represented as

$\frac{dP}{dt} = \frac{1}{5}P (1 - \frac{P}{135} )$

So we will differentiate the equation again we have that

$\frac{d^2 P}{dt^2} = \frac{(1 - \frac{P}{135} )}{5} - \frac{P}{675}$

Now as $t \to [\infty]$

$\frac{d^2 P}{dt^2} \to 0$

$\frac{(1 - \frac{P}{135} )}{5} - \frac{P}{675} = 0$

=> $\frac{(1 - \frac{P}{135} )}{5} = \frac{P}{675}$

=> $P = 67.5$

5 0

3 years ago