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2 years ago

10 points!! plzzzzzzz helpppp :( find the letters ( K,K,T )

Mathematics

2 answers:

zheka24 [161]2 years ago

6 0

Answer:

dude u lie it says 5 points

Step-by-step explanation:

anygoal [31]2 years ago

4 0

Answer:

1) k=90

2) k=135

3) s= 115 t=30 r=35

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In circle O, BC = 17 and DC = 30. The diagram is not drawn to scale.

Aliun [14]

Applying the secant-tangent theorem, the length of diameter, BA is: A. 52.9.

<h3>What is the Secant-Tangent Theorem?</h3>

Considering the image given, the secant-tangent theorem states that:

DC² = (BC)(BA)

Given the following:

BC = 17
DC = 30

Substitute the values into the equation

30² = (17)(BA)

900/17 = BA

BA ≈ 52.9

Length of the diameter, BA is: A. 52.9

Learn more about the secant-tangent theorem on:

brainly.com/question/10732273

#SPJ1

8 0

2 years ago

What is the value of x in the figure shown below?

Kaylis [27]

Answer: 100 degress

Step-by-step explanation:

35+65=100

180-100

Angle = 80 degrees

x=100

8 0

3 years ago

if the length of the rectangle is twice the width and the perimeter of the rectangle is 30 centimeters what is the length of the

Dominik [7]

If the perimeter equals 30 and the length is twice the width, then the length is equal to 20 and the width is equal to 10.

7 0

3 years ago

Suppose that a spherical droplet of liquid evaporates at a rate that is proportional to its surface area: where V = volume (mm3

Alex

Answer:

V = 20.2969 mm^3 @ t = 10

r = 1.692 mm @ t = 10

Step-by-step explanation:

The solution to the first order ordinary differential equation:

$\frac{dV}{dt} = -kA$

Using Euler's method

$\frac{dVi}{dt} = -k *4pi*r^2_{i} = -k *4pi*(\frac {3 V_{i} }{4pi})^(2/3)\\ V_{i+1} = V'_{i} *h + V_{i} \\$

Where initial droplet volume is:

$V(0) = \frac{4pi}{3} * r(0)^3 = \frac{4pi}{3} * 2.5^3 = 65.45 mm^3$

Hence, the iterative solution will be as next:

i = 1, ti = 0, Vi = 65.45

$V'_{i} = -k *4pi*(\frac{3*65.45}{4pi})^(2/3) = -6.283\\V_{i+1} = 65.45-6.283*0.25 = 63.88$

i = 2, ti = 0.5, Vi = 63.88

$V'_{i} = -k *4pi*(\frac{3*63.88}{4pi})^(2/3) = -6.182\\V_{i+1} = 63.88-6.182*0.25 = 62.33$

i = 3, ti = 1, Vi = 62.33

$V'_{i} = -k *4pi*(\frac{3*62.33}{4pi})^(2/3) = -6.082\\V_{i+1} = 62.33-6.082*0.25 = 60.813$

We compute the next iterations in MATLAB (see attachment)

Volume @ t = 10 is = 20.2969

The droplet radius at t=10 mins

$r(10) = (\frac{3*20.2969}{4pi})^(2/3) = 1.692 mm\\$

The average change of droplet radius with time is:

Δr/Δt = $\frac{r(10) - r(0)}{10-0} = \frac{1.692 - 2.5}{10} = -0.0808 mm/min$

The value of the evaporation rate is close the value of k = 0.08 mm/min

Hence, the results are accurate and consistent!

5 0

3 years ago

20.41312<br> (a) The given number truncated to four decimal places is

Ivanshal [37]

Answer:

20.4131

Step-by-step explanation:

You need to have only 4 decimal places after , last decimal being approximated

8 0

3 years ago