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puteri [66]
3 years ago
5

Will give brainliest to right answer

Mathematics
1 answer:
Sladkaya [172]3 years ago
7 0

Answer:

D

Step-by-step explanation:

Area of rectangle: 6 x 4 = 24

Area of triangle: 4(4)/2 = 8

Area of semi-circle: (πr²2)2 = 2π ≈ 6.28

24 + 8 + 6 = 38

You might be interested in
Please do 3 & 4 and show work
Shalnov [3]

Answer:

3. A=B=70°; C=D=110°

4. A=90°; C=48°

Step-by-step explanation:

Same-side angles of a trapezoid add to 180°.

___

3. (2x) +(x+15) = 180

3x + 15 = 180 . . . . simplify

x +5 = 60 . . . . . . . divide by 3

x = 55

A=B=(x+15)° = (55+15)° = 70°

C=D=2x° = 2·55° = 110°

__

4. You are given the values of one of the angles on each side, so the other can be found by subtraction:

A=180°-D = 90°

C = 180° -132° = 48°

8 0
3 years ago
A candle burned at a steady rate. After 37 minutes, the candle was 11.2 inches tall. Eighteen minutes later,
irinina [24]

Answer:

i think the answer is 4.65 inches tall.

Step-by-step explanation:

i attempted it this may be wrong just wait for someone to confirm it.

3 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
What is the answer to this along with steps to doing it?
pashok25 [27]
1/2*(2*8)=3*(-12)
1/2*16=3*(-12)
1/2*16= -36
8= -36
The equality is false because the left hand and right hand side are different
8 0
3 years ago
What is 4k-14+3k=21​
BARSIC [14]

So here. I don't know how to explain so-

5 0
3 years ago
Read 2 more answers
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