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tester [92]
2 years ago
6

Please help me with the questions I have please help me with the question mark you as a brnlist

Mathematics
1 answer:
Harlamova29_29 [7]2 years ago
4 0

Answer:

x = 21

Step-by-step explanation:

You can solve this problem by using similar triangles. You can tell that the two triangles are similar by AAA (since two angles are the same, the third must be too).

Now that we've established that the triangles are similar, you need to identify the corresponding sides:

12 corresponds to 28

9 corresponds to x

In order to find x, you'll need to find the scale factor, which you can find by dividing 28 by 12:

28 ÷ 12 = 7/3

Now that you know that 7/3 is the scale factor, you can multiply it by 9 to find x:

9 × 7/3 = x

x = 63/3

x = 21

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Rectangle measures 15/2 inches by 4/3 inches. What is it’s area?
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Answer:

10 inches squared

Step-by-step explanation:

Area is length times width

A= lw

A= 15/2 x 4/3

A = 60/6

A= 10 in^2

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3 years ago
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2. Find the perimeter and area of a square with a 7-ft side.
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Answer:

Step-by-step explanation:

Perimeter 4L = 28

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3 years ago
Can someone help me with this math homework please!
kobusy [5.1K]

Answer:

It's 2, 1, and y = 2x + 1.

Step-by-step explanation:

You can see the rise is 2 and the run is 1, making the slope = 2, and the y-intercept is 1 because that is where it crosses the y axis. Once you have the slope and y intercept, you can put it in a function, with the form being y=2x+1, the slope being the number before the x and the y-int value being after the x.

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2 years ago
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I need help. if i dont pass this, i wont pass. Will give brainliest!
Tatiana [17]

A the lines are parallel

8 0
2 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
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