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Jobisdone [24]
3 years ago
10

A rectangle has a length of 12.6 m and a width of 6.3 m. A similar rectangle has a width of 2.1 m. What is the length of the sim

ilar rectangle?
Mathematics
1 answer:
Oxana [17]3 years ago
5 0

Step-by-step explanation:

Please see the photo for the method

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inn [45]
-7 1/8 < -7 1/2 is the answer
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What is 2 times 2 divded by3
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The answer is 4/3
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Read 2 more answers
When measuring the length a and the width b of a rectangle, it was found that 5.4&lt; a&lt; 5.5 and 3.6&lt; b&lt; 3.7 (in centim
leva [86]

Answer:

The possible values of the perimeter of the rectangle is between 18 and

18.4 cm ⇒ [18 < perimeter of rectangle < 18.4] cm

Step-by-step explanation:

* Lets explain how to solve the problem

- The perimeter of any quadrilateral is the sum of the length of its

  four sides

- The rectangle is a quadrilateral with each two opposite sides are

  equal in lengths, then its perimeter = 2l + 2w, where l, w are its

  length and width

∵ The the length of the rectangle is a

∵ 5.4 < a < 5.5 cm

∵ The perimeter of the rectangle = 2a + 2b

- Lets multiply the inequality by 2

∴ 5.4 × 2 < a × 2 < 5.5 × 2

∴ 10.8 < 2a < 11 ⇒ (1)

∵ The width of the rectangle is b

∵ 3.6 < b < 3.7 cm

- Lets multiply the inequality by 2

∴ 3.6 × 2 < b × 2 < 3.7 × 2

∴ 7.2 < 2b < 7.4 ⇒ (2)

- <u><em>Add inequalities (1) and (2)</em></u>

∵     10.8 < 2a < 11

  +   7.2 < 2b < 7.4

∴  18 < 2a + 2b < 18.4

∵ 2a + 2b represents the perimeter of the rectangle

∴ The possible values of the perimeter of the rectangle is between

  18 and 18.4 cm ⇒ [18 < perimeter of rectangle < 18.4] cm

7 0
3 years ago
After being treated with chemotherapy, the radius of the tumor decreased by 23%. What is the corresponding percentage decrease i
Crazy boy [7]

Answer:

The volume of the tumor experimented a decrease of 54.34 percent.

Step-by-step explanation:

Let suppose that tumor has an spherical geometry, whose volume (V) is calculated by:

V = \frac{4\pi}{3}\cdot R^{3}

Where R is the radius of the tumor.

The percentage decrease in the volume of the tumor (\%V) is expressed by:

\%V = \frac{\Delta V}{V_{o}} \times 100\,\%

Where:

\Delta V - Absolute decrease in the volume of the tumor.

V_{o} - Initial volume of the tumor.

The absolute decrease in the volume of the tumor is:

\Delta V = V_{o}-V_{f}

\Delta V = \frac{4\pi}{3}\cdot (R_{f}^{3}-R_{o}^{3})

The percentage decrease is finally simplified:

\%V = \left[1-\left(\frac{R_{f}}{R_{o}}\right)^{3} \right]\times 100\,\%

Given that R_{o} = R and R_{f} = 0.77\cdot R, the percentage decrease in the volume of tumor is:

\%V = \left[1-\left(\frac{0.77\cdot R}{R}\right)^{3} \right]\times 100\,\%

\%V = 54.34\,\%

The volume of the tumor experimented a decrease of 54.34 percent.

5 0
3 years ago
(AP Calc) the graph of the function f shown above has three line segments....
I am Lyosha [343]

Answer:

B

Step-by-step explanation:

The average rate of change of f(x) in the closed interval [ a, b ] is

\frac{f(b)-f(a)}{b-a}

Here [ a, b ] = [ - 1, 6 ] and from the graph

f(b) = f(6) = 0

f(a) = f(- 1) = 0, thus

average rate of change = \frac{0-0}{6+1} = \frac{0}{7} = 0

7 0
3 years ago
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