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aksik [14]
3 years ago
13

Which is the median of the following data? 12,8,13,12,15,11,10,7

Mathematics
1 answer:
tensa zangetsu [6.8K]3 years ago
7 0
11.5 beause you take the average of 12 and 11
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How do you find the squared foot of an octagon?
Westkost [7]

Answer:

The four pieces cut off can be arranged into two squares with sides of length b feet and hence the area of the octagon is w 2 - 2 b 2 square feet. Since w = 12 feet, all that is needed is to find b. square feet which is approximately 119 square feet.

Step-by-step explanation:

5 0
3 years ago
areas of four triangles are 40 cm^2, 90 cm^2, 202.5 cm^2 and 455.6 cm^2 are continue in sequence. the enlargement factor is 1.5,
maw [93]

Answer:

9th Triangle

Step-by-step explanation:

Given the area of four triangles:40 cm^2, 90 cm^2, 202.5 cm^2 \:and\: 455.6 cm^2

If the Scale factor is 1.5.

This is a geometric sequence and the common ratio is 1.5 X 1.5.

Therefore, the area of any n triangle can be determined using the function:

A(n)=40\cdot1.5^{2(n-1)}

We want to determine which triangle has an area greater than 15000cm^2.

When A(n)=15000

40\cdot1.5^{2(n-1)}>15000\\1.5^{2(n-1)}>\frac{15000}{40} \\1.5^{2(n-1)}>375\\\text{Change to Logarithm form}\\2(n-1)>Log_{1.5}375\\\text{Applying logarithm change of base to base 10 law}\\2(n-1)>\frac{Log 375}{Log 1.5} \\2n-2>14.62\\2n>14.62+2=16.62\\n>8.31

Therefore, the 9th triangle has an area greater than 15000cm^2.

5 0
3 years ago
Use the theorem in Sec. 28 to show that if f (z) is analytic and not constant throughout a domain D, then it cannot be constant
alexandr1967 [171]

Answer:

The value of f(z) is not constant in any neighbourhood of D. The proof is as explained in the explaination.

Step-by-step explanation:

Given

For any given function f(z), it is analytic and not constant throughout a domain D

To Prove

The function f(z) is non-constant constant in the neighbourhood lying in D.

Proof

1-Assume that the value of f(z)  is analytic and has a constant throughout some neighbourhood in D which is ω₀

2-Now consider another function F₁(z) where

F₁(z)=f(z)-ω₀

3-As f(z) is analytic throughout D and F₁(z) is a difference of an analytic function and a constant so it is also an analytic function.

4-Assume that the value of F₁(z) is 0 throughout the domain D thus F₁(z)≡0 in domain D.

5-Replacing value of F₁(z) in the above gives:

F₁(z)≡0 in domain D

f(z)-ω₀≡0 in domain D

f(z)≡0+ω₀ in domain D

f(z)≡ω₀ in domain D

So this indicates that the value of f(z) for all values in domain D is a constant  ω₀.

This contradicts with the initial given statement, where the value of f(z) is not constant thus the assumption is wrong and the value of f(z) is not constant in any neighbourhood of D.

6 0
3 years ago
What is an equation of the line that passes through the points (0, 2) and ( 5, 4)?
creativ13 [48]

Answer: y = 2/5 x + 2

Step-by-step explanation:

why not do this one too yk

slope = y2 - y1 / x2 - x1

4 - 2 / 5 - 0

slope = 2/5

ill use (5,4) for the y=mx+b initial equation

y = mx+b

4 = (2/5)(5) +b

4 = 2 + b

2 = b

now for the final equation whoop whoop

y = 2/5 x + 2

7 0
3 years ago
Simplify: -14cd-2c^2d^2+15cd+9c^2d^2-15cd
MaRussiya [10]

7c^2d^2-14cd

Hope this helps a bunch!!!!!!!!!!!!

7 0
3 years ago
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