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netineya [11]
3 years ago
9

The sides of a triangle are 10, 14, 18 what tie of triangle is it

Mathematics
1 answer:
leva [86]3 years ago
5 0
Let "a", "b" and "с"  be sides of the triangle ("с" is the longest side). The triangle will be:
right if       a² + b² = c²
аcute if     a² + b² > c²    
obtuse if   a² + b² < c²    

We have a=10, b=14 and c=18

a² + b² = 10² + 14² = 100 + 196 = 296
and
c² = 18² = 324

296 < 324   ⇒  obtuse triangle.
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Find the 30th term of the following sequence. 1, 7, 13, 19, ... 174 175 180 181.
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Checking if the sequence is an arithmetic sequence 7 - 1 = 6 13 – 7 = 6 Therefore the sequence is arithmetic, with a_0 = 1 and d = 6 a_n = a_0 + (n-1)d a_30 = 1 + (29)(6) = 175 Therefore the 30th term is 175.
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4 years ago
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13. Basil Tomlin's online checking account had a balance of $371.07. The balance did not
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Answer:

Step-by-step explanation

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3 years ago
Write the slope intercept form of the equation of the line through the given points
garri49 [273]

Answer:

9) y = -3x - 4

10) y = -1/3x + 14/3

11) y = x - 1

12) y = -7/5x - 18/5

13) y = 3/5x + 9/5

14) y = -2x - 3

Step-by-step explanation:

For this explanation, let's use the last problem as the example. You would use the formula y=mx+b. The first thing you would need to find would be the slope, or m. So, you would find the slope and conclude the answer is -2. After that, you would solve for b and get the answer. Hope this helped!

4 0
3 years ago
Emmett gathered some data to compare the monthly cost of renting a one-bedroom apartment in Dallas and Austin. The data collecte
Goryan [66]

Answer:

The correct options are;

a) The city's data has the higher median monthly apartment cost of the two cities by $155

b) The city's data had the greater variability among monthly apartment costs because the standard deviation was $49.13 more than that of the other set of data

Step-by-step explanation:

The given data for the cost of renting a one-bedroom apartment in Dallas is presented as follows;

$994, $1,322, $1,075, $1,189, $1,172, $1,465, $1,215, $930, $1,090, $1,288

Which can be arranged in increasing order and analyzed to find the interquartile range, mean and standard deviation using Microsoft Excel as follows;

$930, $994, $1,075, $1,090, $1,172, $1,189, $1,215, $1,288, $1,322, and $ 1,465

The first quartile, Q₁ = $1,054.75

The third quartile, Q₃ = $1,296.5

The interquartile range = Q₃ - Q₁ = $241.75

The median = $1,180.5

The average monthly cost = $1,174

The standard deviation of the sample = $159.9597

The given data for the cost of renting a one-bedroom apartment in Austin is presented in increasing order using Microsoft Excel as follows;

$900, $950, $1,100, $1,250, $1,296, $1,375, $1,389, $1,400, $1,450, $1,495

The interquartile range, mean and standard deviation are found using Microsoft Excel as follows;

The first quartile, Q₁ = $1,062.5

The third quartile, Q₃ = $1,412.5

The interquartile range = Q₃ - Q₁ = $1,412.5 - $1,062.5 = $350

The median = $1,335.5

The average monthly cost = $1,260.5

The standard deviation of the sample = $209.0945

The difference in the median cost of the two cities is $1,335.5 - $1,180.5 = $155

Therefore, the Austin's city data has the higher median monthly apartment cost of the two cities by $155

b) The difference in the sample standard deviation of the two cities is $209.0945 - $159.9597 = $49.13476

Therefore, the Austin city data had the greater variability among monthly apartment costs because the standard deviation was $49.13 more than that of the other set of data.

7 0
3 years ago
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Which expression is equivalent
gtnhenbr [62]

Answer:

Option  B is correct.

\frac{81m^2n^5}{8} is equivalent to \frac{(3m^{-1}n^2)^4}{(2m^{-2}n)^3}

Step-by-step explanation:

Given expression: \frac{(3m^{-1}n^2)^4}{(2m^{-2}n)^3}

Using exponents power:

  • (ab)^n = a^nb^n
  • (a^n)^m = a^{nm}
  • a^m \cdot a^n = a^{m+n}

Given: \frac{(3m^{-1}n^2)^4}{(2m^{-2}n)^3}

Apply exponent power :

⇒ \frac{3^4 (m^{-1})^4(n^2)^4}{2^3(m^{-2})^3 n^3}

⇒ \frac{81 m^{-4}n^8}{8m^{-6}n^3} = \frac{81 m^{-4} \cdot m^6 n^8 \cdot n^{-3}}{8}

⇒\frac{81 m^{-4+6} n^{8-3}}{8} = \frac{81 m^2 n^5}{8} = \frac{81m^2 n^5}{8}

Therefore, the expression which is equivalent to  \frac{(3m^{-1}n^2)^4}{(2m^{-2}n)^3} is,  \frac{81m^2 n^5}{8}

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3 years ago
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