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GrogVix [38]
2 years ago
11

Use Pythagorean identities to prove whether ΔLMN is a right, acute, or obtuse triangle. Show all work for full credit.

Mathematics
1 answer:
diamong [38]2 years ago
5 0

Answer:  The given triangle LMN is an obtuse-angled triangle.

Step-by-step explanation:  We are given to use Pythagorean identities to prove whether ΔLMN is a right, acute, or obtuse triangle.

From the figure, we note that

in ΔLMN, LM = 5 units, MN = 13 units  and  LN = 14 units.

We know that a triangle with sides a units, b units and c units (a  > b, c) is said to be

(i) Right-angled triangle if b^2+c^2=a^2,

(ii) Acute-angled triangle if b^2+c^2>a^2,

(iii) Obtuse-angled triangle if b^2+c^2

For the given triangle LMN, we have

a = 14, b = 13 and c = 5.

So,

b^2+c^2=13^2+5^2=169+25=194,\\\\a^2=14^2=196.

Therefore,  b^2+c^2

Thus, the given triangle LMN is an obtuse-angled triangle.

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sveta [45]

Answer:

Hayden's hourly rate is $30

Extra hour rate is 1.5×30=$45

given that he earned $2175 by working 40 regular hours, the number of overtime hours he worked will be given as follows;

let the number of overtime be x

hence total amount earned will be:

(40×30)+(45×x)=2175

1200+45x=2175

45x=2175-1200

45x=975

solving for x we get

x=975/45

x=65/3=21 2/3 hours  

 

Therefore the answer is c.562.50

Step-by-step explanation:

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2 years ago
Brody is purchasing some tools for his workshop. He has a budget of $120 and needs to buy at least 14 tools. Each hammer costs $
STatiana [176]

Answer:

For this case, the first thing we must do is define variables:

x: number of hammers

y: number of wrenches

We write the system of inequations:

10x + 6y <= 120

x + y> = 14

Step-by-step explanation:

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2 years ago
Train A and Train B leave the station at 2 P.M. The graph below shows the distance covered by the two trains. Compare the speeds
adoni [48]

Answer:

Train b is moving faster than a by 45 units an hour

Step-by-step explanation:

8 0
2 years ago
Jar A contains four liters of a solution that is 45% acid. Jar B contains five liters of a solution that is 48% acid. Jar C cont
castortr0y [4]

Answer:

k= 80%

Step-by-step explanation:

Jar A contains 4*0.45 L acid, and 4 L of a solution  of acid.

Jar B contains 5*0.48 L acid., and 5 L of a solution of acid.

Jar C contains 1*k/100 = k/100 acid, and 1 L of a solution.

50% = 0.5

For jar A.

(2/3)*k/100 L acid  is added to jar A.

Now jar A contains   4*0.45 L + (2/3)*k/100 L acid, and it has (4+2/3)L of a solution.

L solute/L solution = 0.5

[4*0.45 L + (2/3)*k/100 L]/(4+2/3)L = 0.5

[1.8 + (2k/300)]/[(12+2)/3] = 0.5

[1.8 + (2k/300)]/[14/3] = 0.5

[1.8 + (2k/300)]= 0.5*(14/3)

(2k/300) = 0.5*(14/3) - 1.8

2k = (0.5*(14/3) - 1.8)*300

k = (0.5*(14/3) - 1.8)*300/2 =80

k= 80%

We also can find k using jar B.

(1/3)k/100 L acid is added  to jar B.

Now jar B contains 5*0.48 L+ (1/3)k/100 L acid, and it has (5+1/3) L of a solution.

L solute/L solution = 0.5

[5*0.48 L+ (1/3)k/100 L ]/(5+1/3)L= 0.5

[5*0.48 + (1/3)k/100 ]/(5+1/3)= 0.5

This equation also gives k=80%

Check.

We can check at least for jar A.

Jar A has 4L solution and 4*0.45=1.8 L acid.

2/3 L of the solution from jar C was added, and now we have 4 2/3 L of solution.

(2/3)* 80%= (2/3)*0.8 acid was added from jar C.

Now we have [1.8 +(2/3)*0.8] L acid in jar A.

L solute/L solution =  [1.8 +(2/3)*0.8] L /(4 2/3) L = 0.5 or 50%  as it is given that jar A has 50% at the end.

7 0
2 years ago
The functions $f$ and $g$ are defined as follows: \[f(x) = \sqrt{\dfrac{x+1}{x-1}}\quad\text{and}\quad g(x) = \dfrac{\sqrt{x+1}}
CaHeK987 [17]

Answer:

The answer is "domain and range are different".

Step-by-step explanation:

Given:

f(x) = \sqrt{\frac{x+1}{x-1}}\\\\g(x) =\frac{\sqrt{x+1}}{\sqrt{x-1}}\\

Solve f(x) to find domain and range:

for element x: R:

⇒ x \leq -1 \ \ \ and \  \ \  x > 1 range:

for  \ \ f(x) \times element : 0 \leq f(x)< 1 \ \ \ or \ \ f(x)>1

g(x) =\frac{\sqrt{x+1}}{\sqrt{x-1}}\\

Solve for domain:

⇒ \ when \ x \ element \ R: \ \ \ x>1  

Solve for range:  

⇒ g(x) \times element  R : g(x)>1

So, the value of the method f(x) and g(x) (range and domain) were different.  

5 0
2 years ago
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