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pentagon [3]
3 years ago
5

Need help with these

Mathematics
1 answer:
Ugo [173]3 years ago
8 0
T = nb is the answer
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PLZ HELP!!! question is in attachment
Triss [41]

Answer:

x = 72.5°

Step-by-step explanation:

x = 1/2(80 + 65)

x = 145/2

x = 72.5

4 0
3 years ago
What is StartFraction one-half EndFraction x is less than or equal to 18.x ≤ 18
Eduardwww [97]

Answer:

x\leq 36

The solution for x is all real numbers less that or equal to 36.

Step-by-step explanation:

Given expression:

\frac{1}{2}x\leq18

To find the solution for x

Solution:

In order to solve for x , we will make sure to isolate the unknown variable x on left side alone by doing math operations on both sides of the inequality.

We have:

\frac{1}{2}x\leq18

Multiplying both sides by 2.

2.\frac{1}{2}x\leq18\times 2

The 2 on left side will cancel each other leaving x alone.

Thus, we have:

x\leq 36

Thus, solution for x is all real numbers less that or equal to 36.

7 0
3 years ago
Read 2 more answers
Question
tigry1 [53]

Answer:

u can ask the tutor and he or she will help u in real time and explain

8 0
3 years ago
The extremes in the proportion 3/4=15/20 are
ollegr [7]

<span>In the form of a/b = c/d in proportions in math, the outer terms are called the extremes and the inner terms are called means. In the given 3/4 = 15/20, the outer terms or the extremes are 3 and 20.</span>

3 0
3 years ago
<img src="https://tex.z-dn.net/?f=%5CLarge%5Cbold%5Cred%7BQuestion%7D" id="TexFormula1" title="\Large\bold\red{Question}" alt="\
liraira [26]

For x such that 0 < x < \frac{\pi}{2}, the mathematical expression is:

\frac{\sqrt{1 \;-cos^2x} }{sinx} + \frac{\sqrt{1 \;-sin^2x} }{cosx} = 1+1=2

<u>Given the following data:</u>

  • \frac{\sqrt{1 \;-cos^2x} }{sinx}

  • \frac{\sqrt{1 \;-sin^2x} }{cosx}

In Trigonometry, you should take note of the following mathematical expression:

sin^2x + cos^2x = 1

Therefore, we can obtain the following:

sin^2x  = 1 - cos^2x   ...equation 1.

cos^2x = 1 - sin^2x    ...equation 2.

Substituting the equations respectively, we have:

\frac{\sqrt{1 \;-cos^2x} }{sinx} + \frac{\sqrt{1 \;-sin^2x} }{cosx} = \frac{\sqrt{sin^2x} }{sinx} + \frac{\sqrt{cos^2x} }{cosx}\\\\

Taking the square roots, we have:

\frac{sinx}{sinx} + \frac{cosx}{cosx} = 1 +1\\\\1+1=2

Therefore, for x such that 0 < x < \frac{\pi}{2}, the expression is:

\frac{\sqrt{1 \;-cos^2x} }{sinx} + \frac{\sqrt{1 \;-sin^2x} }{cosx} = 1+1=2

Read more on trigonometry here: brainly.com/question/4515552

7 0
2 years ago
Read 2 more answers
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