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Serggg [28]
3 years ago
13

Help please, its super duper easy!

Mathematics
2 answers:
Firdavs [7]3 years ago
5 0

Answer:

R would be 36.

Step-by-step explanation:

Subtract the two know angles from 180 to get your missing angle. :)

Hope this helps!

If you don't mind please mark as brainliest! :)

Thanks!!!

tresset_1 [31]3 years ago
4 0

Answer:

36 degrees

Step-by-step explanation:

Okay, so the angles of a triangle are always gonna add up to 180. So, let's add up 68 & 76.

68 + 76 = 144

Now we just have to subtract that from 180

180 - 144 = 36

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8= 8 + 4x +5x<br><br> can anyone help me with this
dedylja [7]

Answer:

0= x

Step-by-step explanation:

8= 8 + 4x +5x

Combine like terms

8 = 8+9x

Subtract 8 from each side

8-8 =8+9x-8

0 = 9x

Divide by 9

0/9 = 9x/9

0= x

3 0
3 years ago
What is the length of the curve with parametric equations x = t - cos(t), y = 1 - sin(t) from t = 0 to t = π? (5 points)
zzz [600]

Answer:

B) 4√2

General Formulas and Concepts:

<u>Calculus</u>

Differentiation

  • Derivatives
  • Derivative Notation

Basic Power Rule:

  1. f(x) = cxⁿ
  2. f’(x) = c·nxⁿ⁻¹

Parametric Differentiation

Integration

  • Integrals
  • Definite Integrals
  • Integration Constant C

Arc Length Formula [Parametric]:                                                                         \displaystyle AL = \int\limits^b_a {\sqrt{[x'(t)]^2 + [y(t)]^2}} \, dx

Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify</em>

\displaystyle \left \{ {{x = t - cos(t)} \atop {y = 1 - sin(t)}} \right.

Interval [0, π]

<u>Step 2: Find Arc Length</u>

  1. [Parametrics] Differentiate [Basic Power Rule, Trig Differentiation]:         \displaystyle \left \{ {{x' = 1 + sin(t)} \atop {y' = -cos(t)}} \right.
  2. Substitute in variables [Arc Length Formula - Parametric]:                       \displaystyle AL = \int\limits^{\pi}_0 {\sqrt{[1 + sin(t)]^2 + [-cos(t)]^2}} \, dx
  3. [Integrand] Simplify:                                                                                       \displaystyle AL = \int\limits^{\pi}_0 {\sqrt{2[sin(x) + 1]} \, dx
  4. [Integral] Evaluate:                                                                                         \displaystyle AL = \int\limits^{\pi}_0 {\sqrt{2[sin(x) + 1]} \, dx = 4\sqrt{2}

Topic: AP Calculus BC (Calculus I + II)

Unit: Parametric Integration

Book: College Calculus 10e

4 0
3 years ago
Which is greater 6.789 or 67.890?
Vlada [557]
67.890, because 67 is greater than 6
3 0
3 years ago
Read 2 more answers
Which of the following relations is a function ?
Effectus [21]

Answer:

Step-by-step explanation:

The relation that is a function is c

6 0
3 years ago
Please help me asap i will mark a branlist number 7
SVETLANKA909090 [29]

Answer:

When doing square roots the opposite is squaring so, to find one that is more than 12 and less that 13 would be finding one that is in between √144 and √169

That leaves your answer as √150, because it is more than √144 (which is 12) and less than √169 (which is 13)

B √150

Hope this helps ;)

6 0
3 years ago
Read 2 more answers
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