Please solve this problem and I’ll give brainliest

Find m\angle Cm∠Cm, angle, C. Round to the nearest degree. Help Course Challenge

astra-53 [7]

Hello,

1. Since Angle C has the longest side for this triangle, it will have the largest degree value.

2. Use the Law of Cosines and inverse properties of “theta” to solve for Angle C. (Ensure that the calculator used is in “degree mode”, not “radian mode”.

c^2 = a^2 + b^2 - 2(a)(b)(cos (C))
15^2 = 11^2 + 14^2 - 2(11)(14)(cos(C))
225 - 317 = -2(11)(14)(cos(C))
-92 / -2(11)(14) = cos(C)
cos(C) becomes ->> cos^-1[92 /-2(11)(14)] = 72.62° ->> to the nearest degree is 73°

The answer for angle C, 73°, is logical because the triangle in the picture represents a 60-60-60 triangle, known as an equilateral triangle.

Good luck to you!

5 0

3 years ago

Read 2 more answers

The formula for finding the perimeter of equilateral triangle is P=3s. how can you transform the formula so that it could be use

Mandarinka [93]

Your answer is D. Hope this helps -John

5 0

3 years ago

The value of which of these expressions is closest to e?

Anna [14]

Answer:

Using a calculator, we can check that e=2.718281828.

Step-by-step explanation:

Lets evaluate each one of our expression the check which one is closest to e:

(1+ \frac{1}{31} )^{31}=2.675686306

(1+ \frac{1}{32})^{32}=2.676990129

(1+ \frac{1}{34} )^{34}=2.679355428

(1+ \frac{1}{33} )^{33}=2.678207651

We can conclude that the value of (1 +1/34) to the power of 34 is the closest to the value of e.

4 0

3 years ago

Read 2 more answers

Angle 1 and angle 2 form a linear pair.if m angle =5x+9 and m angle 2=3x+11,find the measures of both angels.

lbvjy [14]

Answer: ∠1 = 109° ∠2 = 71°

Step-by-step explanation:

If the two angles form a linear pair, then their sum is 180°

∠1 + ∠2 = 180°

(5x + 9) + (3x + 11) = 180

8x + 20 = 180

8x = 160

x = 20

∠1 = 5x + 9

= 5(20) + 9

= 100 + 9

= 109

∠2 = 3x + 11

= 3(20) + 11

= 60 + 11

= 71

CHECK:

∠1 + ∠2 = 180°

109° + 71° = 180°

180° = 180° $\checkmark$

7 0

3 years ago

38. Evaluate f (3x +4y)dx + (2x --3y)dy where C, a circle of radius two with center at the origin of the xy

lina2011 [118]

It looks like the integral is

$\displaystyle \int_C (3x+4y)\,\mathrm dx + (2x-3y)\,\mathrm dy$

where C is the circle of radius 2 centered at the origin.

You can compute the line integral directly by parameterizing C. Let x = 2 cos(t ) and y = 2 sin(t ), with 0 ≤ t ≤ 2π. Then

$\displaystyle \int_C (3x+4y)\,\mathrm dx + (2x-3y)\,\mathrm dy = \int_0^{2\pi} \left((3x(t)+4y(t))\dfrac{\mathrm dx}{\mathrm dt} + (2x(t)-3y(t))\frac{\mathrm dy}{\mathrm dt}\right)\,\mathrm dt \\\\ = \int_0^{2\pi} \big((6\cos(t)+8\sin(t))(-2\sin(t)) + (4\cos(t)-6\sin(t))(2\cos(t))\big)\,\mathrm dt \\\\ = \int_0^{2\pi} (12\cos^2(t)-12\sin^2(t)-24\cos(t)\sin(t)-4)\,\mathrm dt \\\\ = 4 \int_0^{2\pi} (3\cos(2t)-3\sin(2t)-1)\,\mathrm dt = \boxed{-8\pi}$

Another way to do this is by applying Green's theorem. The integrand doesn't have any singularities on C nor in the region bounded by C, so

$\displaystyle \int_C (3x+4y)\,\mathrm dx + (2x-3y)\,\mathrm dy = \iint_D\frac{\partial(2x-3y)}{\partial x}-\frac{\partial(3x+4y)}{\partial y}\,\mathrm dx\,\mathrm dy = -2\iint_D\mathrm dx\,\mathrm dy$

where D is the interior of C, i.e. the disk with radius 2 centered at the origin. But this integral is simply -2 times the area of the disk, so we get the same result: $-2\times \pi\times2^2 = -8\pi$ .

3 0

3 years ago