PLEASE HELP! URGENT. the law of cosines is a2+b2-2abcosC=c2. Find the value of 2abccosC.

PLEASE HELP ASAP I WILL GIVE BRAINLIEST!!! <3

lesantik [10]

1. 42e+18

That Is the answer for one

3 0

2 years ago

Read 2 more answers

Pls help quick i need this

irina [24]

Answer:

See explanation

Step-by-step explanation:

Q1-5.

1. Plane parallel to WXT is ZYU.

2. Segments parallel to $\overline {VU}$ are $\overline {ZY}, \overline {WX}$ and $\overline {ST}$

3. Segments parallel to $\overline {SW}$ are $\overline {VZ}, \overline {YU}$ and $\overline {XT}$

4. Segments skew to $\overline {}$ $\overline {XY}$ are $\overline {SV}$ and $\overline {VZ}$ (not lie in the same plane and not parallel)

5. Segments skew to $\overline {}$ $\overline {VZ}$ are $\overline {WX}$ and $\overline {XT}$ (not lie in the same plane and not parallel)

Q6.

a. $\angle 4$ and $\angle 10$ are the same-side interior angles, transversal k

b. $\angle 8$ and $\angle 11$ are alternate exterior angles, transversal m

c. $\angle 1$ and $\angle 4$ do not form any pair of angles

d. $\angle 2$ and $\angle 12$ are the same-side exterior angles, transversal k

e. $\angle 5$ and $\angle 7$ are corresponding angles, transversal j

f. $\angle 2$ and $\angle 13$ are alternate interior angles, transversal l

Q7.

$m\angle 1=m\angle 7=131^{\circ}$ (as vertical angle with angle 7)

$m\angle 2=180^{\circ}-131^{\circ}=49^{\circ}$ (as supplementary angle with angle 1)

$m\angle 8=49^{\circ}$ (as vertical angle with angle 2)

$m\angle 3=m\angle 1=131^{\circ}$ (as corresponding angles when parallel lines p and q are cut by transversal r)

$m\angle 4=m\angle 2=49^{\circ}$ (as corresponding angles when parallel lines p and q are cut by transversal r)

$m\angle 5=m\angle 7=131^{\circ}$ (as corresponding angles when parallel lines p and q are cut by transversal r)

$m\angle 6=m\angle 8=49^{\circ}$ (as corresponding angles when parallel lines p and q are cut by transversal r)

$m\angle 10=m\angle 16=88^{\circ}$ (as vertical angle with angle 16)

$m\angle 9=180^{\circ}-88^{\circ}=92^{\circ}$ (as supplementary angle with angle 16)

$m\angle 15=92^{\circ}$ (as vertical angle with angle 9)

$m\angle 14=m\angle 16=88^{\circ}$ (as corresponding angles when parallel lines p and q are cut by transversal s)

$m\angle 13=m\angle 15=92^{\circ}$ (as corresponding angles when parallel lines p and q are cut by transversal s)

$m\angle 12=m\angle 10=88^{\circ}$ (as corresponding angles when parallel lines p and q are cut by transversal s)

$m\angle 11=m\angle 9=92^{\circ}$ (as corresponding angles when parallel lines p and q are cut by transversal s)

Q8.

$m\angle 7=m\angle 9=105^{\circ}$ (as vertical angles)

$m\angle 8=180^{\circ}-105^{\circ}=75^{\circ}$ (as supplementary angle with angle 9)

$m\angle 10=m\angle 8=75^{\circ}$ (as vertical angles)

$m\angle 6=m\angle 8=75^{\circ}$ (as alternate interior angles when parallel lines a and b are cut by transversal c)

$m\angle 1=180^{\circ}-75^{\circ}-63^{\circ}=42^{\circ}$ (by angle addition postulate)

$m\angle 3=180^{\circ}-42^{\circ}-63^{\circ}=75^{\circ}$ (by angle addition postulate)

$m\angle 4=m\angle 1=42^{\circ}$ (as vertical angles)

$m\angle 5=m\angle 2=63^{\circ}$ (as vertical angles)

$m\angle 11=m\angle 4=42^{\circ}$ (as alternate interior angles when parallel lines a and b are cut by transversal d)

$m\angle 12=180^{\circ}-42^{\circ}=138^{\circ}$ (as supplementary angles)

$m\angle 13=m\angle 11=42^{\circ}$ (as vertical angles)

$m\angle 14=m\angle 12=138^{\circ}$ (as vertical angles)

3 0

2 years ago

Please help with this AP Calculus question !

melomori [17]

Answer:

C. $\displaystyle \frac{cos(x)}{x} - ln(x)sin(x)$

General Formulas and Concepts:

Calculus

Differentiation

Derivatives
Derivative Notation

Derivative Rule [Product Rule]: $\displaystyle \frac{d}{dx} [f(x)g(x)]=f'(x)g(x) + g'(x)f(x)$

Trig Derivatives

Logarithmic Derivatives

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle f(x) = ln(x)cos(x)$

Step 2: Differentiate

Derivative Rule [Product Rule]: $\displaystyle f'(x) = \frac{d}{dx}[ln(x)]cos(x) + ln(x)\frac{d}{dx}[cos(x)]$
Logarithmic Derivative: $\displaystyle f'(x) = \frac{1}{x}cos(x) + ln(x)\frac{d}{dx}[cos(x)]$
Trig Derivative: $\displaystyle f'(x) = \frac{1}{x}cos(x) + ln(x)[-sin(x)]$
Simplify: $\displaystyle f'(x) = \frac{cos(x)}{x} - ln(x)sin(x)$

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Differentiation

Book: College Calculus 10e

7 0

2 years ago

What is the volume of the pyramid in the diagram? 25 cm h = 15 cm 15 cm 20 cm

charle [14.2K]

Answer:the answer is 2,500 cm^3

Step-by-step explanation:

4 0

2 years ago

A study reported that students who play sports get better grades in school than students who do not play sports. Determine which

Stolb23 [73]

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.

Among the choices provided the incorrect statement is "Playing sports will cause students to get better grades in school."

4 0

2 years ago