Start with x:
Since x and 57º make a straight line, x+57 = 180 (because a line is 180º). This says x=123º

Now for y:
Because lines m and n are parallel, y and 57º are alternate interior angles and have the same measure. y=57º

Finish with z:
Since y and z make a straight line, can you finish that one from here?

6 0

3 years ago

Please answer this in two minutes

NeTakaya

Answer:

Step-by-step explanation:

Use the Pythagorean Thereom:

$r^{2}$ = $9^{2}$ + $12^{2}$

$r^{2}$ = 81+144

$r^{2}$ = 225

$r$ = 15

Please mark me as Brainliest!

8 0

3 years ago

which pair of ratios can form a true properoration? A:1/2,6/10 B:6/8,5/20 C:3/5,6/10 D:3/9,7/80 PLZ HELP!

Goshia [24]

C because

3/5 * 2 both numbers so that .... 3*2= 6 and 5*2= 10 and those two numbers 6/10 = C when you want to solve a proportion just times in by both numerator and denominator so that they equal another number.

7 0

3 years ago

Read 2 more answers

Solve the Anti derivative.

Alex Ar [27]

Answer:

$\displaystyle \int {\frac{1}{9x^2+4}} \, dx = \frac{1}{6}arctan(\frac{3x}{2}) + C$

General Formulas and Concepts:

Algebra I

Factoring

Calculus

Antiderivatives - integrals/Integration

Integration Constant C

U-Substitution

Integration Property [Multiplied Constant]: $\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx$

Trig Integration: $\displaystyle \int {\frac{du}{a^2 + u^2}} = \frac{1}{a}arctan(\frac{u}{a}) + C$

Step-by-step explanation:

Step 1: Define

$\displaystyle \int {\frac{1}{9x^2 + 4}} \, dx$

Step 2: Integrate Pt. 1

[Integral] Factor fraction denominator: $\displaystyle \int {\frac{1}{9(x^2 + \frac{4}{9})}} \, dx$
[Integral] Integration Property - Multiplied Constant: $\displaystyle \frac{1}{9} \int {\frac{1}{x^2 + \frac{4}{9}}} \, dx$

Step 3: Identify Variables

Set up u-substitution for the arctan trig integration.

$\displaystyle u = x \\ a = \frac{2}{3} \\ du = dx$

Step 4: Integrate Pt. 2

[Integral] Substitute u-du: $\displaystyle \frac{1}{9} \int {\frac{1}{u^2 + (\frac{2}{3})^2} \, du$
[Integral] Trig Integration: $\displaystyle \frac{1}{9}[\frac{1}{\frac{2}{3}}arctan(\frac{u}{\frac{2}{3}})] + C$
[Integral] Simplify: $\displaystyle \frac{1}{9}[\frac{3}{2}arctan(\frac{3u}{2})] + C$
[integral] Multiply: $\displaystyle \frac{1}{6}arctan(\frac{3u}{2}) + C$
[Integral] Back-Substitute: $\displaystyle \frac{1}{6}arctan(\frac{3x}{2}) + C$

Topic: AP Calculus AB

Unit: Integrals - Arctrig

Book: College Calculus 10e

7 0

3 years ago