All categories

3 years ago

A/20 + 14/15 = 9/15 Can i get some help?

1 answer:

3 0

Method 1:

$\dfrac{a}{20}+\dfrac{14}{15}=\dfrac{9}{15}\ \ \ \ |multiply\ both\ sides\ by\ 60\\\\60\cdot\dfrac{a}{20}+60\cdot\dfrac{14}{15}=60\cdot\dfrac{9}{15}\\\\3a+4\cdot14=4\cdot9\\\\3a+56=36\ \ \ \ |subtract\ 56\ from\ both\ sides\\\\3a=-20\ \ \ \ |divide\ both\ sides\ by\ 3\\\\\boxed{a=-\frac{20}{3}\to a=-6\frac{2}{3}}$

Method 2.

$\dfrac{a}{20}+\dfrac{14}{15}=\dfrac{9}{15}\ \ \ \ |subtract\ \dfrac{14}{15}\ form\ both\ sides\\\\\dfrac{a}{20}=-\dfrac{5}{15}\\\\\dfrac{a}{20}=-\dfrac{1}{3}\ \ \ \ |multiply\ both\ sides\ by\ 20\\\\\boxed{a=-\dfrac{20}{3}\to a=-6\frac{2}{3}}$

You might be interested in

Solve: a. 5y – 3 = – 18 b. -3x – 9 = 0 c. 4 + 3(z - 8) = -23 d. 1 – 2(y – 4) = 5

oksian1 [2.3K]

Answer:

a. y = -3

b. x = -3

c. z = - 1

d. y = -7

Step-by-step explanation:

a. 5y - 3 = - 18

5y = -18 +3

5y = -15

y = -15/5

y = - 3

b. -3x - 9 = 0

-3x = 9

x = 9/-3

x = -3

c. 4 + 3(z-8)= -23

4 + 3z-24 = -23

3z - 20 = -23

3z = -23 + 20

3z = -3

z = -3/3

z = - 1

d. 1 - 2(y-4) = 5

1 - 2y + 8 = 5

9 - 2y = 5

-2y = 14

y = 14/-2

y = -7

3 0

2 years ago

Help plz!!!!!!!!!!!!!!!!

Nookie1986 [14]

Step-by-step explanation:

what find

I can't understand

5 0

2 years ago

When placed end to end, three red rods are 2 cm longer than than 2 blue rods. Three blue rods are 2 cm longer than four red rods

Oxana [17]

Answer:

Red = 10 cm, blue = 14 cm

Step-by-step explanation:

Let the length of each red rod be r and each blue rod be b

Then we have:

3r - 2b = 2
3b - 4r = 2

Multiply the first equation by 3 and the second one by 2:

9r - 6b = 6
6b - 8r = 4

Add the equations to eliminate one of variables:

9r - 8r = 6 + 4
r = 10 cm

Find the value of b:

3*10 -2b = 2
30 - 2 = 2b
2b = 28
b = 14 cm

5 0

2 years ago

X |Y

Yakvenalex [24]

In all of these cases, y=x+2.

5 0

2 years ago

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

2 years ago